1. A single disturbance that moves from point to point through a medium is called a ___.

a. period b. periodic wave c. wavelength d. pulse

Answer: D
A wave is a continuous and repeating disturbance of a medium and a pulse is a single disturbance.

2. If the particles of the medium are vibrating to and fro in the same direction of energy transport, then the wave is a ____ wave.
a. longitudinal b. sound c. standing d. transverse

Answer: A
In longitudinal waves, particles of the medium vibrate to and from in a direction parallel to the direction of energy
transport. If energy is transmitted along a medium from the east end to the west end, then particles of the medium would
vibrate eastward and westward
A sound wave is a longitudinal wave but not the answer since a wave which exhibits this characteristic is not necessarily
a sound wave.

3. When the particles of a medium are vibrating at right angles to the direction of energy transport, then the wave is a ____ wave.
a. longitudinal b. sound c. standing d. transverse

Answer: D
In transverse waves, particles of the medium vibrate to and from in a direction perpendicular to the direction of energy
transport.

4. A transverse wave is traveling through a medium. See diagram below. The particles of the medium are vibrating _____.

a. parallel to the line joining AD. b. along the line joining CI.
c. perpendicular to the line joining AD. d. at various angles to the line CI.
e. along the curve CAEJGBI.
Answer: A
In transverse waves, particles of the medium vibrate to and from in a direction perpendicular to the direction of energy
transport. In this case, that would be parallel to the line AD.

5. If the energy in a longitudinal wave travels from south to north, the particles of the medium would be vibrating _____.
a. from north to south, only b. both north and south
c. from east to west, only d. both east and west

Answer: B
In longitudinal waves, particles of the medium vibrate to and from in a direction parallel to the direction of energy
transport. If the particles only moved north and not back south, then the particles would be permanently displaced from
their rest position; this is not wavelike.
6. As a pulse travels though a uniform medium, the speed of the pulse ____.
a. decreases b. increases c. remains the same

Answer: C
The speed of a wave or a pulse depends upon the properties of the medium. If the medium is uniform or unchanging,
then the speed is constant.

7. The main factor which effects the speed of a sound wave is the ____.
a. amplitude of the sound wave b. intensity of the sound
c. loudness of the sound d. properties of the medium
e. pitch of the sound

Answer: D
The speed of a wave is dependent upon the properties of the medium and not the properties of the wave.

8. As a wave travels into a medium in which its speed increases, its wavelength would ____.
a. decrease b. increase c. remain the same

Answer: B
As a wave crosses a boundary into a new medium, its speed and wavelength change while its frequency remains the
same. If the speed increases, then the wavelength must increase as well in order to maintain the same frequency.

9. As a wave passes across a boundary into a new medium, which characteristic of the wave would NOT change?
a. speed b. frequency c. wavelength

Answer: B
As a wave crosses a boundary into a new medium, its speed and wavelength change while its frequency remains the
same.
10. What is the amplitude of the wave in the diagram below?

a. 0.03 m. b. 0.04 m. c. 0.05 m. d. 0.06 m.

Answer: A
The amplitude of a wave is measured from rest to crest or from rest to trough; but not from crest to trough. Thus, take the
0.06 m measurement and "halve it" to get the answer.
11. The wavelength of the wave in the diagram above (Question #10) is ____ m.
a. 0.030 b. 0.040 c. 0.060 d. 0.080

Answer: D
The wavelength of a wave is measured as the distance between any two corresponding points on adjacent waves, which
would mean from a crest to the next adjacent crest. Thus, the distance from point b to point d is the wavelength - 0.08 m

12. A wave X meters long passes through a medium with a speed of Y meters per second. The frequency of the wave could be
expressed as
a. Y/X cycles/sec. b. X/Y cycles/sec. c. XY cycles/sec. d. (X + Y) cycles/sec.

Answer: A
Let d = wavelength. Since d*f = v, f = v/d.

Consider the following diagram for Questions #13-#14.

13. How many complete waves are shown in the diagram?

a. 1 b. 2 c. 3 d. 1.5

Answer: D
From point A to point E is one full wave cycle. After point E, the wave begins to repeat itself, but only for one-half of a
cycle. Thus, there are 1.5 waves shown in the diagram.
14. If the distance from point A to point B in the diagram is 60 cm, then the wavelength is ____.
a. 20 cm. b. 40 cm. c. 60 cm. d. 90 cm.

Answer: B
From point A to point E is one full wave cycle. This distance represents two-thirds of the 60 cm from A to G. Thus, the
wavelength is (2/3)*60 cm = 40 cm.

15. The number of cycles of a periodic wave occurring per unit time is defined as a wave's ____.
a. wavelength. b. period. c. amplitude. d. frequency.

Answer: D
This is a basic definition which you should know and be able to apply.

16. A periodic and repeating disturbance in a lake creates waves which emanate outward from its source to produce circular wave
patterns. If the frequency of the source is 2.00 Hz and the wave speed is 5.00m/s then the distance between adjacent wave crests
is ___ meter.
a. 0.200 b. 0.400 c. 1.25 d. 2.50 e. 10.0

Answer: D
Let w=wavelength; then v = w*f. In this problem, it is given that v=5.00 m/s and f = 2.00 Hz. Substitution and algebra
yields w = v / f = 2.50 m.

17. What is the frequency of a wave that has a speed of 0.4 m/s and a wavelength of 0.020 meter?
a. 10 hertz. b. 20 hertz. c. 0.008 hertz. d. 0.5 hertz.

Answer: B
Let w=wavelength; then v = w*f. In this problem, it is given that v=5 m/s and w = .02 m. Substitution and algebra yields
f=v/w=20 Hz.

18. Many wave properties are dependent upon other wave properties. Yet, one wave property is independent of all other wave
properties. Which one of the following properties of a wave is independent of all the others?
a. wavelength b. frequency c. period d. velocity

Answer: D
The speed of a wave is dependent upon the properties of the medium through which it moves, not upon the properties of
the wave itself.

19. A pendulum makes exactly 40 vibrations in 20.0 s. Its period is ____. (Be cautious of the units.)
a. 0.500 Hz. b. 0.500 s. c. 2.00 Hz. d. 2.00 s. e. 8.00 x 102 Hz.

Answer: B
The period is the time for one complete cycle. If the pendulum takes 20 seconds for exactly 40 vibrational cycles, then it
must take 0.500 second for one cycle.

20. A period of 0.005 seconds would be equivalent to a frequency of ____ Hz.

a. 20 b. 50 c. 200 d. 500 e. 2000

Answer: C
The period and the frequency are related by a reciprocal relationship; that is, f = 1/T.

21. TRUE or FALSE:

The number of waves generated per second by a source is called the frequency of the source.
a. True b. False

Answer: A
This is a basic definition which you should know and be able to apply.

22. TRUE or FALSE:

The SI unit for frequency is hertz.
a. True b. False

Answer: A
Know this like the back of your hand.
23. TRUE or FALSE:
Doubling the frequency of a wave source (without altering the medium) doubles the speed of the waves.
a. True b. False

Answer: B
Don't be fooled. Wave speed may equal frequency*wavelength. Yet doubling the frequency only halves the wavelength;
wave speed remains the same. To change the wave speed, the medium would have to be changed.

24. If the frequency of a wave is doubled and if the speed remains constant, its wavelength is ____.
a. quartered. b. halved. c. unchanged. d. doubled.

Answer: B
Wave speed equals frequency*wavelength. So doubling the frequency must halve the wavelength in order for wave
speed to remain the same.

25. Two different ropes with different mass densities are attached to each other. A
pulse is introduced into one end of the rop and approaches the boundary as shown
at the right. At the boundary, a portion of the energy is transmitted into the new
medium and a portion is reflected. Which one of the diagrams below depicts the
possible location and orientation of the pulse shortly after the incident pulse reaches the boundary?

Answer: C
The speeds in the two media can be deduced by the distance of the pulses from the boundary. In A and E, the speed is shown as
fastest on the right, which makes the transmitted medium the less dense. Rule out A and E since a reflected pulse should not invert
when moving from more dense to less dens. Rule out B for just the opposite reasons; the wave is moving from less to more dense
and should invert upon reflection. Rule out D because the transmitted pulse never inverts. That leaves C as the answer.

26. When a pulse reaches a boundary between two different media, it will be____.
a. reflected, only. b. transmitted, only.
c. partly reflected and partly transmitted. d. neither reflected nor transmitted.

Answer: C
This is basic information about the boundary behavior of waves.

27. Diagram P at the right shows a transverse pulse traveling along a dense rope toward its
junction with a less dense rope. Which of the diagrams (A, B, C, D, or E) below depicts the
ropes at the instant that the reflected pulse again passes through its original position marked
X? Consider such features as amplitude and relative speed (i.e., the relative distance of the transmitted and reflected pulses from
boundary).

Answer: E
A, B, and C can be quickly ruled out since it shows the amplitude of the reflected and incident pulse to be the same size.
An incident pulse would give up some of its energy to the transmitted pulse at the boundary, thus making the amplitude of
the reflected pulse less than that of the incident pulse. Rule out D since it shows the reflected pulse moving faster than
the transmitted pulse. This would not happen unless moving from less dense to more dense. This leaves E as the
answer.
28. A wave whose speed in a snakey is 4.4 m/s enters a second snakey. The wavelength changes from 2.0 m to 3.0 m. The wave
in the second snakey travels at approximately ____.
a. 1.5 m/s. b. 2.2 m/s. c. 2.9 m/s. d. 4.4 m/s. e. 6.6 m/s.

Answer: E
This is another boundary behavior question with a mathematical slant to it. The frequency of the incident and transmitted
waves are always the same. Thus, use f =v/w to find the frequency of the incident wave - 2.2 Hz. The frequency of the
transmitted wave is >also 2.2 Hz, the wavelength is 3.0 m, and so the speed is f*w = 6.6 m/s.

29. The diagram at the right shows a disturbance moving through a rope towards the right. If this
disturbance meets a similar disturbance moving to the left, then which one of the diagrams below depict a
pattern which could NEVER appear in the rope?

Answer: D
WOW! Tough one. Draw a second wave to the right of the wave which is given. Then visually move the wave to the left.
Visualize in your mind the shape of the resultant as interference occurs. It will never look like D. If you still don't get it,
take a break and watch some TV.
30. A 2.0-meter long rope is hanging vertically from the ceiling and attached to a vibrator. A single pulse is observed to travel to the
end of the rope in 0.50 s. What frequency should be used by the vibrator to maintain three whole waves in the rope?
a. 0.75 Hz b. 1.3 Hz c. 4.0 Hz d. 6.0 Hz e. 8.0 Hz

Answer: D
The given info allows you to determine the speed of the wave: v=d/t=2 m/0.5 s) = 4 m/s. If there are 3 waves in a 2-meter
long rope, then each wave is 2/3-meter long. Now find frequency with the equation v=f*w where v=4 m/s and w=0.667 m.
Proper algebra yields 6 Hz as the answer.

31. A standing wave experiment is performed to determine the speed of waves in a rope. The standing wave pattern shown below
is established in the rope. The rope makes exactly 90 complete vibrational cycles in one minute. The speed of the waves is ____
m/s.

a. 3.0 b. 6.0 c. 180 d. 360 e. 540

 Answer: B
If there are exactly 90 vibrations in 60.0 seconds, then there is a frequency of 1.5 Hz. The diagram shows 1.5 waves in
6.0-meters of rope; thus, the wavelength is 4.0 meters. Now use the equation v=f*w to calculate the speed of the wave.
Proper substitution yields 6.0 m/s.

32. Consider the standing wave pattern shown below. A wave generated at the left end of the medium undergoes reflection at the
fixed end on the right side of the medium. The number of antinodes in the diagram is _____.

a. 3 b. 5 c. 6 d. 7 e. 12

Answer: C
An antinode is a point on the medium which oscillates from a large + to a large - displacement. Count the number of
these points - there are 6 - but do not count them twice.

33. A node is a point located along the medium where there is always ___.
a. a double crest b. c. constructive interference
d. destructive interference e. a double rarefaction
Answer: D
A node is a point along the medium of no displacement. The point is not displaced because destructive interference
occurs at this point.

34. TRUE or FALSE:

Constructive interference of waves occurs when two crests meet.
a. True b. False
Answer: A
Yes! Or when a trough meets a trough or whenever two waves displaced in the same direction (such as both up or both
down) meet.

35. Which phenomenon is produced when two or more waves passing simultaneously through the same medium meet up with one
another?
a. refraction b. diffraction c. interference d. reflection

Answer: C
Interference is the meeting of two or more waves when passing along the same medium - a basic definition which you
should know and be able to apply.

36. Two pulses are traveling in opposite directions along the same medium as shown in the diagram at the right. Which diagram
below best depicts the appearance of the medium when each pulse meets in the middle?

Answer: D
When a crest is completely overlapped with a trough having the same amplitude, destructive interference occurs.
Complete cancellation takes place if they have the same shape and are completely overlapped.
37. TRUE or FALSE:
A vibrating object is necessary for the production of sound.

a. True b. False

Answer: A
Absolutely! If you don't believe it, then think of some sounds - voice, guitar, piano, tuning fork, chalkboard screech, etc. -
and consider what the vibrational source is. All sounds have a vibrating object of some kind as their source.
38. Which one of the following CANNOT transmit sound?
a. Liquid air b. Gaseous oxygen c. Liquid water
d. Solid steel e. Perfect vacuum

Answer: E
Sound is a mechanical wave and as such requires a medium in order to move through space. For this reason, sound
cannot move through a vacuum.
Part A: TRUE/FALSE
1. Which of the following statements are TRUE of sound waves? Identify all that apply.
a. A sound wave is a mechanical wave.
b. A sound wave is a means of transporting energy without transporting matter.
c. Sound can travel through a vacuum.
d. A sound wave is a pressure wave; they can be thought of as fluctuations in pressure with respect to time.
e. A sound wave is a transverse wave.
f. To hear the sound of a tuning fork, the tines of the fork must move air from the fork to one's ear.
g. Most (but not all) sound waves are created by a vibrating object of some type.
h. To be heard, a sound wave must cause a relatively large displacement of air (for instance, at least a cm or more) around an
observer's ear.

