Multiple Choice Questions

1.Which of the following properties of wave does not change with change in
medium?
(a) Frequency (b) Wavelength

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(c) Velocity (d) Amplitude
2. A tuning fork of frequency 580 Hz is employed to produce transverse waves on a
long rope.The distance between nearest crust found to be20 cm. The velocity of
wave is
(a) 58 m/s (b)580m/s (c) 20m/s (d) 116m/s
3. A 5.5 meter length of string has a mass of 0.035 kg. If the tension in the string is
77 N, the speed of a wave on string is
(a) 110 m/s (b) 165 m/s (c) 77 m/s (d) 102m/s
4. If the density of oxygen is 16 times that of the hydrogen, what will be the
corresponding ratio of their velocities of sound waves?
(a) 1:4 (b)16:1 (c) 4:1 (d) 1:16
5. Transverse waves are generated in uniform steel wires A and B by attaching their
free ends to a vibrating source of frequency500 Hz. The diameter of wire A is half
that oof B and the tension of wire A is half that of wire B. The velocities of waves in
wire A & B are in the ratio.
(a) 1:2 (b)2:1 (c) 1:√2 (d) √2 :1
6.The temperature at which speed of sound in air becomes double of its value at
27°C:
(a) 54°C (b) 327°C (c) 927°C (d) -123°C
7.A travelling harmonic wave is represented by the equation y(x,t)=10 -3 sin(50+2x)
where x &y are in meters and t is in seconds. Which of the following is a correct
statement about the wave?
(a) The wave is propogating along the -ve X axis with a speed 25m/s.
(b) The wave is propogating along the +ve X axis with a speed 100m/s.
(c) The wave is propogating along the -ve X axis with a speed 100m/s.
(d) The wave is propogating along the +ve X axis with a speed 25m/s.
8.The phase difference between two waves, represented by
Y1=10-6sin(100t+(x/50) +0.5) m
Y2=10-6 cos(100t+(x/50) m
where x is in metres and t is in second , is approximately
(a) 1.07 rad (b) 2.07 rad (c) 0.5 rad (d) 1.5 rad
9. A wave y=a sin (ωt – kx) , on a string meets with another wave producing a node
at x=0. Then the equation of the unknown wave is
(a) y = a sin (ωt + kx) (b) y = -a sin (ωt + kx)
(c) y = a sin (ωt – kx) (d) y= -a sin (ωt – kx)
10. In an experiment with sonometer a tuning fork of frequency 256 Hz resonates
with a length of 25 cm and another tuning fork resonates with a length of 16 cm.
Tension of the string remaining constant, the frequency of the second tuning fork is
(a) 163.84 Hz (b) 400 Hz (c) 320 Hz (d) 204.8 Hz
11. Five organ pipes are described below. Which one has the highest fundamental
frequency?
(a) A 2.3 m pipe with one end open and the other end closed
(b) A 3.3 m pipe with one end open and the other closed
(c) A 1.6 m pipe with both ends open
(d) A 3.0 m pipe with both ends open
12. An organ pipe open at one end is vibrating in first overtone and is in resonance
with another pipe open at both ends and vibrating in third harmonic. The ratio of
length of two pipes is
(a) 1:2 (b) 4:1 (c) 8:3 (d) 3:8
13.An organ pipe, open at both ends produces 5 beats per second when vibrated

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with a source of frequency 200 Hz. The second harmonic of the same pipe produces
10 beats per second with a source of frequency 420 Hz.The frequency of source is
(a) 195 Hz (b) 205 Hz (c) 190 Hz (d) 210 Hz
14.A fork produces 6 beats per sec with another fork of frequency 384. If the prongs
of first fork are slightly filed 4 beats per second are produced. The frequency of the
first fork after filing is
(a) 390 Hz (b) 378 Hz (c) 380 Hz (d) 388 Hz
15. A stationary wave is set up in a resonance air column of a glass tube partially
filed with water by holding a tuning fork near the open end , the open end of the tube
is
(a) always a node
(b) always an antinode
(c) sometimes a node and sometimes an antinode
(d) neither a node nor an antinode
Answers Of MCQs

1.a 2.d 3.d 4.b 5.d

6.c 7.a 8.a 9.b 10.b
11.c 12.a 13.b 14.c 15.b

MCQs Answers (Brief Explanation):

2. ANS (d)
V= ν λ = 580 X 0.20 = 116m/s
3. ANS (a)
m = 0.035/5.5 Kg/m , T =77N
V=√T/m =√77X5.5/0.035 =110m/s
4. ANS (a)
Vo/VH= √dH/dO =√1/16 =1:4
5. ANS (d)
VA/VB =√TA/D2A X √D2B/TB = √2 : 1
6. ANS (c)
V2/V1 = √ T2/T1 V2 = 2V1 , T1 = 300 K
2V1/ V1 = √ T2/300
T2 = 927°C
8 ANS (a)
Y1=10-6sin(100t+(x/50) +0.5) m
Y2=10-6 cos(100t+(x/50) m = 10-6sin(100t+(x/50) +π/2)
Δφ= π/2- 0.05 =3.14/2 – 0.5= 1.57 – 0.5 = 1.07 rad
1. ANS (b)
Y = Y1 + Y2 = a sin (ωt – kx) -a sin (ωt + kx) = -2a cos ωt sin kx
At x =0, y=0, a node is formed.
2. ANS (b)
ν2/ ν1 =L1/L2 ; ν2 =25/16 X256 =400 Hz
3. ANS (c)
Fundamental frequency of a closed organ pipe ν =V/4L
Fundamental frequency of a open organ pipe ν =V/2L
ν will be maximum for an open organ pipe of length 1.6 m
4. ANS (a)
First overtone of a closed pipe =3ν = 3V/4L

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Third harmonic of an open pipe =3ν’ =3V/2L’
As two pipes are in resonance L / L’ =1:2
5. ANS (b)
Fundamental frequency of open pipe,
f=200+- 5=195 Hz or 205 Hz
Second harmonics of open pipe,
2f = 420+-10= 410 Hz or 430 Hz
f = 205 Hz or 215 Hz
6. ANS (c)
Let frequency of the tuning fork be f
f = 384+- 6= 390 or 378 Hz
On filing the frequency increases & beats produced are 4 so un known frequency is
378 Hz and on filing 380 Hz

Assertions And Reasons

Directions:
In the following questions, a statement of assertion(A) is followed by a statement of
reason(R). Mark the correct choice as:
(a) If both assertion and reason are true and the reason is the correct explanation of
the assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of
the assertion.
(c) If assertion is true but reason is false.
(d) If the assertion and reason both are false.
16. Assertion : Sound would travel faster on a hot summer day than on a cold winter
day.
Reason : Velocity of sound is directly proportional to the square of its absolute
temperature.
17.Assertion : The change in air pressure effect the speed of sound.
Reason : The speed of sound in a gas is inversely proportional to square root of
pressure.
18. Assertion : In the case of a stationary wave, a person hear a loud sound at the
nodes as compared to the antinodes.
Reason : In a stationary wave all the particles of the medium vibrate in phase.
19. Assertion : To hear distinct beats, difference in frequencies of two sources
should be less than 10.
Reason : More the number of beats per sec more difficult to hear them.
20. Assertion : Solids can support both longitudinal and transverse waves but only
longitudinal waves can propagate in gases.
Reason : For the propagation of transverse waves, medium must also necessarily
have the property of rigidity.
ANSWERS

16.c 17.d 18.c 19.b 20.a

Case Study Based Questions

PARAGRAPH 1
If during the propagation of a wave through a medium, the particles of the medium

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vibrate simple harmonically about their mean positions, then the wave is said to be
plane progressive wave. For mathematical description of a travelling wave, we need
a function of both position x and time t. If we wish to describe a sinusoidal travelling
wave the corresponding function must also be sinusoidal. If the position of the
constituents of the medium is denoted by x, the displacement from the equilibrium
position may be denoted by y. A sinusoidal travelling wave is then described by:
y (x ,t)= a Sin(ωt – kx + φ0) The term φ0= initial phase, a= amplitude, k= angular
wave number, ω=angular frequency ,(ωt- kx + φ0)= phase angle. If the wave is
travelling along -ve x direction, then y (x ,t)= a Sin(ωt +kx + φ0)