Answer: ABD
a. TRUE - A sound wave transports its energy by means of particle interaction. A sound wave cannot travel through a vacuum. This makes
sound a mechanical wave.
b. TRUE - Absolutely! Particles do not move from the source to the ear. Particles vibrate about a position; one particle impinges on its
neighboring particle, setting it in vibrational motion about its own equilibrium position.
c. FALSE - Only electromagnetic waves can travel through a vacuum; mechanical waves such as sound waves require a particle-interaction to
transport their energy. There are no particles in a vacuum.
d. TRUE - As particles move back and forth longitudinally, there are times when they are very close within a given region and other times that
they are far apart within that same region. The close proximity of particles produces a high pressure region known as a compression; the
distancing of particles within a region produces a low pressure region known as a rarefaction. Over time, a given region undergoes oscillations in
pressure from a high to a low pressure and finally back to a high pressure.
e. FALSE - Never! Waves are either longitudinal or transverse. Longitudinal waves are those in which particles of the medium move in a
direction parallel to the energy transport. And that is exactly how particles of the medium move as sound passes through it.
f. FALSE - It is the disturbance that moves from the tuning fork to one's ear. the particles of the medium merely vibrate back and forth about the
same location, never really moving from that location to another location. This is true of all waves - they transport energy without actually
transporting matter.
g. FALSE - All sound waves are created by vibrating objects of some sort.
h. FALSE - Quite surprisingly to many, most sounds which we are accustomed to hearing are characterized by particle motion with an
amplitude on the order of 1 mm or less.

2. Which of the following statements are TRUE of sound intensity and decibel levels? Identify all that apply.
a. The intensity of a sound wave has units of Watts/meter.
b. When a sound wave is said to be intense, it means that the particles are vibrating back and forth at a high frequency.
c. Intense sounds are characterized by particles of the medium vibrating back and forth with a relatively large amplitude.
d. Intense sounds are usually perceived as loud sounds.
e. The ability of an observer to hear a sound wave depends solely upon the intensity of the sound wave.
f. From the least intense to the most intense, humans have a rather narrow range of intensity over which sound waves can be
heard.
g. The intensity of sound which corresponds to the threshold of pain is one trillion times more intense than the sound which
corresponds to the threshold of hearing.
h. Two sounds which have a ratio of decibel ratings equal to 2.0. This means that the second sound is twice as intense as the
first sound.
i. Sound A is 20 times more intense than sound B. So if Sound B is rated at 30 dB, then sound A is rated at 50 dB.
j. Sound C is 1000 times more intense than sound D. So if sound D is rated at 80 dB, sound C is rated at 110 dB.
k. A machine produces a sound which is rated at 60 dB. If two of the machines were used at the same time, the decibel rating
would be 120 dB.
l. Intensity of a sound at a given location varies directly with the distance from that location to the source of the sound.
m. If the distance from the source of sound is doubled then the intensity of the sound will be quadrupled.
n. If the distance from the source of sound tripled, then the intensity of the sound will be increased by a factor of 6.

Answer: CDGJ
a. FALSE - Intensity is a power/area relationship and as such the units are typically Watts/meter2. The Watt is a unit of power and
the meter2 is a unit of area.
b. FALSE - Intense sounds are simply sounds which carry energy outward from the source at a high rate. They are most
commonly sound waves characterized by a high amplitude of movement. While frequency does effect one's perception of the
loudness of a sound, it does not effect the intensity of a sound wave.
c. TRUE - An intense sound is the result of a large vibration of the source of sound that sets particles of the medium in motion
with a high amplitude of movement about their usual rest position.
d. TRUE - Loudness is more of a subjective response to sound, dependent in part upon the quality of an observer's ears. Intensity
is an objective characteristic of sound that can actually be measured in Watts/meter2. However, intense sounds will always be
observed to be louder by an observer than less intense sounds.
e. FALSE - Not only must the sound be intense enough to cause an audible disturbance of the mechanisms of the ear, it also must
fall within the human frequency range of 20 Hz to 20000 Hz.
f. FALSE - Humans actually have a phenomenal range of intensities to which they are sensitive to. The intensity of the sound at
the threshold of pain is one-trillion times more intense than the sound at the threshold of hearing. That's quite a range.
g. TRUE - The threshold of pain has an intensity of 1 W/m2 and the threshold of hearing has an intensity of 1.0 x 10-12 W/m2.
That's a ratio of one trillion.
h. FALSE - No! Since the decibel scale is based on a logarithmic function, this is simply not the case.
i. FALSE - Two sounds separated by 20 dB on the decibel scale have intensity ratios of 100:1. If one sound is 20 times more
intense than another sound, then it is 13 dB higher on the decibel scale [ that comes from 10*log(20) ].
j. TRUE - Always remember that a decibel rating is based on the logarithmic function. A sound which is 1000 times (103 times)
more intense than another sound is 3 bels or 30 dBels greater on the decibel scale.
k. FALSE - Two machines would produce twice the intensity; but when converted to the logarithmic scale of decibels, there decibel
rating would differ by 3 dB.
l. FALSE - Intensity varies inversely with distance from the source. To be more specific, it varies inversely with the square of the
distance.
m. FALSE - If the distance from the source is doubled, then the intensity is decreased by a factor of four.
n. FALSE - If the distance from the source is tripled, then the intensity is decreased by a factor of nine.

3. Which of the following statements are TRUE of the speed of sound? Identify all that apply.
a. The speed of a sound wave depends upon its frequency and its wavelength.
b. In general, sound waves travel fastest in solids and slowest in gases.
c. Sound waves travel fastest in solids (compared to liquids and gases) because solids are more dense.
d. The fastest which sound can move is when it is moving through a vacuum.
e. If all other factors are equal, a sound wave will travel fastest in the most dense materials.
f. A highly elastic material has a strong tendency to return to its original shape if stressed, stretched, plucked or somehow
disturbed.
g. A more rigid material such as steel has a higher elasticity and therefore sound tends to move through it at high speeds.
h. The speed of sound moving through air is largely dependent upon the frequency and intensity of the sound wave.
i. A loud shout will move faster through air than a faint whisper.
j. Sound waves would travel faster on a warm day than a cool day.
k. The speed of a sound wave would be dependent solely upon the properties of the medium through which it moves.
l. A shout in a canyon produces an echo off a cliff located 127 m away. If the echo is heard 0.720 seconds after the shout, then
the speed of sound through the canyon is 176 m/s.
m. The speed of a wave within a guitar string varies inversely with the tension in the string.
n. The speed of a wave within a guitar string varies inversely with the mass per unit length of the string.
o. The speed of a wave within a guitar string will be doubled if the tension of the string is doubled.
p. An increase in the tension of a guitar string by a factor of four will increase the speed of a wave in the string by a factor of
two.
q. An increase in the linear mass density of a guitar string by a factor of four will increase the speed of a wave in the string by a
factor of two.

Answer: BFGJKNP
a. FALSE - The speed of a wave is calculated by the product of the frequency and wavelength. However, it does not depend upon
the frequency and the wavelength. An alteration in the frequency or the wavelength will not alter the speed.
b. TRUE - For the same material, speed is greatest in materials in which the elastic properties are greatest. Despite the greater
density of solids, the speed is greatest in solids, followed by liquids, followed by gases.
c. FALSE - Sound waves travel faster in solids because the particles of a solid have a greater elastic modulus. That is to say that a
disturbance of a particle from its rest position in a solid leads to a rapid return to its rest position and as such an ability to rapidly
transmit the energy to the next particle.
d. FALSE - Sound is a mechanical wave which moves due to particle interaction. There are no particles in a vacuum so sound can
not move through a vacuum.
e. FALSE - Sound waves (like all waves) will travel slower in more dense materials (assuming all other factors are equal).
f. TRUE - This is the definition of elasticity. Elasticity is related to the ability of the particles of a material to return to their original
position if displaced from it.
g. TRUE - A more rigid material is characterized by particles which quickly return to their original position if displaced from it.
Sound moves fastest in such materials.
h. FALSE - The speed of sound through a material is dependent upon the properties of the material, not the characteristics of the
wave.
i. FALSE - A loud shout will move at the same speed as a whisper since the speed of sound is independent of the characteristics of
the sound wave and dependent upon the properties of the material through it is moving.
j. TRUE - The speed of sound through air is dependent upon the temperature of the air.
k. TRUE - This is a big principle. Know it.
l. FALSE - Speed is distance traveled per time. For this case, the sound travels a distance of 254 m (to the cliff and back) in 0.720
seconds. That computes to 353 m/s.
m. FALSE - For a guitar string, the equation for the speed of waves is v = SQRT (Ftens/mu). From the equation, it is evident that
an increase in tension will result in an increase in the speed; they are directly related.
n. TRUE - For a guitar string, the equation for the speed of waves is v = SQRT (Ftens/mu). From the equation, it is evident that an
increase in mass per unit length (mu) will result in an decrease in the speed; they are inversely related.
o. FALSE - The speed of a wave in a guitar string varies directly with the square root of the tension. If the tension is doubled, then
the speed of sound will increase by a factor of the square root of two.
p. TRUE - The speed of a wave in a string is directly related to the square root of the tension in the string. So the speed will be
changed by the square root of whatever factor the tension is changed.
q. FALSE - An increase in the linear mass density by a factor of four will decrease the speed by a factor of 2. The speed is
inversely related to the square root of the linear density.

4. Which of the following statements are TRUE of the frequency of sound and the perception of pitch? Identify all that apply.
a. A high pitched sound has a low wavelength.
b. A low-pitched sound is a sound whose pressure fluctuations occur with a low period.
c. If an object vibrates at a relatively high frequency, then the pitch of the sound will be low.
d. The frequency of a sound will not necessarily be the same as the frequency of the vibrating object since sound speed will be
altered as the sound is transmitted from the object to the air and ultimately to your ear.
e. Two different guitar strings are used to produce a sound. The strings are identical in terms of material, thickness and the
tension to which they are pulled. Yet string A is shorter than string B. Therefore, string A will produce a lower pitch.
f. Both low- and high-pitched sounds will travel through air at the same speed.
g. Doubling the frequency of a sound wave will halve the wavelength but not alter the speed of the wave.
h. Tripling the frequency of a sound wave will decrease the wavelength by a factor of 6 and alter the speed of the wave.
i. Humans can pretty much hear a low-frequency sound as easily as a high-frequency sound.
j. Ultrasound waves are those sound waves with frequencies less than 20 Hz.

Answer: AFG
a. TRUE - High pitch corresponds to a sound with high frequency and therefore low wavelength.
b. FALSE - Low pitched sound have a low frequency. Frequency is inversely related to period. So low pitched sounds have a high
period. That is, the time for the vibrations to undergo one complete cycle is large for a low frequency (or low pitch) sound.
c. FALSE - Pitch is a subjective response of the ear to sound. Frequency is an objective measure of how often the sound
undergoes an oscillation from high to low pressure. The two are related in the sense that a sound with a high frequency will be
perceived as a sound with a high pitch.
d. FALSE - As waves (of any type) are transmitted from one medium to another, the speed and the wavelength can be altered,
but the frequency will not be changed. Thus, the frequency of the source is the frequency of the sound waves which impinge upon
the ear.
e. FALSE - The strings are identical in terms of their properties; this means that waves travel at the same speed through each. Yet
string A is shorter than string B, so the wavelengths of waves are shortest in string A. As such, the frequencies are greatest for
string A and it is observed to produce sounds of higher pitch.
f. TRUE - The speed at which waves travel through air is dependent upon the properties of the air and not the properties of the
wave.
g. TRUE - Frequency and wavelength are inversely related; doubling one will halve the other. Yet the speed of a wave is
independent of each.
h. FALSE - Tripling the frequency of a sound wave will make wavelength decrease by a factor of 3 but not alter the speed of a
wave.
i. FALSE - The response of the ear to sound is dependent in part on the frequency of the sound. A higher pitch sound of the same
intensity is generally heard to be louder than a lower pitch sound of the same intensity.
j. FALSE - Ultrasound waves are waves which have a frequency beyond the human range of audible frequencies - above 20000
Hz.

5. Which of the following statements are TRUE of standing wave patterns? Identify all that apply.
a. A standing wave pattern is formed as a result of the interference of two or more waves.
b. When a standing wave pattern is established, there are portions of the medium which are not disturbed.
c. A standing wave is really not a wave at all; it is a pattern resulting from the interference of two or more waves which are
traveling through the same medium.
d. A standing wave pattern is a regular and repeating vibrational pattern established within a medium; it is always characterized
by the presence of nodes and antinodes.
e. An antinode on a standing wave pattern is a point which is stationary; it does not undergo any displacement from its rest
position.
f. For every node on a standing wave pattern, there is a corresponding antinode; there are always the same number of each.
g. When a standing wave pattern is established in a medium, there are alternating nodes and antinodes, equally spaced apart
across the medium.

Answer: ABCDG
a. TRUE - Interference results when two (or more) waves interfere to produce a regular and repeating pattern of nodes and
antinodes. The presence of the nodes standing still along the medium at the same position is what gives it its name of "standing"
wave.
b. TRUE - Standing wave patterns are characterized by the presence of nodes - points of no disturbance.
c. TRUE - This is correct. When a standing wave pattern is being observed, it is an interference pattern - a pattern of the medium
resulting from two or more waves interfering to produce the very visible pattern.
d. TRUE - This is probably a good definition of a standing wave pattern.
e. FALSE - Nodes are points which are stationary and undergoing no displacement. The antinodal positions of a standing wave
pattern undergo oscillation from a maximum positive displacement to a maximum negative displacement.
f. FALSE -This would be a true statement for standing wave patterns formed in closed-end resonance air columns. Yet for guitar
strings and open end air columns, there is always one more node than antinode.
g. TRUE - This is a good way to describe what is seen in the pattern.

6. Which of the following statements are TRUE of the concept of resonance? Identify all that apply.
a. A musical instrument can play any frequency imaginable.
b. All musical instruments have a natural frequency or set of natural frequencies at which they will vibrate; each frequency
corresponds to a unique standing wave pattern.
c. The result of two objects vibrating in resonance with each other is a vibration of larger amplitude.
d. Objects which share the same natural frequency will often set each other into vibrational motion when one is plucked,
strummed, hit or otherwise disturbed. This phenomenon is known as a forced resonance vibration.
e. A vibrating tuning fork can set a second tuning fork into resonant motion.
f. The resonant frequencies of a musical instrument are related by whole number ratios.

Answer: BCDE(mostly)F
a. FALSE - An instrument which is truly musical can only play a specific set of frequencies, each one corresponding to a standing
wave pattern with which that instrument can vibrate. (Of course, one could make a case that by modifying properties of the
instrument, small adjustments could be made in the speed at which the waves might move and thus allow the instrument to
produce about any frequency imaginable.)
b. TRUE - The frequencies at which an instrument would naturally vibrate are known as its harmonics. Each frequency
corresponds to a unique standing wave pattern.
c. TRUE - Resonance results in a big vibration because two waves are now interfering in a regular manner to produce a resultant
wave with a large amplitude of vibration.
d. TRUE - This is a good definition of resonance vibrations.
e. TRUE - This can happen provided that the two tuning forks have the same natural frequency and that they are
somehow connected (for instance by air).
f. TRUE - The frequencies at which an instrument would naturally vibrate are known as its harmonics. The frequency of each
harmonic is a whole number multiple of the fundamental frequency. As such, every frequency in the set of natural frequency is
related by whole number rations.