QUESTIONS
21.If equation of a sound wave is y=0.0015 sin(62.8x+314t) then its wavelength will
be
(a) 0.1 unit (b) 0.2 unit (c) 0.3 unit (d) 2 unit
22.The displacement of a wave travelling in the X- direction is given by
Y=10-4sin (600t - 2x + π/3)
Where x is expressed in metres & t in seconds. The speed of wave motion is
(a) 300 (b)600 (c) 1200 (d) 200
23.The equation of the wave is given by y=10 sin( 2πt/30 + α)
If displacement is 5 cm at t=0 then the total phase at t=7.5 s will be
(a) π/3 rad (b) π/2 rad (c) 2π/5 rad (d) 2π/3 rad
24.The equation of progressive wave is y=5sin(100 πt – 0.4 πx),where y & x are in m
and t in s.
(1) The amplitude wave is 5m. (2) The wavelength of the wave is 5 m.
(3) The frequency of the wave is 50 Hz. (4) The velocity of the wave is 250 m/s
Which of the above statements are correct
(a) (1),(2) &(3) (b) (2) & (3) (c) (1) & (4) (d) All are correct
PARAGRAPH 2
In a string, for example, a wave travelling in one direction will get reflected at one
end, which in turn will travel and get reflected from the other end. This will go on until
there is a steady wave pattern set up on the string. Such wave patterns are called
standing waves or stationary waves. To see this mathematically, consider a wave
travelling along the positive direction of x-axis and a reflected wave of the same
amplitude and wavelength in the negative direction of x-axis
Y1 = A sin (ωt – kx) (Incident wave)
Y2 = A sin (ωt + kx) (reflected wave)
The stationary wave formed by the superposition of the incident and reflected wave

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will be
Y = Y1+ Y2 = 2A cos kx sin ωt)
Thus, in this wave pattern, the amplitude varies from point-to-point, but each element
of the string oscillates with the same angular frequency ω or time period. There is no
phase difference between oscillations of different elements of the wave. The string
as a whole vibrates in phase with differing amplitudes at different points. The wave
pattern is neither moving to the right nor to the left.
The points at which the amplitude is zero (i.e., where there is no motion at all) are
nodes; the points at which the amplitude is the largest are called antinodes.

QUESTIONS
25.The transverse displacement of a string (clamped at its both ends) is given by
y (x,t) = 0.06 sin (2πx/3) cos (120πt)
All the points on the string between two consecutive nodes vibrate with
(a) same frequency (b) same phase
(c) same energy (d) different amplitude
26. Which of the following statements are true for a stationary wave?
(a) Every particle has a fixed amplitude which is different from the amplitude of its
nearest particle.
(b) All the particles cross their mean position at the same time.
(c) All the particles are oscillating with same amplitude.
(d) There is no net transfer of energy across any plane.
(e) There are some particles which are always at rest
27.The equation of stationary wave in a stretched string is given by y = 5 sin ( πx/3)
cos (40πt). The separation between two adjacent nodes is
(a)1.5cm (b) 3 cm (c) 6 cm (d) 4 cm
28. In a resonance tube we get
(a) Stationary longitudinal waves (b) Stationary transverse waves

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(c) Progressive longitudinal waves (d) Progressive transverse waves
Answers Case Study Questions

21.a 22.a 23.d 24.d

25.a,b &d 26 a,b,d &e 27.b 28.a

Explanation:
21 ANS (a)
From the equation
2π/λ = 62.8
λ = 0.1 unit

22 ANS (a)
From the equation ω = 600 rad/s, k= 2 rad/m
V= ω/k =600/2 = 300 m/s

23 ANS (d)
At t= 0, 5=10 sinα , sinα = ½ or α = π/6
Total phase at t= 7.5s, φ =2π X7.5 /30 + π/6= 2π/3

24 ANS (d)
y=5 sin (100 πt – 0.4 πx),
y= a Sin (ωt – kx) , a= 5m, ω=100π rad/s, k= 0.4 π m-1
λ = 2 π/k = 5m
ν = ω/2 π = 50 Hz
V= νλ= 50 X 5 =250 m/s

27 ANS K= π/3 , λ = 2 π/k = 6 cm

The separation between two adjacent nodes is 3 cm.

1 Mark Questions
29. Sound of maximum intensity is heard successively at an interval of 0.2 second on
sounding two tuning fork to gather. What is the difference of frequencies of two tuning
forks?
ANS. The beat period is 0.2 second so that the beat frequency is f b=1/0.2=5Hz
Therefore, the difference of frequencies of the two tuning forks is 5 Hz.

30. Why do tuning forks have two prongs?

ANS. The tuning fork has two prongs because the two prongs of a tuning fork produce
resonant vibrations that help to keep the vibrations going for longer.
31. Velocity of sound increases on a cloudy day. Why?
ANS. Velocity of sound increases on a cloudy day, because the air is wet on a
cloudy day, it contains a lot of moisture, the density of air is lower, and because
velocity is inversely proportional to density, velocity increases.
32. A sitar wire is replaced by another wire of same length and material but of three
times the earlier radius. If the tension in the wire remains the same, by what factor
will the frequency change?
ANS. Frequency will become 1/3. Since frequency

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 = √1/πr2ρ
33. A tuning fork A, marked 512 Hz, produces 5 beats per second, where sounded
with another unmarked tuning fork B. If B is loaded with wax the number of beats is
again 5 per second. What is the frequency of the tuning fork B when not loaded?
ANS. Frequency of tunning fork B is 517 Hz.
34. Why longitudinal waves are called pressure waves?
ANS. Longitudinal waves are called pressure waves because the propagation of
longitudinal waves through a medium consists of the variations in the volume and the
pressure of the air, these variations in volume and air pressure result in the formation
of compressions and rarefactions.
35. At what temperatures (in °C) will the speed of sound in air be 3 times its value at
0°C?
ANS. , V2= 3 V1
T2=9 x 273 = 2457 K
36. When we start filling an empty bucket with water, the pitch of the sound produced
goes on changing?
ANS. A bucket may be regarded as a pipe closed at one end. It produces a note of
frequency, ν= V/4L, where Vis the velocity of sound in air and L is the length of air
column, which is equal to depth of water level from open end. As bucket is filled with
water, the value of L decreases. Hence pitch of the sound produced goes on
changing.
37. A sonometer wire is vibrating in resonance with a tuning fork. Keeping the
tension applied same, the length of the wire is doubled. Under what conditions would
the tuning fork still be is resonance with the wire?
ANS. Wire of twice the length vibrates in its second harmonic. Thus if the tuning fork
resonates at L, it will resonate at 2L
38. Solids can support both longitudinal and transverse waves, but only longitudinal
waves can propagate in gases, why?
ANS. This is due to the fact that gases have only the bulk modulus of elasticity
whereas solids have both, the shear modulus as well as the bulk modulus of
elasticity.

2 Marks Questions
39. A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched
string is 20.0 m. If the transverse jerk is struck at one end of the string, how long
does the disturbance take to reach the other end?
ANS. m kg / m , T = 200 N
Speed of transverse jerk is v=√T/m =40m/s
Time taken by jerk to reach the other end = Distance / speed =0.5 s.
40. A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is
resonantly excited by a source of 1237.5 Hz? (sound velocity in air = 330 m s –1)
ANS. Fundamental frequency or first harmonics = 412.5 Hz
Thus answer is third harmonics.
41.Give four differences between transverse & longitudinal waves.
ANS. Page 280 NCERT Physics part II
42. Given below are some examples of wave motion. State in each case if the wave
motion is transverse, longitudinal or a combination of both:
(a) Motion of a kink in a longitudinal spring produced by displacing one end of the

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spring sideways.
(b) Waves produced in a cylinder containing a liquid by moving its piston back and
forth.
(c) Waves produced by a motorboat sailing in water.
(d) Ultrasonic waves in air produced by a vibrating quartz crystal.
ANS. (a) Transverse and longitudinal
(b) Longitudinal
(c) Transverse and longitudinal
(d) Longitudinal
43. Figure shows two vibrating modes of an air column. Find the ratio of frequencies
of two modes.