7. Which of the following statements are TRUE of the harmonics and standing wave patterns in guitar strings? Identify all that
apply.
a. The fundamental frequency of a guitar string is the highest frequency at which the string vibrates.
b. The fundamental frequency of a guitar string corresponds to the standing wave pattern in which there is a complete
wavelength within the length of the string.
c. The wavelength for the fundamental frequency of a guitar string is 2.0 m.
d. The wavelength for the second harmonic played by a guitar string is two times the wavelength of the first harmonic.
e. The standing wave pattern for the fundamental played by a guitar string is characterized by the pattern with the longest
possible wavelength.
f. If the fundamental frequency of a guitar string is 200 Hz, then the frequency of the second harmonic is 400 Hz.
g. If the frequency of the fifth harmonic of a guitar string is 1200 Hz, then the fundamental frequency of the same string is 6000
Hz.
h. As the frequency of a standing wave pattern is tripled, its wavelength is tripled.
i. If the speed of sound in a guitar string is 300 m/s and the length of the string is 0.60 m, then the fundamental frequency will
be 180 Hz.
j. As the tension of a guitar string is increased, the fundamental frequency produced by that string is decreased.
k. As the tension of a guitar string is increased by a factor of 2, the fundamental frequency produced by that string is decreased
by a factor of 2.
l. As the linear density of a guitar string is increased, the fundamental frequency produced by the string is decreased.
m. As the linear density of a guitar string is increased by a factor 4, the fundamental frequency produced by the string is
decreased by a factor of 2.

Answer: EFLM
a. FALSE - The fundamental frequency is the lowest possible frequency which an instrument will play.
b. FALSE - For a guitar string, the standing wave pattern for the fundamental frequency is one in which there is one-half
wavelength within the length of the string.
c. FALSE - The wavelength for the fundamental frequency is two times the length of the string (not 2.0 m).
d. FALSE - The wavelength of the second harmonic is one-half the length of the wavelength of the fundamental (the frequency of
the second harmonic is twice the frequency of the fundamental).
e. TRUE - The fundamental frequency is the lowest possible frequency and the longest possible wavelength with which an
instrument will vibrate.
f. TRUE -The frequency of the nth harmonic is n times larger than the frequency of the fundamental or first harmonic.
g. FALSE - The frequency of the fundamental would be 240 Hz if the frequency of the fifth harmonic is 1200 Hz.
h. FALSE - If the frequency is tripled, then the wavelength is one-third as much.
i. FALSE - The fundamental frequency would be 250 Hz. The wavelength of the fundamental is two times the length of the string -
1.2 m. And the frequency of the fundamental is the speed divided by the wavelength of the fundamental.
j. FALSE - As the tension of a guitar string is increased, the speed of vibrations in the string is increased and the frequency will be
increased.
k. FALSE - If the tension in a guitar string is increased by a factor of 2, then the speed of vibrations in the string will be increased
by a factor of the square root of 2 (1.41) and the frequency will be increased by a factor of 1.41.
l. TRUE - If the linear density of a guitar string is increased, then the speed of vibrations in the string will be decreased and the
frequency will be decreased.
m. TRUE - If the linear density of a guitar string is increased by a factor of 4, then the speed of vibrations in the string will be
decreased by a factor of the square root of 4 (2.0) and the frequency will be decreased by a factor of 2.

8. Which of the following statements are TRUE of the harmonics and standing wave patterns in air columns? Identify all that apply.
a. The speed of the waves for the various harmonics of open-end air columns are whole number multiples of the speed of
the wave for the fundamental frequency.
b. Longer air columns will produce lower frequencies.
c. The pitch of a sound can be increased by shortening the length of the air resonating inside of an air column.
d. An open end of an air column allows air to vibrate a maximum amount whereas a closed end forces air particles to behave as
nodes.
e. Open-end air columns have antinodes positioned at each end while closed-end air columns have nodes positioned at each
end.
f. Closed-end air columns can only produce odd-numbered harmonics.
g. Open-end air columns can only produce even-numbered harmonics.
h. A closed-end air column that can play a fundamental frequency of 250 Hz cannot play 500 Hz.
i. An open-end air column that can play a fundamental frequency of 250 Hz cannot play 750 Hz.
j. A closed-end air column has a length of 20 cm. The wavelength of the first harmonic is 5 cm.
k. An open-end air column has a length of 20 cm. The wavelength of the first harmonic is 10 cm.
l. Air column A is a closed-end air column. Air column B is an open-end air column. Air column A would be capable of playing
lower pitches than air column B.
m. The speed of sound in air is 340 m/s. An open-end air column has a length of 40 cm. The fundamental frequency of this air
column is approximately 213 Hz.
n. The speed of sound in air is 340 m/s. A closed-end air column has a length of 40 cm. The fundamental frequency of this air
column is approximately 213 Hz.
o. If an open-end air column has a fundamental frequency of 250 Hz, then the frequency of the fourth harmonic is 1000 Hz.
p. If a closed-end air column has a fundamental frequency of 200 Hz, then the frequency of the fourth harmonic is 800 Hz.

Answer: BCDFHLNO
a. FALSE - It is the frequency (not the speed) of the various harmonics which are whole number multiples of the fundamental
frequency.
b. TRUE - Assuming that two air columns are of the same type (both open- or both closed), the standing wave patterns of the
longer air column would have longer wavelengths and thus lower frequencies and pitch.
c. TRUE - As the length of an air column is shortened, the wavelengths are decreased and the frequencies are increased.
d. TRUE - This is exactly the case and is clearly portrayed in the standing wave patterns which are constructed for air columns.
e. FALSE - A closed-end air column is an air column with one end open and one end closed. It is the closed end which is
characterized by a nodal position but the open end is characterized by an antinode.
f. TRUE - Closed-end air columns produce a first, third, fifth, seventh, etc. harmonic - all odd numbers.
g. FALSE - Open-end air columns can produce all harmonics - first, second, third, fourth, etc.
h. TRUE - If a closed-end air column has a fundamental frequency of 250 Hz, then the other frequencies in the set of natural
frequencies are 750 Hz, 1250 Hz, 1750 Hz, etc. It can only have odd-numbered harmonics.
i. FALSE - If an open-end air column has a fundamental frequency of 250 Hz, then the other frequencies in the set of natural
frequencies are 500 Hz, 750 Hz, 1000 Hz, 1250 Hz, etc. An open-end air column can produce all the harmonics.
j. FALSE - The wavelength of the first harmonic of a closed-end air column is four times the length of the air column - 80 cm.
k. FALSE - The wavelength of the first harmonic of an open-end air column is two times the length of the air column - 40 cm.
l. TRUE (sort of) - For the same length (the sort of part), a closed-end air column would have standing waves which are longer
and therefore have frequencies which are lower.
m. FALSE - If the length of this open-end air column is 40 cm, then the wavelength of the fundamental is 0.80 m. The frequency
of the fundamental is (340 m/s)/(0.8 m) = 425 Hz.
n. TRUE - If the length of this closed-end air column is 40 cm, then the wavelength of the fundamental is 1.60 m. The frequency
of the fundamental is (340 m/s)/(1.6 m) = 213 Hz.
o. TRUE - The frequency of the fourth harmonic is four times the frequency of the first harmonic.
p. FALSE - A closed-end air column cannot have a fourth harmonic; there are only odd-numbered frequencies.

9. Which of the following statements are TRUE of sound interference and beats? Identify all that apply.
a. Beats result when two sounds of slightly different frequencies interfere.
b. Beats are characterized by a sound whose frequency is rapidly fluctuating between a high and a low pitch.
c. Two sounds with a frequency ratio of 2:1 would produce beats with a beat frequency of 2 Hz.
d. Two tuning forks are sounding out at slightly different frequencies - 252 Hz and 257 Hz. A beat frequency of 5 Hz will be
heard.
e. A piano tuner is using a 262 Hz tuning fork in an effort to tune a piano string. She plucks the string and the tuning fork and
observes a beat frequency of 2 Hz. Therefore, she must lower the frequency of the piano string by 2 Hz.

Answer: AD
a. TRUE - This is the number one criteria for the formation of audible beats.
b. FALSE - Beats are characterized by sounds which are rapidly oscillating between high and low levels of loudness due to
fluctuations in the amplitude of the resulting wave.
c. FALSE - Two sounds with a frequency difference (not ratio) of 2 Hz will produce a bear frequency of 2 Hz.
d. TRUE - The beat frequency is the frequency at which the amplitude of the oscillations increase and decrease. This beat
frequency is always the difference in frequency of the two sounds which interfere to create the beats.
e. FALSE - She must either lower or increase the frequency of the piano string by 2 Hz.
Part B: Multiple Choice
10. What type of wave is produced when the particles of the medium are vibrating to and fro in the same direction of wave
propagation?
a. longitudinal wave. b. sound wave. c. standing wave. d. transverse wave.

Answer: A
This is the definition of a longitudinal wave. A longitudinal wave is a wave in which particles of the medium vibrate to and fro in a
direction parallel to the direction of energy transport.

11. When the particles of a medium are vibrating at right angles to the direction of energy transport, the type of wave is described
as a _____ wave.
a. longitudinal b. sound c. standing d. transverse
Answer: D
This is the definition of a transverse wave. A transverse wave is a wave in which particles of the medium vibrate to and fro in a
direction perpendicular to the direction of energy transport.

12. A transverse wave is traveling through a medium. See diagram below. The particles of the medium are moving.

a. parallel to the line joining AD. b. along the line joining CI.
c. perpendicular to the line joining AD. d. at various angles to the line CI.
e. along the curve CAEJGBI.

Answer: A
In transverse waves, particles of the medium vibrate to and fro in a direction perpendicular to the direction of energy transport. In
this case, that would be parallel to the line AD.

13. If the energy in a longitudinal wave travels from south to north, the particles of the medium ____.
a. move from north to south, only. b. vibrate both north and south.
c. move from east to west, only. d. vibrate both east and west.

Answer: B
In longitudinal waves, particles of the medium vibrate to and from in a direction parallel to the direction of energy transport. If the
particles only moved north and not back south, then the particles would be permanently displaced from their rest position; this is
not wavelike.
14. The main factor which effects the speed of a sound wave is the ____.
a. amplitude of the sound wave b. intensity of the sound wave
c. loudness of the sound wave d. properties of the medium
e. pitch of the sound wave

Answer: D
The speed of a wave is dependent upon the properties of the medium and not the properties of the wave.

15. As a wave travels into a medium in which its speed increases, its wavelength ____.
a. decreases b. increases c. remains the same
Answer: B
As a wave crosses a boundary into a new medium, its speed and wavelength change while its frequency remains the same. If the
speed increases, then the wavelength must increase as well in order to maintain the same frequency.

16. As a wave passes across a boundary into a new medium, which characteristic of the wave would NOT change?
a. speed b. frequency c. wavelength

Answer: B
As a wave crosses a boundary into a new medium, its speed and wavelength change while its frequency remains the same. This is
true of all waves as they pass from one medium to another medium.

17. The ____ is defined as the number of cycles of a periodic wave occurring per unit time.
a. wavelength b. period c. amplitude d. frequency

Answer: D
This is a basic definition which you should know and be able to apply.

18. Many wave properties are dependent upon other wave properties. Yet, one wave property is independent of all other wave
properties. Which one of the following properties of a wave is independent of all the others?
a. wavelength b. frequency c. period d. velocity

Answer: D
The speed (or velocity) of a wave is dependent upon the properties of the medium through which it moves, not upon the
properties of the wave itself.

19. Consider the motion of waves in a wire. Waves will travel fastest in a ____ wire.
a. tight and heavy b. tight and light c. loose and heavy d. loose and light

Answer: B
The speed of a wave in a wire is given by the equation
v = SQRT (Ftens/mu)
where Ftens is the tension of the wire and a measure of how tight it is pulled and mu is the linear density of the wire and a measure
of how light it is on a per meter basis. Tighter wires allow for faster speeds. Light wires allow for faster speeds.

20. TRUE or FALSE:

The SI unit for frequency is hertz.
a. True b. False

Answer: A
Know this like the back of your hand (assuming you know the back of your hand well).

21. TRUE or FALSE:

Doubling the frequency of a sound source doubles the speed of the sound waves which it produces.
a. True b. False

Answer: B
Don't be fooled. Wave speed may equal frequency*wavelength. Yet doubling the frequency only halves the wavelength;
wave speed remains the same. To change the wave speed, the medium would have to be changed.

22. A sound wave has a wavelength of 3.0 m. The distance between the center of a compression and the center of the next
adjacent rarefaction is ____.
a. 0.75 m. b. 1.5 m. c. 3.0 m. d. 6.0 m.
e. impossible to calculate without knowing frequency.

Answer: B
The wavelength of a wave is measured as the distance between any two corresponding points on adjacent wave. For a
sound wave, that would be from compression to the next adjacent compression. If that distance is 3.0 meters, then the
distance from compression to the next adjacent rarefaction is 1.5 m.

23. Which one of the following factors determines the pitch of a sound?
a. The amplitude of the sound wave
b. The distance of the sound wave from the source
c. The frequency of the sound wave
d. The phase of different parts of the sound wave
e. The speed of the sound wave
Answer: C
The pitch of a sound wave is related to the frequency of the sound wave.
24. A certain note is produced when a person blows air into an organ pipe. The manner in which one blows on a organ pipe (or
any pipe) will effect the characteristics of the sound which is produced. If the person blows slightly harder, the most probable
change will be that the sound wave will increase in ____.
a. amplitude b. frequency c. pitch d. speed e. wavelength

Answer: A
If you put more energy into the wave - i.e., blow harder - then the amplitude of the waves will be greater. Energy and
amplitude are related.

25. A vibrating object with a frequency of 200 Hz produces sound which travels through air at 360 m/s. The number of meters
separating the adjacent compressions in the sound wave is ____.
a. 0.90 b. 1.8 c. 3.6 d. 7.2 e. 200

Answer: B
Let w=wavelength; then v = w*f. In this problem, it is given that v=360 m/s and f = 200 Hz. Substitution and algebra
yields w = v/f = 1.8 m. The question asks for the wavelength - i.e., the distance between adjacent compressions.

26. Consider the diagram below of several circular waves created at various times and locations. The diagram illustrates ____.

a. interference b. diffraction c. the Doppler effect. d. polarization

Answer: C
The Doppler effect or Doppler shift occurs when a source of waves is moving with respect to an observer. The observer observes a
different frequency of waves than that emitted by the source. This is due to the fact that the waves are compressed together into
less space in the direction in which the source is heading.

27. In the diagram above, a person positioned at point A would perceive __________ frequency as the person positioned at point
B.
a. a higher b. a lower c. the same

Answer: A
The Doppler effect or Doppler shift occurs when a source of waves is moving with respect to an observer. The observer observes a
different frequency of waves than that emitted by the source. If the source and observer are approaching, then the observed
frequency is higher than the emitted frequency. If the source and observer are moving away from each other, the observer
observes a lower frequency than the emitted frequency.

28. A girl moves away from a source of sound at a constant speed. Compared to the frequency of the sound wave produced by the
source, the frequency of the sound wave heard by the girl is ____.
a. lower. b. higher. c. the same.

Answer: A
The Doppler effect or Doppler shift occurs when a source of waves is moving with respect to an observer. The observer
observes a different frequency of waves than that emitted by the source. If the source and observer are moving away,
then the observed frequency is lower than the emitted frequency.

29. An earth-based receiver is detecting electromagnetic waves from a source in outer space. If the frequency of the waves are
observed to be increasing, then the distance between the source and the earth is probably ____.
a. decreasing. b. increasing. c. remaining the same.

Answer: A
The Doppler effect or Doppler shift occurs when a source of waves is moving with respect to an observer. The observer
observes a different frequency of waves than that emitted by the source. If the source and observer are approaching,
then the observed frequency is higher than the emitted frequency. If the source and observer are approaching, then the
distance between them is decreasing.

30. As two or more waves pass simultaneously through the same region, ____ can occur.
a. refraction b. diffraction c. interference d. reflection

Answer: C
Interference is the meeting of two or more waves when passing along the same medium - a basic definition which you
should know and be able to apply.

31. TRUE or FALSE:

If two crests meet while passing through the same medium, then constructive interference occurs.
a. True b. False
Answer: A
Yes! Or when a trough meets a trough or whenever two waves displaced in the same direction - both up or both down -
meet.

32. A node is a point along a medium where there is always ____.

a. a crest meeting a crest b. a trough meeting a trough c. constructive interference
d. destructive interference e. a double rarefaction.

Answer: D
A node is a point along the medium of no displacement. The point is not displaced because destructive interference
occurs at this point.

33. TRUE or FALSE:

It is possible that one vibrating object can set another object into vibration if the natural frequencies of the two objects are the
same.
a. True b. False

Answer: A
Yes! This is known as resonance. Resonance occurs when a vibrating object forces another object into vibration at the
same natural frequency. A basic definition of a commonly discussed phenomenon.

34. An object is vibrating at its natural frequency. Repeated and periodic vibrations of the same natural frequency impinge upon
the vibrating object and the amplitude of its vibrations are observed to increase. This phenomenon is known as ____.
a. beats b. fundamental c. interference d. overtone e. resonance

Answer: E
Resonance occurs when a vibrating object forces another object into vibration at the same natural frequency and thus
increase the amplitude of its vibrations. A basic definition of a commonly discussed phenomenon.

35. A standing wave experiment is performed to determine the speed of waves in a rope. The standing wave pattern shown below
is established in the rope. The rope makes 90.0 complete vibrational cycles in exactly one minute. The speed of the waves is ____
m/s.

a. 3.0 b. 6.0 c. 180 d. 360 e. 540

Answer: B
Ninety vibrations in 60.0 seconds means a frequency of 1.50 Hz. The diagram shows 1.5 waves in 6.0-meters of rope;
thus, the wavelength (w) is 4 meters. Now use the equation v=f*w to calculate the speed of the wave. Proper
substitution yields 6.0 m/s.

36. Standing waves are produced in a wire by vibrating one end at a frequency of 100. Hz. The distance between the 2nd and the
5th nodes is 60.0 cm. The wavelength of the original traveling wave is ____ cm.
a. 50.0 b. 40.0 c. 30.0 d. 20.0 e. 15.0

Answer: B
The frequency is given as 100. Hz and the wavelength can be found from the other givens. The distance between
adjacent nodes is one-half a wavelength; thus the 60.0-cm distance from 2nd to 5th node is 1.50 wavelengths. For this
reason, the wavelength is 40.0 cm.

37. Consider the standing wave pattern shown below. A wave generated at the left end of the medium undergoes reflection at the
fixed end on the right side of the medium. The number of antinodes in the diagram is

a. 3.0 b. 5.0 c. 6.0 d. 7.0 e. 12

Answer: C
An antinode is a point on the medium which oscillates from a large + to a large - displacement. Count the number of
these points - there are 6 - but do not count them twice.
38. The standing wave pattern in the diagram above is representative of the ____ harmonic.
a. third b. fifth c. sixth d. seventh e. twelfth

Answer: C
If there are six antinodes in the standing wave pattern, then it is the sixth harmonic.

39. The distance between successive nodes in any standing wave pattern is equivalent to ____ wavelengths.
a. 1/4 b. 1/2 c. 3/4 d. 1 e. 2.

Answer: B
Draw a standing wave pattern or look at one which is already drawn; note that the nodes are positioned one-half of a
wavelength apart. This is true for guitar strings and for both closed-end and open-end resonance tubes.

40. A vibrating tuning fork is held above a closed-end air column, forcing the air into resonance. If the sound waves created by the
tuning fork have a wavelength of W, then the length of the air column could NOT be ____.
a. 1/4 W b. 2/4 W c. 3/4 W d. 5/4 W e. 7/4 W

Answer: B
Review your diagrams for the standing wave patterns in closed end air columns; note that resonance occurs when the
length of the air column is 1/4, 3/4, 5/4, 7/4, ... of a wavelength. Because these possible resonant lengths are
characterized by an odd-numbered numerator, it is said that closed-end air columns only produce odd harmonics.

41. TRUE or FALSE:

A vibrating tuning fork is held above an air column, forcing the air into resonance. The length of the air column is adjusted to
obtain various resonances. The sound waves created by the tuning fork have a wavelength of W. The difference between the
successive lengths of the air column at which resonance occurs is 1/2 W.
a. True b. False

Answer: A
True! Observe the standing wave patterns and the length-wavelength relationships which we have discussed for both open- and
closed-end tubes. In each case, resonance occurs at lengths of tubes which are separated by one-half wavelength; e.g.,
Closed: .25*W, .75*wW 1.25*W, 1.75*W... Open: .5*W, 1.0*W, 1.5*W, 2.0*W, ...

42. TRUE or FALSE:

An organ pipe which is closed at one end will resonate if its length is equal to one-half of the wavelength of the sound in the pipe.
a. True b. False

Answer: B
It will resonate if the length is equal to the one-fourth (or three-fourths, or five-fourths or ...) the wavelength of the
sound wave.

43. A 20.0-cm long pipe is covered at one end in order to create a closed-end air column. A vibrating tuning fork is held near its
open end, forcing the air to vibrate in its first harmonic. The wavelength of the standing wave pattern is ____.
a. 5.0 cm b. 10.0 cm c. 20.0 cm d. 40.0 cm e. 80.0 cm

Answer: E
This is a closed-end air column. If you draw the standing wave pattern for the first harmonic, you will notice that the
wavelength is four times the length of the air column. Thus take the length of 20.0 cm and multiply by 4.

44. A stretched string vibrates with a fundamental frequency of 100. Hz. The frequency of the second harmonic is ____.
a. 25.0 Hz b. 50.0 Hz c. 100. Hz d. 200. Hz e. 400. Hz

Answer: D
The frequency of the nth harmonic is n times the frequency of the first harmonic where n is an integer. Thus, f 2 = 2*f1 =
2*100. Hz = 200. Hz.

45. A 40.0-cm long plastic tube is open at both ends and resonating in its first harmonic. The wavelength of the sound which will
produce this resonance is ____.
a. 10.0 cm b. 20.0 cm c. 40.0 cm d. 80.0 cm e. 160 cm

Answer: D
For an open-end air column, the length of the column is 0.5*wavelength. This becomes evident after drawing the
standing wave pattern for this harmonic. Then, plug in 40.0 cm for length and calculate the wavelength.

46. The diagrams below represent four different standing wave patterns in air columns of the same length. Which of the columns
is/are vibrating at its/their fundamental frequency? Include all that apply.
Answer: CD
The fundamental frequency is the lowest possible frequency for that instrument, and thus the longest possible
wavelength. For open tubes, there would be anti-nodes on each end and a node in the middle. For closed end tubes,
there would be a node on the closed end, an anti-node on the open end, and nothing in the middle. Diagram C is the
third harmonic for a closed end tube and diagram D is the second harmonic for an open-end tube.

47. The diagrams above (Question #46) represent four different standing wave patterns in air columns of equal length. Which of
the columns will produce the note having the highest pitch?
a. A b. B c. C d. D
e. All column produce notes having the same pitch

Answer: D
Just look at the wave patterns and notice that the shortest wavelength is in diagram D and so it must have the highest
frequency or pitch.

48. An air column closed at one end filled with air resonates with a 200.-Hz tuning fork. The resonant length corresponding to the
first harmonic is 42.5 cm. The speed of the sound must be ____.
a. 85.0 m/s b. 170. m/s c. 340 m/s d. 470. m/s e. 940. m/s

Answer: C
Draw the standing wave pattern for the first harmonic of a closed-end tube to assist with the length-wavelength relation.
Then, L=0.425 m so w=1.70 m. Since f is given as 200. Hz, the speed can be calculated as f*w or 200. Hz*1.70 m. The
speed of sound is 340. m/s.
49.TRUE or FALSE:
A violinist plays a note whose fundamental frequency is 220 Hz. The third harmonic of that note is 800 Hz.
a. True b. False

Answer: B
The frequency of the nth harmonic is n times the frequency of the first harmonic where n is an integer. Thus, f 3 = 3*f1 =
3*220 Hz = 660 Hz.

50. In order for two sound waves to produce audible beats, it is essential that the two waves have ____.
a. the same amplitude b. the same frequency
c. the same number of overtones d. slightly different amplitudes
e. slightly different frequencies

Answer: E
Beats occur whenever two sound sources emit sounds of slightly different frequencies. Perhaps you recall the
demonstration in class with the two tuning forks of slightly different frequencies.

51. TRUE or FALSE:

Two tuning forks with frequencies of 256 Hz and 258 Hz are sounded at the same time. Beats are observed; 2 beats will be heard
in 2 s.
a. True b. False

Answer: B
Beats occur whenever two sound sources emit sounds of slightly different frequencies. The beat frequency is just the
difference in frequency of the two sources. In this case, the beat frequency would be 2.0 Hz, which means that 2 beats
would be heard every 1 second or 4 beats every 2 seconds.

52. A tuning fork of frequency 384 Hz is sounded at the same time as a guitar string. Beats are observed; exactly 30 beats are
heard in 10.0 s. The frequency of the string in hertz is ____.
a. 38.4 b. 354 or 414 c. 369 or 399 d. 374 or 394 e. 381 or 387

Answer: E
Beats occur whenever two sound sources emit sounds of slightly different frequencies. The beat frequency is just the
difference in frequency of the two sources. In this case, the beat frequency is given as 3.00 Hz, which means that the
second source must have a frequency of either 3.00 Hz above or 3.00 Hz below the first source - either 381 Hz or 387
Hz.

Part C: Problem-Solving and Computational Problems

53. Determine the decibel rating of the following intensities of sound.
a. I = 1.0 x 10-5 W/m2
b. I = 1.0 x 10-2 W/m2
c. I = 6.1 x 10-6 W/m2
d. I = 2.2 x 10-4 W/m2
e. A sound which is 4 times more intense than the sound in part d.
f. A sound which is 7 times more intense than the sound in part d.
g. A sound which is 10 times more intense than the sound in part d
h. A sound which is 100 times more intense than the sound in part d.
i. The sound of an orchestra playing a movement pianissimo at 7.5 x 10-6 W/m2 (very softly)
j. The sound of an orchestra playing a movement fortissimo at 2.5 x 10-4 W/m2 (very loudly)

Answers: See below

The equation which relates the intensity of a sound wave to its decibel level is:
dB = 10 * log( I / 1.0 x 10-12 W/m2)
where I = intensity of the sound in units of W/m2. Using this equation, parts a - d can be computed in straightforward
fashion.
a. dB = 10 * log( 1 x 10-5 W/m2 / 1.0 x 10-12 W/m2) = 70. dB
b. dB = 10 * log( 1 x 10-2 W/m2 / 1.0 x 10-12 W/m2) = 100. dB
c. dB = 10 * log( 6.1 x 10-6 W/m2 / 1.0 x 10-12 W/m2) = 68 dB
d. dB = 10 * log( 2.2 x 10-4 W/m2 / 1.0 x 10-12 W/m2) = 83 dB
e. The intensity is 4 x (2.2 x 10-4 W/m2 ) or 8.8 x 10-4 W/m2 ; so the decibel rating is 89 dB.
f. The intensity is 7 x (2.2 x 10-4 W/m2 ) or 1.54 x 10-3 W/m2 ; so the decibel rating is 92 dB.
g. The intensity is 10 x (2.2 x 10-4 W/m2 ) or 2.2 x 10-3 W/m2 ; so the decibel rating is 93 dB.
h. The intensity is 100 x (2.2 x 10-4 W/m2 ) or 2.2 x 10-2 W/m2 ; so the decibel rating is 103 dB.
i. dB = 10 * log( 7.5 x 10-6 W/m2 / 1.0 x 10-12 W/m2) = 69 dB
j. dB = 10 * log( 2.5 x 10-4 W/m2 / 1.0 x 10-12 W/m2) = 84 dB

54. A machine produces a sound with an intensity of 2.9 x 10-3 W/m2. What would be the decibel rating if four of these machines
occupy the same room?
Answer: 101 dB
Four of these machines would be four times as intense as one machine - that would be an intensity of 1.16 x 10-2 W/m2 .
The decibel rating is
dB = 10 * log( 1.16 x 10-2 W/m2 / 1.00 x 10-12 W/m2) = 101 dB

55. The sound in the United Center during a Chicago Bulls basketball game in 1998 was seven times as intense as it is today. If the
decibel rating today is 89 dB, then what was the intensity rating in 1998?