ANS. For mode (b) L = 3λ/4 or λ= 4L/3 , ν1 = V/ λ= 3V/4L

For mode (c) L = 5λ/4 or ν2 = V/ λ= 5V/4L
Hence ν1 : ν2 =3:5
44. Estimate the speed of sound in air at standard temperature and pressure. The
mass of 1 mole of air is 29.0 ×10–3 kg.
ANS. We know that 1 mole of any gas occupies 22.4 litres at STP. Therefore,
density of air at STP is: ρ o = (mass of one mole of air)/ (volume of one mole of air at
STP)
29.0 X10-3 kg /22.4X 10-3 m 3 = 1.29 kg m–3
45. What are stationary waves? State the necessary condition for formation of
stationary waves.
ANS. Definition of stationary wave
Condition for formation of stationary waves
(i) A stationary wave cannot be formed from two independent waves travelling in
opposite directions.
(ii) A stationary wave can be produced in a finite medium which has its boundaries
46. Two sitar strings A and B playing the note ‘Dha’ are slightly out of tune and
produce beats of frequency 5 Hz. The tension of the string B is slightly increased and
the beat frequency is found to decrease to 3 Hz. What is the original frequency of B if
the frequency of A is 427 Hz ?
ANS. Increase in the tension of a string increases its frequency. If the original
frequency of B (νB ) were greater than that of A (νA ), further increase in νB should
have resulted in an increase in the beat frequency. But the beat frequency is found
to decrease. This shows that νB < νA . Since νA – νB = 5 Hz, and νA = 427 Hz, we get
νB = 422 Hz.
47.On the basis of dimensional considerations, write the formula for speed of
transverse waves on a stretched string.
ANS. Derive it using dimensions.
48. An incident wave is represented by y(x,t) = 20 sin(2x-4t). Write the expression for

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reflected wave:
(i) from a rigid boundary
(ii) from a open boundary
ANS. (i) The wave reflected from a rigid boundary is y= 20 sin (2x + 4t)
(ii) The wave reflected from an open boundary is y = -20 sin (2x+4t)

3 Marks Questions
49. Use the formula v=√γ P/ρ
to explain why the speed of sound in air
(a) is independent of pressure,
(b) increases with temperature,
(c) increases with humidity.
ANS. (a) We have,
v=√γ P/ρ
Where, Density,
ρ=Mass/Volume=M/V
M = Molecular weight of gas
V = Volume of gas
Hence, equation (i) becomes: v=√γ PV/M ……(ii)
Now ideal gas equation for n=1 is: PV=RT
For constant T, PV=Constant
Both M and γ are constants, v=Constant
Hence, the speed of sound is independent of the change in the pressure of the gas at
a constant temperature.
(b) increases with temperature,
Ans: We have, v=√γ P/ρ ……(i)
Now ideal gas equation for n=1 is: PV=RT
P=RT/V……(ii)
Substituting (ii) in (i), we get: v=√γ RT/Vρ = √γ RT/M……(iii)
Where Mass M=ρ V is a constant & γ and R are also constants.
We get from equation (iii),
V ∝√T
.Hence, the sound speed in a gas is directly proportional to the square root of the
gaseous medium’s temperature, i.e., the sound speed increases with rise in the
gaseous medium’s temperature and vice versa.
(c) increases with humidity.
Ans: Let v m and v d are the sound speed in moist air and dry air respectively and ρ
m and ρ d are the densities of moist air and dry air respectively.
We have ,v=√γ P/ρ
The speed of sound in moist air is: v m=√γ P/ρm ……(i)
The speed of sound in dry air is v d=γ P/ρ d ……(ii)
On dividing equations (i) and (ii), we get:
v m/v d=√ρd/ρm……(iii)
However, the presence of water vapour decreases the density of air, i.e.,
ρ d< ρ m ⇒vm>vd
Hence, the speed of sound in moist air is higher than it is in dry air. Thus, in a gaseous

275
medium, the sound speed increases with humidity.
50. Write Newton’s formula for speed of sound in a gas. Why &what correction was
applied by Laplace in this formula?
ANS. NEWTON’S FORMULA FOR SPEED OF SOUND IN A GAS
Assumption : Sound waves travel through gas under isothermal conditions. Thus
temperature of gas remains constant. We have already seen that the sound waves
travel in the form of compressions and rarefactions of small volume elements of air.
The elastic property that determines the stress under compressional strain is the bulk
modulus of the medium .
Thus, the general formula for longitudinal waves in a medium is: v = √B /ρ
We can estimate the speed of sound in a gas in the ideal gas approximation. For an
ideal gas, the pressure P, volume V and temperature T are related by
PV = N kB T --------(i)
where N is the number of molecules in volume V, kB is the Boltzmann constant and T
the temperature of the gas (in Kelvin).
Therefore, for an isothermal change it follows from Eq.(1) that