Answer: 5.56 x 10-3 W/m2

This problem involves finding the intensity level in the stadium today and then multiplying it by seven. The conversion
from dB to intensity in W/m2 is done using the usual equation.
dB = 10 * log( I / 1 x 10-12 W/m2)
The decibel level is substituted into the equation and then each side is divided by 10.
89 dB = 10 * log( I / 1 x 10-12 W/m2)
8.9 = log( I / 1 x 10-12 W/m2)
Now to solve for I, take the invlog of each side of the equation.
invlog(8.9) = invlog( log( I / 1 x 10-12 W/m2) )
7.94 x 108 = I / 1 x 10-12 W/m2
Now multiply both sides of the equation by 1 x 10-12 W/m2 to solve for I.
7.94 x 10-4 W/m2 = I
Now multiply by 7 since the intensity was seven times greater.
I = 5.56 x 10-3 W/m2

56. A sound has an intensity of 8.0 x10-3 W/m2 at a distance of 2.0 m from its source. What is the intensity at a distance of ...
a. ... 4.0 m from the source?
b. ... 6.0 m from the source?
c. ... 8.0 m from the source?
d. ... 24.0 m from the source?
e. ... 46.1 m from the source?

Answers: See below.

This problem tests your understanding of the inverse square law: the intensity of a sound varies inversely with the
square of the distance from the source of the sound.
I = k • (1/R2)
a. As the distance is doubled (4 m is two times 2 m), the intensity level is reduced by a factor of 4. The new intensity is
8.0 x10-3 W/m2/ 4 = 2.0 x10-3 W/m2.
b. As the distance is tripled (6 m is three times 2 m), the intensity level is reduced by a factor of 9. The new intensity is
8.0 x10-3 W/m2/ 9 = 8.9 x10-4 W/m2.
c. As the distance is quadrupled (8 m is four times 2 m), the intensity level is reduced by a factor of 16. The new
intensity is 8.0 x10-3 W/m2/ 16 = 5.0 x10-4 W/m2.
d. As the distance is increased by a factor of 12 (24 m is 12 times 2 m), the intensity level is reduced by a factor of 144.
The new intensity is 8.0 x10-3 W/m2/ 144 = 1.4 x10-5 W/m2.
e. As the distance is increased by a factor of 23.05 (46.1 m is 23.05 times 2 m), the intensity level is reduced by a factor
of 531. The new intensity is 8.0 x10-3 W/m2/ 531 = 1.5 x10-5 W/m2.
57. Ben Stupid is sitting 2.0 m in front of the speakers on the stage at the Twisted Brother concert. The decibel rating of the sound
heard there is 110 dB. What would be the decibel rating at a location of ...
a. ... 4.0 m from the speaker?
b. ... 6.0 m from the speaker?
c. ... 20.0 m from the speaker?
Answer: See answers below.
This problem is similar to the last problem in that it tests your understanding of the inverse square law. There is however
a slight twist. The decibel rating is given and since the decibel scale is logarithmic, we must first find the intensity level
before we do the division. A decibel rating of 110 db is equivalent to an intensity level of 1011 times the threshold of
hearing (1 x 10-12 W/m2). So the intensity level at 2.0 m from the speaker is 0.1 W/m2. Now the inverse square law is
applied; the intensity of a sound varies inversely with the square of the distance from the source of the sound.
I = k • (1/R2)
a. As the distance is doubled (4 m is two times 2 m), the intensity level is reduced by a factor of 4. The new intensity is
0.1 W/m2/ 4 = 0.025 W/m2 or 2.5 x10-2 W/m2. This converts to a decibel rating of 104 dB.
b. As the distance is tripled (6 m is three times 2 m), the intensity level is reduced by a factor of 9. The new intensity is
0.1 W/m2/ 9 = 0.011 W/m2 or 1.1 x10-2 W/m2. This converts to a decibel rating of 100.5 dB.
c. As the distance is increased by a factor of 10 (20 m is 10 times 2 m), the intensity level is reduced by a factor of 100.
The new intensity is 0.1 W/m2/ 100 = 0.0010 W/m2 or 1.0 x10-3 W/m2. This converts to a decibel rating of 90 dB.

58. Use the Doppler equation for a moving source to calculate the observed frequency for a 250.-Hz source of sound if it is moving
with a speed of ____ . (Assume that the speed of sound in air is 340. m/s.)
a. 30. m/s towards the observer.
b. 30. m/s away from the observer.
c. 300. m/s towards the observer.
d. 300. m/s away from the observer.
e. 320. m/s towards the observer.
f. 335 m/s towards the observer.

Answers: See below.

The Doppler equation for determining the observed frequency for a moving source is:
fobserved = vsound / (vsound ± vsource) •fsource
The + sign is used if the source moves away from the observer
The - sign is used if the source moves towards the observer.

If applied to this situation, vsound is 340 m/s and fsource is 250 Hz.
a. fobserved = (340. m/s)/(340. m/s - 30. m/s) • (250 Hz) = (340. m/s)/(310. m/s) • (250. Hz) = 274 Hz
b. fobserved = (340. m/s)/(340. m/s + 30. m/s) • (250 Hz) = (340. m/s)/(370. m/s) • (250. Hz) = 230. Hz
c. fobserved = (340. m/s)/(340. m/s - 300. m/s) • (250 Hz) = (340. m/s)/(40. m/s) • (250. Hz) = 2130 Hz (2125 Hz)
d. fobserved = (340. m/s)/(340. m/s + 300. m/s) • (250 Hz) = (340. m/s)/(640. m/s) • (250. Hz) = 133 Hz
e. fobserved = (340. m/s)/(340. m/s - 320. m/s) • (250 Hz) = (340. m/s)/(20. m/s) • (250. Hz) = 4250 Hz
f. fobserved = (340. m/s)/(340. m/s - 335 m/s) • (250 Hz) = (340. m/s)/(5 m/s) • (250. Hz) = 1.70 x 104 Hz

59. Shirley Yackin is holding the phone cord in her hand. It is stretched to a length of 2.4 m and has a mass of 1.8 kg. If the
tension in the phone cord is 2.5 N, then determine the speed of vibrations within the cord.

Answer: 1.8 m/s

The telephone cord acts as a guitar string. As such the speed of waves in the cord is given by the equation
v = SQRT(Ftens/ mu)
where Ftens is the tension in the cord in Newtons and mu is the linear density in kg/m. The tension is given and the mu
value can be calculated:
mu = mass/length = (1.8 kg) / (2.4 m) = 0.75 kg/m
Now substitute mu and Ftens into the equation and solve for speed:
v = SQRT[(2.5 N) / (0.75 kg/m)] = SQRT(3.33) = 1.8 m/s (1.8257 ... m/s)

60. (Referring to problem #59.) With what frequency must Shirley vibrate the cord up and down in order to produce the second
harmonic within the cord?

Answer: 0.763 Hz

If Shirley is producing the second harmonic, then there will be one full wavelength in the length of the cord.
Since the length of the cord is 2.4 m, the wavelength is 2.4 m as well (see diagram at right). With the wavelength and speed
(problem #59) known, the frequency can be calculated. (Note that the unrounded number from problem #59 is used i this
calculation.)
f = v / lambda = (1.8257 m/s) / (2.4 m) = 0.76 Hz (0.7607... Hz)
61. (Referring to problem 60.) If Shirley maintains this same frequency and wishes to produce the fourth harmonic, then she will
have to alter the speed of the wave by changing the tension. Assuming the same mass density as in #59, and the same frequency
as in #60, to what tension must the cord be pulled to produce the fourth harmonic?

Answer: 0.63 N

In this case, Shirley is changing the medium. So she is producing the fourth harmonic in a different
medium (different properties) using the same frequency as in #60. For the fourth harmonic, there are two full
wavelengths inside the length of the cord (see diagram at right); so the length of the wave ( wavelength) is one-half the
length of the cord - 1.2 m. With frequency and wavelength known, the speed can be computed:
v = f • lambda = (0.7607 Hz) • (1.2 m) = 0.91287... m/s.
Now with speed and mu known, the tension can be calculated from the equation:
v = SQRT(Ftens/ mu)
First, perform algebra to manipulate the equation into a form with Ftens by itself. This is done by squaring both sides of
the equation and multiplying by mu.
v2 = Ftens/ mu
mu • v2 = Ftens
Now substitute and solve:
Ftens = (0.75 kg/m) • ( 0.91287... m/s)2 = 0.63 N (0.6250 N)

62. A guitar string has a mass of 32.4 g and a length of 1.12 m. The string is pulled to a tension of 621 N. Determine the speed at
which vibrations move within the string.

Answer: 147 m/s

This problem is very similar to question #59. The speed of waves in the guitar string is given by the equation
v = SQRT(Ftens/ mu)
where Ftens is the tension in the string in Newtons and mu is the linear density in kg/m. The tension is given and the mu
value can be calculated (one will have to be careful of units):
mu = mass/length = (0.0324 kg) / (1.12 m) = 0.028928... kg/m
Now substitute mu and Ftens into the equation and solve for speed:
v = SQRT [ (621 N)/(0.028928... kg/m) ] = SQRT(21467) = 147 m/s (146.515... m/s)

63. (Referring to problem #62.) Stan Dingwaives is playing this guitar. If Stan leaves the string "open" and uses its full length to
produce the first harmonic, then what frequency will Stan be playing?

Answer: 65.4 Hz
For the first harmonic, there is only one-half of a wave inside of the length of the string. So the length of the wave is two times
longer than the length of the string (see diagram at right). Thus, the wavelength is 2.24 m. Now with speed and wavelength
known, the frequency can be calculated.
f = v /lambda = (146.515... m/s) / (2.24 m) = 65.4 Hz (65.4085... Hz)

64. (Referring to problem #62 and #63.) If Stan wishes to increase the frequency by a factor of 1.2599, then how far (in cm) from
the end of the string must he "close" the string (i.e., where must he press his finger down to change the length and produce the
desired frequency)? Use the same mass density and speed as in problem #62.

Answer: 0.23 m
For the same speed (same properties of the medium), frequency and wavelength are inversely related. Increasing the
frequency by a factor of 1.2599 will be the result of decreasing the wavelength by the same factor. So the new
wavelength will be
lambda = (2.24 m) / 1.2599 = 1.78 m
Since the wavelength is twice the effective length of the string, the string must have an effective length of 0.89 m or 89
cm. So, Stan must place his finger a distance of 0.23 m or 23 cm (1.12 m - 0.89 m) from the end of the string.

65. A guitar string has a fundamental frequency of 262 Hz. Determine the frequency of the ...
a. ... second harmonic.
b. ... third harmonic.
c. ... fifth harmonic.
d. ... eighth harmonic.

Answers:
The frequencies of the various harmonics of an instrument are whole number rations of the fundamental frequency. The frequency
of the second harmonic is two times the fundamental frequency; the frequency of the third harmonic is three times the
fundamental frequency and so on. The frequencies of the harmonics can be found utilizing the formula
fn = n * f 1
a. f2 = 2 * f1 = 2 * 262 Hz = 524 Hz
b. f3 = 3 * f1 = 3 * 262 Hz = 786 Hz
c. f5 = 5 * f1 = 5 * 262 Hz = 1310 Hz
d. f8 =8 * f1 = 8 * 262 Hz = 2.10x 103 Hz

66. Determine the speed of sound through air if the temperature is ...
a. ... 0 degrees Celsius.
b. ... 12 degrees Celsius.
c. ... 25 degrees Celsius.
d. ... 40 degrees Celsius.

Answers:
There are numerous equations for computing the speed of sound through air based on the temperature of air. A common
equation found in books is
v = 331 m/s * SQRT (1 + T/273)
where T is the Celsius temperature. The following answers were found using this equation.
a. v = 331 m/s * SQRT (1 + 0/273) = 331 m/s
b. v = 331 m/s * SQRT (1 + 12/273) = 338 m/s
c. v = 331 m/s * SQRT (1 + 25/273) = 346 m/s
d. v = 331 m/s * SQRT (1 + 40/273) = 354 m/s
An alternative equation which is comonly used is v = 331 m/s + (.60 m/s/C)*T where T is the temperature is
degrees celsius. Using this equation yields nearly the same same speed values - 331 m/s, 338 m/s, 346 m/s, and 355
m/s repsectively.

67. A wind chime is an open-end air column. Determine the fundamental frequencies of a 62.5-cm chime when the temperature is
... .
a. ... 12 degrees on a cold autumn evening.
b. ... 25 degrees on a summer evening.
c. ... 40 degrees during a hot summer day.

Answers:

The frequency of the sound produced by a wind chime is related to the

speed of air in the wind chime and the wavelength of the standing wave pattern of the resonating air column. The speed
of the wave in air depends on the properties of air (temperature); these values were just computed in problem #66. The
wavelength (lambda) of the resonating air column can be determined using a good diagram accompanied by the length
of the air column. As depicted in the diagram at the right, the wavelength of the wave for the fundamental frequency is
two times the length of the open-end air column. So in each of these cases, the wavelength is 2*62.5 cm = 125 cm =
1.25 m. With speed and wavelength known, the frequency values can be computed.
a. f1 = v / lambda = (338 m/s) / (1.25 m) = 271 Hz
b. f1 = v / lambda = (346 m/s) / (1.25 m) = 277 Hz
c. f1 = v / lambda = (354 m/s) / (1.25 m) = 284 Hz
68. An organ pipe has a length of 2.45 m and is open at both ends. Determine the fundamental frequency of the pipe if the
temperature in the room is 25 degrees Celsius.

Answer: 70.6 Hz
For an open-end air column, the wavelength of the fundamental's standing wave pattern is two times the length of the
air column; this relationship is depicted in the diagram at the right. So the wavelength of the wave is 4.90 m. The speed
of the sound wave in air is dependent upon temperature. This speed was calculated in problem #66; it is 346 m/s. The
frequency of the fundamental can now be calculated:
f1 = v / lambda
f1 = (346 m/s) / (4.90 m) = 70.6 Hz

69. (Referring to problem #68.) Determine the fundamental frequency of the pipe if it is closed at one end.

Answer: 35.3 Hz
The easy way to solve this problem is to recognize that by changing an open end to a closed end has the effect of lengthening the
wave by a factor of two. If the wavelength is twice as large, then the frequency is twice as small. Thus, divide the original answer
of 70.6 Hz (question #68) by 2.

Alternatively, one could repeat the entire process of re-computing the wavelength from a diagram of the standing wave of the
fundamental in a closed-end air column. For the fundamental, the wavelength is four times the length of the air column. In this
case, the wavelength would be 9.80 m. Now divide the speed of the sound wave (346 m/s) by the wavelength value to obtain the
frequency.

70. The auditory canal of the outer ear acts as a closed end resonator which has a natural frequency of around 3500 Hz. This canal
serves to amplify sounds with frequencies around this value, thus making us more sensitive to such frequencies. If the speed of
waves inside the canal is 350 m/s, then what is the estimated length of the canal?
Answer: 2.5 cm
For closed end resonators, the standing wave patterns of the natural frequencies are characterized by a node at the
closed end and an antinode at the open end. For the first harmonic (and it must be assumed that the 3500 Hz
corresponds to the first harmonic), the wavelength is four times the length of the air column. The strategy in this
problem involves determining the wavelength (lambda) from the speed and the frequency and then determining the
length from the wavelength-length relationship.
lambda = v / f = (350 m/s) / (3500 Hz) = 0.1 m
L = 0.25 * lambda = 0.25 * (0.1 m) = 0.025 m = 2.5 cm

71. Determine the frequency of the lowest three harmonics at which a closed-end air column would sound out at 25 degrees
Celsius if its length is 135 cm.

Answer: 64.0 Hz, 192 Hz, 320 Hz

The fundamental frequency of a closed-end air column is characterized by a standing wave pattern in which there is a
node at the closed end and an antinode at the open end. There is one-fourth of a wavelength present within the length
of the tube; thus, the wavelength is four times the length. In the case of this air column, the wavelength of the first
harmonic is
lambda1 = 4*135 cm = 540 cm = 5.40 m
The speed of sound waves in air at 25 degree Celsius was computed in problem #66 to be 346 m/s. The frequency of
the first harmonic can now be calculated from the speed and the wavelength.
f1 = v / lambda1 = (346 m/s) / (5.40 m) = 64.0 Hz
The next two lowest frequencies are the third and the fifth harmonics. (Recall the closed-end air columns do not have
even-numbered harmonics.) The frequencies of these harmonics are multiples of the first harmonic frequency.
f3 = 3 * f1 = 3 * 64.0 Hz = 192 Hz
f5 = 5 * f1 = 5 * 64.0 Hz = 320 Hz

72. Suppose that a sound is produced in a helium-filled air column rather than an air-filled air column. By what factor will this
change in medium alter the frequency of the sound. (GIVEN: vair = 331 m/s; vHe = 970 m/s)
Answer: 2.93
The frequency of a resonating air column is dependent upon the speed of sound in that air column; the relationship is a
direct relationship. If the air is replaced by another gas that transmits sound waves with a different speed, then the
frequency will be increased by the same factor by which the speed is increased. In this problem, the speed of sound is
increased by a factor of
Factor = (970 m/s) / (331 m/s) = 2.93
Thus, the frequency is increased by a factor of 2.93.
73. An organ pipe is used to produce the lowest note audible to the human ear - 20.-Hz. If the temperature is 25 °C, then how
long is the organ pipe? (First decide whether it will produce this low note as a closed- or as an open-end air column.)

Answer: 4.3 m
Harmonics produced by closed-end air columns are characterized by longer wavelengths than the harmonics of open-end
air columns of the same length. Thus, a closed-end air column produces lower frequencies. For this reason, we'll assume
that the organ pipe in this problem is a closed-end air column. (Note, your answer will be two times larger if you
assumed this to be an open-end air column.)
At a temperature of 25°C, sound waves travel at 346 m/s (see problem #66). The frequency and speed can be used to
determine the wavelength (lambda):
lambda = v / f = (346 m/s) / (20. Hz) = 17.3 m
The wavelength of the first harmonic of a closed-end air column is four times the length of the column; so the length of
the column is one-fourth the wavelength.
L = 0.25 * 17.3 m = 4.3 m

74. Determine the length of an open-end air column which would produce a 262 Hz frequency on a balmy day when the
temperature is 12 degrees.

Answer: 64.5 cm
The speed of sound waves at 12 °C is 338 m/s (see problem #66). The wavelength (lambda) of a 262 Hz sound at this
temperature is
lambda = v / f = (338 m/s) / (262 Hz) = 1.29 m
For an open-end air column, the wavelength of the first harmonic is two times the length of the air column. So the length
is one-half the wavelength:
L = 0.5 * 1.29 m = 0.645 m = 64.5 cm

75. A 440.-Hz tuning fork is held above the open end of a water-filled pop bottle and resonance is heard. The length of the pop
bottle (bottom to top) is 28.2 cm. If the speed of sound is 345 m/s, then to what height is the pop bottle filled with water?

Answer: 19.6 cm
The wavelength (lambda) of a 440 Hz sound can be determine from the wave equation (v = f*lambda)
lambda = v / f = (345 m/s) / (440. Hz) = 0.784 m
For a closed-end air column, the wavelength of the first harmonic is four times the length of the air column. So the
length is one-fourth the wavelength:
L = 0.25 * 0.784 m = 0.196 m = 19.6 cm
If the pop bottle is 28.2 cm tall, then it should be filled with 8.6 cm of water in order for there to remain 19.6 cm of air.
Part A: Multiple Choice
1. Which of the following statements are true statements about interference?
a. Interference occurs when two (or more) waves meet while traveling along the same medium.
b. Interference can be constructive or destructive.
c. Interference of two waves at a given location results in the formation of a new wave pattern which has a greater amplitude
than either of the two interfering waves.
d. The meeting of a trough of one wave with a trough of another wave results in destructive interference.
e. The only way for two waves to interfere constructively is for a crest to meet a crest or a trough to meet a trough.
f. It is only a theory that light can interfere destructively; the theory is based on the assumption that light is a wave and most
waves exhibit this behavior. Experimental evidence supporting the theory has not yet been observed.

Answer: AB
A - True: This is the definition of interference - "the meeting of two or more waves along the same medium."
B - True: These are the two possible types of interference.
C - False: When interference occurs, there are two possible results: a resulting wave with a greater displacement than either of the
original waves (constructive interference) or a resulting wave with a smaller displacement than one or both of the original waves
(destructive interference)
D - False: This is an example of constructive interference leading to a resulting wave with a greater displacement than the
individual wave; a "super-trough" would be formed.
E - False: Crest meeting crest and trough meeting trough are examples of constructive interference. These special cases result in
the formation of antinodal points - points of maximum displacement. But more generally, constructive interference will occur
anytime a wave with a "positive" (up or right or ...) displacement meets another wave with a "positive" displacement OR when a
wave with a "negative" (down or left or ...) displacement meets another wave with a "negative" displacement. When the
displacements of the two interfering waves are in the same direction at a given point, then constructive interference occurs at that
point.
F - False: There is plenty of experimental and observable evidence that light undergoes destructive interference. The best evidence
from our studies in class are the dark fringes of a two-point interference pattern. These dark fringes are the result of the
destructive interference of light.

2. Which of the following statements are true statements about two-point light source interference patterns?
a. Two-point light source interference patterns consist of alternating nodal and antinodal lines.
b. If projected onto a screen, two-point light source interference patterns would be viewed as alternating bright and dark spots
with varying gradients of light intensity in between.
c. As the distance between the sources is decreased, the distance between the nodal and antinodal lines is decreased.
d. As the wavelength of the laser light is decreased, the distance between the nodal and antinodal lines is decreased.
e. A nodal point would be formed if a trough of one wave meets a trough of another wave.
f. Antinodal points are points where the medium is undergoing no vibrational motion.
g. Suppose point P is a point where a wave from one source travels a distance of 2.5 wavelengths before meeting up with a
wave from another source which travels a distance of 3.5 wavelengths. Point P would be a nodal point.
h. Suppose point Q is a point where a wave from one source travels a distance of 2 wavelengths before meeting up with a wave
from another source which travels a distance of 3.5 wavelengths. Point Q would be a nodal point.
i. Suppose point R is a point where a wave from one source travels a distance of 2 wavelengths before meeting up with a wave
from another source which travels a distance of 3 wavelengths. Point R would be a nodal point.
j. If the path difference for points on the first nodal line is 4 cm, then the wavelength would be 6 cm. (NOTE: the first nodal line
is considered to be the first nodal line to the left or right from the central antinodal line.)

Answer: ABDH
A - True: This is exactly what we have observed through computer animations, video segments, transparency overlays, and the
actual experiment.
B - True: This is exactly what we observed when we performed Young's experiment.
C - False: The equation relating the variables of Young's experiment can be rearranged to the following form:
y = m • L • W / d ... (where W=wavelength).
Now one notices that y is inversely related to d. So if the slit separation distance (d) is decreased, the distance between nodal and
antinodal lines (related to y) would be increased.
D - True: Young's equation is often written as
W = y • d / (m • L) ... (where W=wavelength).
From the equation, one notices that wavelength (W) is directly related to y. So if the wavelength (W) is decreased, the distance
between nodal and antinodal lines (related to y) would be decreased.
E - False: Antinodal points are points of maximum displacement; for a light interference pattern, these are the brightest points.
F - False: Nodal points are points of no displacement or no disturbance; for a light light interference pattern, these are the darkest
points.
G - False: In this case the path difference is 1 wavelength; when two waves traveling to the same point have a difference in
distance traveled of 1 wavelength, then a crest of one wave would meet up with a crest of the second wave. This condition leads
to constructive interference and an antinodal point is formed.
H - True: In this case the path difference is 1.5 wavelengths; when two waves traveling to the same point have a difference in
distance traveled of 1.5 wavelengths, then a crest of one wave would meet up with a trough of the second wave. This condition
leads to destructive interference and a nodal point is formed.
I - False: In this case the path difference is 1 wavelength; when two waves traveling to the same point have a difference in
distance traveled of 1 wavelength, then a crest of one wave would meet up with a crest of the second wave. This condition leads
to constructive interference and an antinodal point is formed.
J - False: The first nodal line is designated as m = 0.5; the path difference is 4 cm. Substituting into the equation PD =
m•Wavelength and solving for wavelength yields a value of 8 cm.

3. Which of the following statements are true statements about nodal and antinodal points in light interference patterns?
a. Antinodes result from constructive interference.
b. Nodes result from destructive interference.
c. The nodal points on an interference pattern are positioned along lines; these lines are called nodal lines.
d. The central line on the interference pattern is a nodal line.
e. Points on nodal lines would be represented by bright spots if projected onto a screen.
f. The path difference for points on the central antinodal line would be 0.
g. The path difference for points on the first antinodal line would be 1 cm.
h. (This question presumes that the interference pattern is a water interference pattern.) If the path difference for points on the
first antinodal line is 5 cm, then the path difference for points on the second antinodal line would be 7 cm.
i. (This question presumes that the interference pattern is a water interference pattern.) If the path difference for points on the
first antinodal line is 5 cm, then the path difference for points on the third antinodal line would be 15 cm.
j. (This question presumes that the interference pattern is a water interference pattern.) If the path difference for points on the
first antinodal line is 6 cm, then the path difference for points on the second nodal line would be 9 cm. (NOTE: the second
nodal line is considered to be the second nodal line to the left or right from the central antinodal line.)
k. (This question presumes that the interference pattern is a water interference pattern.) If the path difference for points on the
first nodal line is 4 cm, then the path difference for points on the third nodal line would be 12 cm. (NOTE: the third nodal line
is considered to be the third nodal line to the left or right from the central antinodal line.)
Answer: ABCFIJ
A - True: An antinode is a point where a crest meets a crest or a trough meets a trough; both are examples of constructive
interference.
B - True: A node is a point where a crest meets a trough; this is an example of destructive interference and leads to a location of
no displacement.
C - True: Nodal points all lie along lines. Question #21 illustrates this well.
D - False: The central line - that is, the line extending outward from the midpoint between the two sources - is a line upon which
antinodes are formed; it is called an antinodal line. Question #21 illustrates this well.
E - False: Nodal lines are formed as a result of destructive interference. If projected onto a screen, the nodal points would appear
as the darkest points on the interference pattern.
F - True: The path difference for points on the central antinodal line would be given be the equation: PD = m•W where
W=wavelength and m=0 (for the central antinodal line). Substituting into this equation yields PD = 0•W which would be 0.
G - False: The path difference for points on the first antinodal line would be given be the equation: PD = m•W where
W=wavelength and m=1 (for the first antinodal line). So the path difference for the first antinodal line would always be 1•W; but it
would only be 1 cm for the case in which the wavelength is 1 cm.
H - False: The first antinodal line is numbered as the m=1 line. The path difference relates to the wavelength (W) by the equation
PD = m•W. Substituting m=1 and PD=5 cm into this equation yields a wavelength value of 5 cm. The second antinodal line is
numbered as the m=2 line. Re-using the equation for this line with m=2 and W=5 cm yields a path difference of 10 cm.
I - True: The first antinodal line is numbered as the m=1 line. The path difference relates to the wavelength (W) by the equation
PD = m•W. Substituting m=1 and PD=5 cm into this equation yields a wavelength value of 5 cm. Re-using the equation for the
third antinodal line with m=3 and W=5 cm yields a path difference of 15 cm.
J - True: The logic on this question is similar to the above question. The first antinodal line is numbered as the m=1 line. The path
difference relates to the wavelength (W) by the equation PD = m•W. Substituting m=1 and PD=6 cm into this equation yields a
wavelength value of 6 cm. The second nodal line is numbered as m=1.5. Re-using the equation for the second nodal line with
m=1.5 and W=6 cm yields a path difference of 9 cm.
K - False: The first nodal line is numbered the m=0.5 line. If the path difference for a point on this line is 4 cm, then the
wavelength is 8 cm (using the PD = m•W equation). The third nodal line is numbered as the m=2.5 line. Using the same equation
to find the path difference yields a value of 20 cm.

4. Which of the following statements are true statements about Thomas Young's experiment?
a. Young's experiment provided evidence that light exhibits particle-like behavior.
b. Young's experiment depends upon the use of white light from two sources.
c. The two sources of light in Young's experiment could be two different light bulbs.
d. For Young's equation to be geometrically valid, the distance from the sources to the screen must be much greater than the slit
separation distance.
e. For Young's equation to be geometrically valid, the wavelength of the light must be much greater than the slit separation
distance.
f. Thomas Young measured the distance from an antinodal point (of known number) to each of the two sources, computed a
path difference and calculated the wavelength of light.
g. Thomas Young was able to determine the wavelength of a light wave.