V∆P + P∆V = 0
Hence, substituting in Eq. (14.16), we have B = P Therefore, the speed of a
longitudinal wave in an ideal gas is given by,
v = √P/ρ -----------(ii)
This relation was first given by Newton and is known as Newton’s formula.
According to Newton’s formula for the speed of sound in a medium, we get for the
speed of sound in air at STP,
V = √((1.013 X 105)/ 1.293) = 280 m/s
The result shown is about 15% smaller as compared to the experimental value of 331
m/ s
If we examine the basic assumption made by Newton that the pressure variations in
a medium during propagation of sound are isothermal, we find that this is not correct.
It was pointed out by Laplace that the pressure variations in the propagation of sound
waves are so fast that there is little time for the heat flow to maintain constant
temperature. These variations, therefore, are adiabatic and not isothermal.
For adiabatic processes the ideal gas satisfies the relation
PV γ = constant .Differentiating both sides , we get
P(γ V γ –1 )dV + V γ dP =0

γ =B
where γ is the ratio of two specific heats, Cp /Cv .
The speed of sound is, therefore given by,
v = √ γ P/ρ ---------(III)
This modification of Newton’s formula is referred to as the Laplace correction. For air
γ = 7/5. Now using Eq. (III) to estimate the speed of sound in air at STP, we get a
value 331.3 m s–1 , which agrees with the measured speed.
51. A wave travelling along a string is described by,
y(x, t) = 0.005 sin (80.0 x – 3.0 t),
in which the numerical constants are in SI units (0.005 m, 80.0 rad/m , and 3.0
rad/s). Calculate (a) the amplitude, (b) the wavelength, and (c) the period and
frequency of the wave.

276
 Also, calculate the displacement y of the wave at a distance x = 30.0 cm and time
t = 20 s ?
ANS. On comparing this displacement equation with Eq. y (x, t) = a sin (kx – ωt), we
find
(a) the amplitude of the wave is 0.005 m = 5 mm.
(b) the angular wave number k and angular frequency ω are k = 80.0 m–1 and ω = 3.0
s–1
We, then, relate the wavelength λ to k through Eq
λ = 2π/k
λ = 2π/180 = 7.85 cm
(c) Now, we relate T to ω by the relation T = 2π/ω
T= 2π/3 = 2.09 s and frequency, v = 1/T = 0.48 Hz
The displacement y at x = 30.0 cm and time t = 20 s is given by
y = (0.005 m) sin (80.0 × 0.3 – 3.0 × 20)
= (0.005 m) sin (–36 + 12π)
= (0.005 m) sin (1.699)
= (0.005 m) sin (97°) = 5 mm
52. A pipe, 30.0 cm long, is open at both ends. Which harmonic mode of the pipe
resonates a 1.1 kHz source? Will resonance with the same source be observed if one
end of the pipe is closed ? Take the speed of sound in air as 330 m /s.
ANS The first harmonic frequency is given by
ν1 = v/ λ1= v/2L (open pipe) where L is the length of the pipe.
The frequency of its nth harmonic is: νn = n v /2L, for n = 1, 2, 3, ... (open pipe)
First few modes of an open pipe are shown .
For L = 30.0 cm, v = 330 m/ s , νn = n 330 (m s ) /0.6 (m) = 550 n s–1
Clearly, a source of frequency 1.1 kHz will resonate at v2 , i.e. the second harmonic.
53. A transverse harmonic wave on a string is described by
y(x,t)=3.0sin(36t+0.018x+π/4)
Where x and y are in cm and t in s. The positive direction of x is from left to right.
(a) Is this a travelling wave or a stationary wave?
If it is travelling, what are the speed and direction of its propagation?
ANS The given equation is the equation of a travelling wave, moving from right to left
because it is an equation of the type
y(x,t)=Asin(ωt+kx+ϕ)
−1
Here, A=3.0cm, ω=36rad , k=0.018cm and ϕ=π/4
Speed of wave propagation is given by v=ω/k = 20ms−1
(b) What are its amplitude and frequency?
Ans: Amplitude of wave, A=3.0cm
Frequency of wave, ν=ω/2π=36/2π=5.7Hz
(c) What is the initial phase at the origin?
Ans: Initial phase at origin, ϕ=π/4 rad
(d) What is the least distance between two successive crests in the wave?
Ans: Least distance between two successive crests in the wave,
⇒λ=2π/k=2π/0.018=349cm=3.49m

Long Questions 5 Marks

54. Discuss the formation of harmonics in a stretched string. Show that in the case
of stretched string the first four harmonics are 1:2:3:4.
ANS. Consider a wave travelling along the positive direction of x-axis and a reflected

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wave of the same amplitude and wavelength in the negative direction of x-axis.
we get: y1 (x, t) = a sin (kx – ωt)
y2 (x, t) = a sin (kx + ωt)
The resultant wave on the string is, according to the principle of superposition:
y (x, t) = y1 (x, t) + y2 (x, t)
= a [sin (kx – ωt) + sin (kx + ωt)]
Using the familiar trignometric identity Sin (A+B) + Sin (A–B) = 2 sin A cos B
we get, y (x, t) = 2a sin kx cos ωt
The amplitude of this wave is 2a sin kx. Thus, in this wave pattern, the amplitude
varies from point-to-point, but each element of the string oscillates with the same
angular frequency ω or time period..
The points at which the amplitude is zero (i.e., where there is no motion at all) are
nodes;
the points at which the amplitude is the largest are called antinodes.
Let us determine these normal modes for a stretched string fixed at both ends. the
positions of nodes (where the amplitude is zero) are given by
sin kx = 0 . which implies kx = nπ; n = 0, 1, 2, 3, ...
Since, k = 2π/λ , we get x =n λ/2 ; n = 0, 1, 2, 3,
the distance between any two successive nodes is λ/ 2.
In the same way, the positions of antinodes (where the amplitude is the largest) are
given by the largest value of sin kx , sin k x = 1
which implies kx = (n + ½) π ; n = 0, 1, 2, 3, ...
With k = 2π/λ, we get x = (n + ½) 2 λ ; n = 0, 1, 2, 3, ...
Again the distance between any two consecutive antinodes is λ/2 .
. Taking one end to be at x = 0, the boundary conditions are that x = 0 and x = L are
positions of nodes. The x = 0 condition is already satisfied. The x = L node condition
requires that the length L is related to λ by L = n λ/2 ; n = 1, 2, 3, ..
Thus, the possible wavelengths of stationary waves are constrained by the relation
λ = 2L/ n ; n = 1, 2, 3, …
with corresponding frequencies vn = n v/2L , for n = 1, 2, 3, (14.42)
We have thus obtained the natural frequencies - the normal modes of oscillation of
the system. The lowest possible natural frequency of a system is called its
fundamental mode or the first harmonic. For the stretched string fixed at either end it
is given by v = v/ 2L , corresponding to n = 1 of Eq. (14.42). Here v is the speed of
wave determined by the properties of the medium
. The n = 2 frequency is called the second harmonic;
n = 3 is the third harmonic and so on. We can label the various harmonics by the
symbol ν n ( n = 1, 2, ...).

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55.(a) Discuss the various modes of vibration in a closed end organ pipe. Show that
in a closed end organ pipe the harmonics are in the ratio of 1:3:5…..
(b) A meter-long tube open at one end, with a movable piston at the other end,
shows resonance with a fixed frequency source (a tuning fork of frequency 340Hz )
when the tube length is 25.5cmor 79.3cm. Estimate the speed of sound in air at the
temperature of the experiment.
ANS (a) Normal modes of a closed organ pipe

First mode of vibration, there is one node at closed end and one antinode at the
open end.
L= λ1/4 or λ1 = 4L
Frequency v1 = 1/4L√ γ P/ρ = v Fundamental frequency
Second mode of vibration, two nodes & two antinodes
L= 3λ2/4 λ2 = 4L /3
Frequency v2 = 3/4L√ γ P/ρ = 3 v =First overtone or third harmonics
Third mode of vibration, three nodes & three antinodes
Frequency v3 = 5/4L√ γ P/ρ = 5v = Second overtone or fifth harmonics
(b)Frequency of the turning fork, v=340Hz
Because one end of the supplied pipe is connected to a piston, it will behave as a pipe
with one end closed and the other open, as illustrated in the diagram.
Such a system produces odd harmonics. The fundamental note in a closed pipe is
given by the relation:
l1=λ/4
Where,
Length at the pipe, l1=25.5cm=0.255m

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∴λ=4l1=4×0.255m=1.02m
The speed of sound is given by the relation:
v=vλ=340×1.02=346.8m/s