Answer: DG
a. - False: Young's experiment supports the wave-nature of light. Waves interfere and Young's experiment provided clear evidence
that light undergoes interference.
b. - False: There are two requirements for the light which is utilized in Young's experiment: the two light sources must be coherent
and monochromatic. Monochromatic means that the light sources must provide light of the same wavelength (and a single
wavelength); using a white light bulb would produce light of many wavelengths. Second, coherent means that the light from the
two sources must be vibrating together, experiencing a crest at the same time and a trough at the same time. Using two light
bulbs (as opposed to a single light source shining on a double slit) would likely result in incoherent light.
c. - False: If two light bulbs emitting monochromatic light of the same color were used, one of the two requirements would be met.
Yet there would still be the problem of incoherence. See explanation to part b.
d. - True: There are two geometric requirements for Young's experiment: the screen distance (L) must be much greater than the
slit separation distance (d) and the slit separation distance must be much greater than the wavelength. That is L >>> d and d
>>> W.
e. - False: Vice versa; d >>> W. See explanation to part d.
f. - False: Thomas Young used the equation W = y•d/m•L. Measurement of y, d, m, and L is much more practical since the size of
these quantities is much larger. The error introduced in the measurement would not overwhelm the precision of the wavelength
measurement. On the other hand, a measurement of the path difference would be very difficult since the only way to achieve this
measurement is to measure the two distances. Given the fact that the slits are so close together, these two distances are so nearly
identical that the error introduced in the measurement of one distance would overwhelm the actual difference in distance between
the two measurements. That's why Young had to derive the equation W = y•d/m•L.
g. - True: Measuring the wavelength of a visible light wave was one of the main outcomes of Young's experiment.

5. Light which is vibrating in a single plane is referred to as _____ light

a. electromagnetic b. transverse c. unpolarized d. polarized

Answer: D
Unpolarized light is light whose vibrations are in a multitude of directions. To simplify matters, unpolarized light is light which can
be thought of as vibrating in a vertical and a horizontal plane. If one of these planes of vibration is removed, then light would be
vibrating in a single plane and said to be "polarized."

6. Light which is vibrating in a variety of planes is referred to as _____ light

a. electromagnetic b. transverse c. unpolarized d. polarized

Answer: C
Unpolarized light is light whose vibrations are in a multitude of directions.

7. Light usually vibrates in multiple vibrational planes. It can be transformed into light vibrating in a single plane of vibration. The
process of doing this is known as ____.
a. translation b. interference c. polarization d. refraction

Answer: C
Polarization is defined as the process of transforming unpolarized light (light whose vibrations are in a multitude of planes) into
polarized light (light which can be thought of as vibrating in a single plane).

8. Light is passed through a Polaroid filter whose transmission axis is aligned horizontally. This will have the effect of ____.
a. making the light one-half as intense and aligning the vibrations into a single plane.
b. aligning the vibrations into a single plane without any effect on its intensity.
c. merely making the light one-half as intense; the vibrations would be in every direction.
d. ... nonsense! This will have no effect on the light itself; only the filter would be effected.
Answer: A
Polaroid filters have the effect of polarizing light - that is, aligning their vibrations into a specific plane. They can be
thought of as performing this feat by removing the vibrations which occur within a plane perpendicular to
the transmission axis.

9. Light is passed through a Polaroid filter whose transmission axis is aligned horizontally. It then passes through a second filter
whose transmission axis is aligned vertically. After passing through both filters, the light will be ______.
a. polarized b. unpolarized
c. entirely blocked d. returned to its original state.

Answer: C
The first filter serves the role of blocking one-half the light; the horizontal vibrations would emerge from the filter and the vertical
vibrations would be blocked. The second filter would allow the vertical vibrations to pass through if there were any. However, since
the vertical vibrations have already been filtered out, there is no light remaining after the second filter is used. This combination of
two filters serves to block all the light.

10. Which of the following are effective methods of polarization? Include all that apply.
a. Passing light through a Polaroid filter.
b. Reflection of light off a nonmetallic surface.
c. Passing light from water to air.
d. Passing light through a birefringent material such as Calcite.
e. Turning the light on and off at a high frequency.
f. Interfering light from one source with a second source.
Answer: AB
The use of a filter, the reflection of light off nonmetallic surfaces and the use of a birefringent material are all means of polarizing
light. Refraction at an air-water surface would change the speed and the direction of light but would not have any effect upon its
vibrational orientation. Turning a light on and off at a high frequency would only annoy or impress those present in the room. And
light interference could create a pattern of bright and dark spots but would not have any effect upon light's vibrational orientation.

11. Consider the three pairs of sunglasses to the right. Which pair of glasses is capable of
eliminating the glare from a road surface? (The transmission axes are shown by the straight
lines.)

Answer: C
When light reflects off a road surface, a portion of the light vibrations becomes oriented in a
plane which is horizontal to the road surface. This polarization often leads to an annoying glare.
The glare can be reduced by blocking the polarized light. Since the light is polarized horizontally
(assuming a horizontal road way - a good assumption), the sunglasses should be capable of
blocking horizontal light and allowing the vertical vibrations to be transmitted. Selecting
sunglasses C would make accomplish this feat.

12. TRUE or FALSE:

White and black are actual colors of light.
a. TRUE b. FALSE

Answer: B
Black is the absence of all light. Things appear black when they do not reflect or emit light. White is the presence of all colors of
visible light. Objects appear white when they reflect or emit all wavelengths of visible light (or at least three wavelengths - Red,
Blue and Green - in equal intensity).

13. The three primary colors of light are ____.

a. white, black, gray b. blue, green, yellow
c. red, blue, green d. red, blue, yellow
e. ... nonsense! There are more than three primary colors of light.

Answer: C
Yes, you must know this one! It forms the basis of most of our logic and reasoning about color, light and the appearance
of objects.

14. The three secondary colors of light are ____.

a. cyan, magenta, green b. cyan, magenta, and yellow
c. orange, yellow, violet d. red, blue, yellow
e. ... nonsense! There are more than three secondary colors of light.

Answer: B
The secondary colors of light are those colors which are formed when two primary colors are mixed in equal amounts. Mixing blue
and green light results in cyan light. Mixing red and blue light results in magenta light. And mixing red and green light results
in yellow light.

15. Combining red and green light (with equal intensity) makes ____ light; combining red and blue light (with equal intensity)
makes ____ light; and combining blue and green light (with equal intensity) makes ____ light. Choose the three colors in
respective order.
a. brown, purple, aqua b. brown, magenta, yellow
c. yellow, magenta, brown d. yellow, magenta, cyan

Answer: D
You must know this for it forms the foundation of much of our reasoning. To assist in recalling the three primary colors
of light, three secondary colors of light, and the means by which adding primaries form secondaries, develop some form
of graphical reminder such as a color wheel or a diagram like those at the right.

16. Demonstrate your understanding of color addition by completing the following color
equations. Select colors from the Color Table at the right.
a. Red + Blue = _____
b. Red + Green = _____
c. Green + Blue = _____
d. Red + Blue + Green = _____
e. Blue + Yellow = _____
Answer: See table above.
A. Magenta is a secondary color of light formed by combining red light with blue light in equal amounts. Refer to graphic in
previous question.
B. Yellow is a secondary color of light formed by combining red light with green light in equal amounts. Refer to graphic in previous
question.
C. Cyan is a secondary color of light formed by combining green light with blue light in equal amounts. Refer to graphic in previous
question.
D. White light is formed when all three primary colors of light are combined in equal amounts.
E. Yellow light is a combination of red and green light. So combining blue with yellow light is like combining blue light with red and
green light. The result of combining these three primary colors of light is to produce white light.

17. Demonstrate your understanding of color subtraction by completing the following color
equations. Select colors from the Color Table at the right.
a. White - Blue = _____
b. White - Red = _____
c. White - Green = _____
d. White - Blue - Green = _____
e. White - Yellow = _____
f. Red + Green - Green = _____
g. Yellow - Green = _____
h. Yellow - Red = _____
i. White - Magenta = _____
j. White - Cyan = _____
k. Yellow + Blue - Cyan = _____
l. Yellow + Cyan + Magenta = _____
m. Yellow + Cyan - Magenta = _____
n. Yellow + Cyan - Blue - Red = _____
Answer: See table above.
Each of these questions is best answered by first converting any secondary color of light into
a mix of two primary colors of light. Then "do the arithmetic." If the result of the
"arithmetic" is a combination of two primary colors, translate the combo into a secondary
color of light. Here it goes:
a. White - Blue = R+G+B - B = R+G = Yellow
b. White - Red = R+B+G - R = G+B = cyan
c. White - Green = R+G+B - G = R+B = magenta
d. White - Blue - Green = R+G+B - B - G = R = red
e. White - Yellow = R+G+B - R+G = B = blue
f. Red + Green - Green = R + G - G = R = red
g. Yellow - Green = R+G - G = R = red (Note the similarity to part f.)
h. Yellow - Red = R+G - R = G = green
i. White - Magenta = R+G+B - R+B = G = green
j. White - Cyan = R+G+B - G+B = R = red
k. Yellow + Blue - Cyan = R+G + B - G+B = R = red
(Note the similarity to part j: R+G + B is the same as white; so this question is White -
Cyan.)
l. Yellow + Cyan + Magenta = R+G + B+G + R+B = R+R+G+G+B+B = white + white (that
is very bright white since there is double the red, green and blue added together)
m. Yellow + Cyan - Magenta = R+G + B+G - R+B = G+G = green
n. Yellow + Cyan - Blue - Red = R+G + G+B - B - R = G+G = green

18. Sunsets often have a reddish-orange color associated with them. This is attributable to the phenomenon of _____.
a. polarization b. diffraction c. dispersion d. refraction

Answer: B
Sunsets are the result of the longer wavelengths of light diffracting around atmospheric particles and reaching our eyes, giving the
reddish-orange appearance. More detail about the phenomenon can be accessed using the Useful Web Link below.

19. A filter serves the function of ____.

a. subtracting color(s) from the light which is incident upon it
b. adding color(s) to the light which is incident upon it
c. removing nicotine from light so that we can live longer lives
d. confusing physics students who are studying color, causing them to live shorter lives
Answer: A
Filters can be thought of as absorbing one or more of the primary colors of light which are incident upon it, allowing remaining
colors to be transmitted. For instance, a green filter will absorb all wavelengths except for green light. In this sense, filters subtract
colors from the mix of incident light, allowing only selected colors to pass through.

20. Express your understanding of filters by answering the following questions. Choose the
best answer(s) from the Color Table shown at the right.
a. A red filter is capable of transmitting ____ light (if it is incident upon the filter).
b. A blue filter is capable of transmitting ____ light (if it is incident upon the filter).
c. A green filter is capable of transmitting ____ light (if it is incident upon the filter).
d. A red filter will absorb ____ light (if it is incident upon the filter).
e. A blue filter will absorb ____ light (if it is incident upon the filter).
f. A yellow filter will absorb ____ light (if it is incident upon the filter).
g. A magenta filter will absorb ____ light (if it is incident upon the filter).
h. A white object is illuminated with white light and viewed through a green filter. The object will appear _____.
i. A white object is illuminated with white light and viewed through a blue filter. The object will appear _____.
j. A white object is illuminated with white light and viewed through a cyan filter. The object
will appear _____.
k. A blue object is illuminated with white light and viewed through a green filter. The object
will appear _____.
l. A cyan object is illuminated with white light and viewed through a cyan filter. The object
will appear _____.
m. A cyan object is illuminated with white light and viewed through a green filter. The object will appear _____.
n. A yellow object is illuminated with white light and viewed through a green filter. The object will appear _____.
o. A yellow object is illuminated with white light and viewed through a magenta filter. The object will appear _____.
p. A yellow object is illuminated with yellow light and viewed through a yellow filter. The object will appear _____.
q. A yellow object is illuminated with yellow light and viewed through a blue filter. The object will appear _____.
r. A yellow object is illuminated with blue light and viewed through a yellow filter. The object will appear _____.
s. A blue object is illuminated with blue light and viewed through a yellow filter. The object will appear _____.
t. A yellow object is illuminated with yellow light and viewed through a red filter. The object
will appear _____.
u. A yellow object is illuminated with yellow light and viewed through a green filter. The
object will appear _____.
v. A yellow object is illuminated with green light and viewed through a yellow filter. The
object will appear _____.
w. A yellow object is illuminated with green light and viewed through a green filter. The
object will appear _____.
x. A yellow object is illuminated with green light and viewed through a red filter. The object will appear _____.
y. A yellow object is illuminated with green light and viewed through a cyan filter. The object will appear _____.
z. A red object is illuminated with yellow light and viewed through a cyan filter. The object will appear _____.
Answer: See sentences above.
Parts a-g target your understanding of the ability of filters to subtract colors of light from the mix of incident light that strikes it. A
filter will absorb its complementary color of light. So a yellow filter absorbs blue light since blue is across from it on the color
wheel. Whatever light is not absorbed will be transmitted; so yellow filters transmit red and green light (if incident upon it), also
known as yellow light.
a. Red filters absorb cyan light (the complementary color of red). If white light (red + blue + green) shines on a red filter and cyan
(blue + green) light is absorbed, all that is left to be transmitted is red light.
b. Blue filters absorb yellow light (the complementary color of blue). If white light (red + blue + green) shines on a blue filter and
yellow (red + green) light is absorbed, all that is left to be transmitted is blue light.
c. Green filters absorb magenta light (the complementary color of green). If white light (red + blue + green) shines on a green
filter and magenta (red + blue) light is absorbed, all that is left to be transmitted is green light.
d. Red filters absorb its complementary color - cyan. So this question could be answered as cyan. And since cyan light consists of
blue + green light, this question could also be answered as blue + green.
e. Blue filters absorb its complementary color - yellow. So this question could be answered as yellow. And since yellow light
consists of red + green light, this question could also be answered as red + green.
f. Yellow filters absorb its complementary color - blue. So this question must be answered as blue.
g. Magenta filters absorb its complementary color - green. So this question must be answered as green.

Parts h - z target your understanding of color subtraction for both pigments and filters. In each question, there is light incident
upon an object. This light can be broken down into primary colors. Some light might be subtracted from this incident mix by either
the object or the filter. The only possible color of light that could ultimately pass through the filter and effect the appearance of the
object would be one of the primary colors in the incident light. For instance, suppose that an object is illuminated with yellow light
(which is a combination of red and green primary colors of light. The object could appear yellow (if neither red nor green are
subtracted away), or red (if green light subtracted is taken away) or green (if red light is subtracted away) or black (if both red and
green light is subtracted away).
In the explanations below, each question will be approached by identifying the primary colors of light in the incident mix (the light
used to illuminate the object) and then primaries will be successively subtracted away by the pigments in the object and by the
filter. Here it goes:
h. RGB light (white light) hits a white object; white objects do not subtract (i.e., absorb) any colors; so RGB reflects off the object
and heads towards a green filter. Green filters would subtract R and B (when present) and allow G to pass through. So RGB -
nothing - GB = R = red.
i. RGB light (white light) hits a white object; white objects do not subtract (i.e., absorb) any colors; so RGB reflects off the object
and heads towards a blue filter. Blue filters would subtract R and B (when present) and allow B to pass through. So RGB - nothing
- RG = B = blue.
j. RGB light (white light) hits a white object; white objects do not subtract (i.e., absorb) any colors; so RGB reflects off the object
and heads towards a cyan filter. Cyan filters would subtract R (when present) and allow G to pass through. So RGB - nothing - R =
GB = cyan.
k. RGB light (white light) hits a blue object; blue objects subtract (i.e., absorb) R and G light (when present); so B light reflects off
the object and heads towards a green filter. Green filters would subtract R and B (when present) and allow G to pass through; blue
light is present so it will be subtracted. So RGB - GB - B = nothing = black.
l. RGB light (white light) hits a cyan object; cyan objects subtract (i.e., absorb) R light (when present); so GB light reflects off the
object and heads towards a cyan filter. Cyan filters would subtract R (when present) and allow GB to pass through. So RGB - R =
GB = cyan.
m. RGB light (white light) hits a cyan object; cyan objects subtract (i.e., absorb) R light (when present); so GB light reflects off the
object and heads towards a green filter. Green filters would subtract RB (when present) and allow G to pass through; B is present
so it will be subtracted. So RGB - R - B = G = green.
n. RGB light (white light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so RG light reflects off
the object and heads towards a green filter. Green filters would subtract RB (when present) and allow G to pass through; R is
present so it will be subtracted. So RGB - B - R = G = green.
o. RGB light (white light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so RG light reflects off
the object and heads towards a magenta filter. Magenta filters would subtract G (when present) and allow RB to pass through; G is
present so it will be subtracted. So RGB - B - G = R = red.
p. RG light (yellow light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so RG light reflects off
the object and heads towards a yellow filter. Yellow filters would subtract B (when present) and allow RG to pass through. So RG -
nothing - nothing = RG = yellow.
q. RG light (yellow light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so RG light reflects off
the object and heads towards a blue filter. Blue filters would subtract RG (when present) and allow B to pass through; R and G are
both present so they will be subtracted. So RG - nothing - RG = nothing = black.
r. B light (blue light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so no light light reflects off
the object and it wouldn't matter what type of filter is used. This object will appear black. So B - B - nothing = nothing = black.
s. B light (blue light) hits a blue object; blue objects subtract (i.e., absorb) RG light (when present); so B light reflects off the
object and heads towards a yellow filter. Yellow filters would subtract B (when present) and allow RG to pass through (if present);
neither R nor G are present and the B gets subtracted. So B - nothing - B = nothing = black.
t. RG light (yellow light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so RG light reflects off
the object and heads towards a red filter. Red filters would subtract GB (when present) and allow R to pass through (if present); G
is present so it gets subtracted. So RG - nothing - G = R = red.
u. RG light (yellow light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so RG light reflects off
the object and heads towards a green filter. Green filters would subtract RB (when present) and allow G to pass through (if
present); R is present so it gets subtracted. So RG - nothing - R = G = green.
v. G light (green light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so G light reflects off the
object and heads towards a yellow filter. Yellow filters would subtract B (when present) and allow RG to pass through (if present).
So G - nothing - nothing = G = green.
w. G light (green light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so G light reflects off the
object and heads towards a green filter. Green filters would subtract RB (when present) and allow G to pass through (if present).
So G - nothing - nothing = G = green.
x. G light (green light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so G light reflects off the
object and heads towards a red filter. Red filters would subtract GB (when present) and allow R to pass through (if present); G is
present so it gets subtracted. So G - nothing - G = nothing = black.
y. G light (green light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so G light reflects off the
object and heads towards a cyan filter. Cyan filters would subtract R (when present) and allow GB to pass through (if present). So
G - nothing - nothing = G= green.
z. G light (green light) hits a red object; red objects subtract (i.e., absorb) GB light (when present). G is present so it gets
subtracted and it wouldn't matter what filter is used to view this object; there is no light reflecting off the object so it will appear
black. So G - G - nothing = nothing = black.

Part B: Diagramming, Analysis, Calculations

21. Two point sources are vibrating together (in phase) at the
same frequency to produce a two-point source interference pattern. The diagram at the right depicts the two-point source
interference pattern. The crests are represented by thick lines and the troughs by thin lines. Several points on the pattern are
marked by a dot and labeled with a letter. Use the diagram to answer the following questions.
a. Which of the labeled points are antinodal points?
b. Which of the labeled points are nodal points?
c. Which of the labeled points are formed as a result of constructive interference?
d. Which of the labeled points are located on the central antinodal line?
e. Which of the labeled points are located on the first antinodal line?
f. Which of the labeled points are located on the second antinodal line?
g. Which of the labeled points are located on the third antinodal line?
h. Which of the labeled points are located on the first nodal line (using the notation that the first nodal line is the nodal line directly
to the left or the right of the central antinodal line)?
i. Which of the labeled points are located on the second nodal line (using the notation that the second nodal line is the second
nodal line directly to the left or the right of the central antinodal line)?
j. Which of the labeled points are located on the third nodal line (using the notation that the third nodal line is the third nodal line
directly to the left or the right of the central antinodal line)?

Answers: See diagram above for a visual representation of many of the answers.
a. ACDEHLMOPQ are all antinodal points which lie on antinodal lines. They are formed as a result of either a crest meeting a
crest (two thick lines) or a trough meeting a trough (two thin lines). At the moment in time depicted in the diagram, point A is not
a crest-crest or a trough-trough interference point. However, as time progresses and the circular waves continue their motion
outwards from the source, point A (and all points on the antinodal lines) will be locations of crest-crest (or trough-trough)
interference. At some instant in time, all points falling upon antinodal lines will be locations of constructive interference.
b. BFGJKNRS are all nodal points. They are formed as a result of a crest (thick line) meeting a trough (thin line).
c. CDEHLMOPQ (and possibly A) are all formed by constructive interference. They are all antinodal points and as such, they are
the result of constructive interference. Point A is certainly not an antinodal point; however it is likely the result of constructive
interference - two waves meeting with their displacement in the same direction (just not two crests or two troughs).
d. CL are on the central antinodal line. The central antinodal line is marked on the diagram above as AN0. It is the line which
extends from the midpoint of the line connecting the two sources.
e. EHPQ are on the first antinodal line. They lie on the first antinodal line to the left or right of the central antinodal line. See
diagram above.
f. ADMO are on the second antinodal line. They lie on the second antinodal line to the left or right of the central antinodal line.
See diagram above.
g. There is no third antinodal line. See diagram above.
h. FG are on the first nodal line. They lie on the first nodal line to the left or right of the central antinodal line. See diagram above.
i. BKR are on the second nodal line. They lie on the second nodal line to the left or right of the central antinodal line. See diagram
above.
j. JNS are on the third nodal line. They lie on the third nodal line to the left or right of the central antinodal line. See diagram
above.

22. Consider the interference pattern at the right. (The crests are represented by thick lines
and the troughs by thin lines.) If the distance from S1 to point A is 49.5 cm and the distance
from S2 to point A is 60.5 cm, then what is the wavelength?

Answer: W = 11.0 cm
Given: S1A = 49.5 cm and S2A = 60.5 cm and m=1 (the point is on the first antinodal line to the right of center)
Find: W (wavelength)
Strategy: Find the path difference (PD) from the two distances and then use the PD = m • W equation to calculate the
wavelength.
PD = | S2A - S1A | = | 60.5 cm - 49.5 cm | = 11.0 cm
Now substitute into the path difference-wavelength equation and solve for wavelength (W):
11.0 cm = 1 • W
W = 11.0 cm
An alternative strategy involves recognizing from the diagram that point A is a distance of 4.5 wavelengths from point S 1. Thus, the
distance 49.5 cm equals 4.5 • W. Solving for W yields 11.0 cm. The same strategy can be used for the distance from S2 to point A,
yielding the same answer.

23. Consider the interference pattern at the right. (The crests are represented by
thick lines and the troughs by thin lines.) If the distance from S1 to point B is 50.4
cm and the distance from S2 to point A is 34.5 cm, then what is the wavelength?

Answer: 6.28 cm
Given: S1B = 50.4 cm and S2B = 34.5 cm and m=2.5 (the point is on the third nodal
line to the right of center)
Find: W (wavelength)
Strategy: Find the path difference (PD) from the two distances and then use the PD
= m • W equation to calculate the wavelength.
PD = | S2B - S1B | = | 34.5 cm - 50.4 cm | = 15.7 cm
Now substitute into the path difference-wavelength equation and solve for
wavelength (W):
15.7 cm = 2.50 • W
W = 6.28 cm
An alternative strategy involves recognizing from the diagram that point B is a
distance of 8 wavelengths from point S1. Thus, the distance 50.4 cm equals 8 • W.
Solving for W yields 6.28 cm. The same strategy can be used for the distance from
S2 to point B, yielding the same answer.

24. Two point sources are vibrating in phase to produce an interference pattern. The wavelength of the waves is 7.60 cm. Point C
is a point on the third nodal line. The distance from S1 (the nearest source) to point C is 65.6 cm. Determine the distance from
S2 to point C.
Answer: 84.6 cm
Given: S1C = 65.6 cm and W = 7.6 cm and m=2.5 (third nodal line)
Find: S2C
Strategy: Find the path difference (PD) using the equation PD = m • W. The path difference signifies the difference in distance
from the sources to the nearest point. The S 2C distance is larger than the S1C distance by an amount equal to the path difference.
First find the path difference:
PD = m • W = 2.50 • 7.60 cm = 19.0 cm
Now add the path difference to the S1C distance to determine S2C.
S2C = S1C + PD = 65.6 cm + 19.0 cm = 84.6 cm

25. Consider the interference pattern at the right. (The crests are represented by
thick lines and the troughs by thin lines.) The distance from S1 to point D is 47.2 cm.
What is the wavelength? What is the distance from S2 to point D? (HINT: Use the
diagram.)

Answer: wavelength = 7.87 cm; S2D = 59.0 cm

Given: S1D = 47.2 cm and m = 1.50 (the second nodal line to the left of the central
antinodal line)
Find: W and S2D
Strategy: Since neither a wavelength or a path difference is given or implicitly
stated, the diagram will have to be used to determine the wavelength. The
wavelength will then be used to determine the path difference and the path
difference will be used to find the S2D distance.
From the diagram, it is observed that the point D is exactly 6 full wavelengths from
S1. So S1D = 6.00 • W. Substituting and solving for W yields the following:
47.2 cm = 6.00 • W
W = (47.2 cm) / 6.00 = 7.87 cm
Now the path difference can be found using the relationship PD = m • W where m =
1.50 and W = 7.87 cm. Substituting and solving for PD yields
PD = m • W = 1.50 • (7.87 cm) = 11. 8 cm
This means that the point S2 is 11.8 cm further from the point D than S1's distance
from point D. So adding 11.8 cm to 47.2 cm yields 59.0 cm.
26. Laser light is directed towards a pair of slits which are 2.50 x 10-2 mm apart. The light shines on a screen 8.20 meters away
and an interference pattern is observed. A point on the 3rd antinode is observed to be 39.6 cm away from the central antinode.
What is the wavelength of the laser light in units of nanometers? (1 m = 109 nm)

Answer: 402 nm
Given: d = 2.50 x 10-2 mm; L = 8.20 m; m = 3; y = 39.6 cm
Find: wavelength (W)
Strategy: Substitute into Young's equation and solve for W; be very careful with units - in fact, first perform conversions to get all
quantities in unit of meters. Once the W is calculated, convert it to nanometers.
First the conversions of all given quantities to meters yields:
d = 2.50 x 10-5 m; L = 8.20 m; y = 0.396 m
Now substitute into Young's equation:
W = y • d / (m • L) = (0.396 m) • (2.50 x 10-5 m) / [(3) • (8.20 m)] = 4.02 x 10-7 m
Now convert to meters using the conversion factor: (1•109 nm/1 m). This yields 402 nm as the answer.

27. This same laser light (from #26) is reflected off of the grooves in a compact disc. The disc is 4.5 meters from the screen where
its interference pattern is projected. Antinode 1 is found to be 1.2 meters from the central antinode. What is the spacing between
the "grooves" of the C.D.?

Answer: 1.5 x 10-6 m

Given: W = 4.02 x 10-7 m; L = 4.5 m; m = 1; y = 1.2 m
Find: d
Strategy: Use Young's equation to solve for the unknown quantity.
Rearrange Young's equation to produce an equation with d expressed in terms of the known quantities.
d=m•L•W/y
Substitute and solve
d = (1) • (4.5 m) • (4.02 x 10-7 m) / (1.2 m) = 1.5 x 10-6 m

28. Different colors of paper are illuminated with different primary colors of light. Determine the colors of light absorbed by the
paper (if any), the colors of light reflected by the paper (if any), and the appearance of the paper.

Color of Color
Colors Absorbed Colors Reflected Appearance
Light of Paper
a. White White None RGB White
Cyan
b. White None GB Cyan
(= GB)
Yellow
c. White None RG Yellow
(= RG)
None
d. Red Yellow R Red
(B if present)
R
e. Red Blue None Black
(G if present)
 f. Red Cyan R None Black
None
g. Red Red R Red
(GB if present)
Magenta B
h. Red R Red
(= RB) (G if present)
Yellow G
i. Red R Red
(= RG) (B if present)
Cyan
j. Red GB None Black
(= GB)
Cyan G
k. Blue B Blue
(= GB) (R if present)
Yellow
l. Blue RG None Black
(= RG )
Yellow R
m. Green G Green
(= RG ) (B if present)
Yellow
n. Cyan R G green
(= RG )
Yellow
o. Magenta G R Red
(= RG)

Answer: See table above

In the first column, if a secondary color of light is shown, it is translated into the equivalent primaries. These primaries will strike
the paper and may or may not be absorbed. The color which a paper pigment will absorb is the complementary color; this color is
typically expressed in terms of the equivalent primary colors of light. The subtraction process is then done to determine what
primary color of light is reflected. This/these reflected primaries determine the color appearance of the paper. They are added (if
there are more than 2) to determine the resulting appearance. As an example of the entire process, consider row i:
Row i: Yellow light is equivalent to red and green (RG). Red paper contains pigments capable of absorbing both green and blue
light if present. Only green light is present, so it is absorbed. So the subtraction process is
RG - G = R
Red light is reflected; this gives the paper the appearance of red.
The same process can be performed for all other parts of this question.