56.What are beats? Prove that the number of beats per second is equal to the
difference between the frequencies of the two superimposing waves..
ANS Beats is an interesting phenomenon arising from interference of waves. When
two harmonic sound waves of close (but not equal) frequencies are heard at the
same time,. We hear audibly distinct waxing and waning of the intensity of the
sound, with a frequency equal to the difference in the two close frequencies
Let us consider two harmonic sound waves of nearly equal angular frequency ω1
and ω2 and fix the location to be x = 0 for convenience. Eq. (14.2) with a suitable
choice of phase (φ = π/2 for each) and, assuming equal amplitudes, gives
s1 = a cos ω1 t and s2 = a cos ω2 t
The resultant displacement is, by the principle of superposition,
s = s1 + s2 = a (cosω1 t + cos ω2 t)
Using the familiar trignometric identity for cos A + cos B,

we get
s = [2 a cos ωb t ] cos ωa t

where ωb = and ωa =
The amplitude is the largest when the term cos ωb t takes its limit +1 or –1.
In other words, the intensity of the resultant wave waxes and wanes with a
frequency which is 2ωb = ω1 –ω2 . Since ω = 2πν, the beat frequency νbeat, is given
by
vbeat = ν1 – ν2

HOT QUESTIONS (CCT)

57. A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long
uniform string is vibrating in its second harmonics and it resonates with the
fundamental frequency of the pipe. If the tension in the pipe is 50 N and speed of
sound is 320 m/s, the mass of string is
(a) 5 grams (b) 10 grams (c) 20 grams (d) 40 grams
ANS (b)
For hollow pipe, fundamental frequency is f = V/4l = 320/4X0.8
For string in 2nd harmonic
f = 1/l√T/m =1/0.5 √50X 0.5/m
Equating & solving m= 10 g

58. A wave travelling along the x-axis is described by the equation y(x, t) =
0.005 cos(αx –β t). If the wavelength and the time period of the wave are 0.08 m
and 2.0 s, respectively, then α and β in appropriate units are
(a) α = 12.50π, β = π/2.0
(b) α = 25.00π, β = π
(c) α = 0.08/π, β = 2.0/π
(d) α = 0.04/π, β =1.0/π
Answer: (b) α = 25.00π, β = π
The wave travelling along the x-axis is given by
y(x,t) = 0.005 cos(αx – βt)

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Therefore,α = k = 2π/λ As λ = 0.08 m
α = 2π/0.08 = π/0.04 ⇒ α = (π/4 ) x 100 = 25.00π
ω = β ⇒2π/2 =π
∴ α = 25.00π
β=π

59.The length of a sonometer wire is 0.75 m and density 9 X 10 3 Kg/m3. It can bear a
stress of 8.1 X 108 N /m2 without exceeding the elastic limit. What is the fundamental
frequency that can be produced in the wire? (IIT 1990)
ANS. Let a be the area of cross section of wire then fundamental frequency ,
ν = 1/2L√T/m = 1/2L √Stress X a/a X 1 X ρ = 200 Hz.

SELF ASSESSMENT
1.An open organ pipe produces a note of frequency 5/2 Hz at 150C, calculate the
length of pipe. Velocity of sound at 00C is 335 m/s.
2. An incident wave is represented by Y(x, t)=20sin(2x-4t).Write the expression for
reflected wave (i) From a rigid boundary (ii) From an open boundary.
3. Explain why
(i) in a sound wave a displacement node is a pressure antinode and vice- versa
(ii) The shape of pulse gets- distorted during propagation in a dispersive medium.
4. The equation of a plane progressive wave is y=10 sin2π(t-0.005 x), where y & x
are in cm & t in second. Calculate the amplitude, frequency, wavelength & velocity of
the wave.
5. (i) A steel rod 100 cm long is clamped at its middle. The fundamental frequency of
longitudinal vibrations of the rod is given to be 2.53 kHz. What is the speed of sound
in steel?
(ii) A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is
resonantly exited by a 430 Hz source? Will this same source be in resonance with
the pipe if both ends are open? (Speed of sound = 340 m/s).
6. a) Explain the formation of beats.
b) When a tuning fork of unknown frequency is sounded with another tuning fork
whose frequency is 384 Hz, 6 beats per second are produced. When wax is attached
to the first fork, then on sounding it with the sound, 4 beats per second are produced.
Determine the unknown frequency

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