Wave Motion

Wave Motion

𝜕𝑦 𝜕𝑦
= −𝑣
𝜕𝑡 𝜕𝑥
Particle velocity = − Wave velocity × slope
y

x
 Example
(a) Transverse and longitudinal Given below are some examples of wave motion. State in each case if the wave
(b) Longitudinal motion is transverse, longitudinal or a combination of both:
(c) Transverse and longitudinal (a) Motion of a kink in a longitudinal spring produced by displacing one end of the
(d) Longitudinal 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.
Solution
𝐴𝑛𝑠. 𝑎 5 mm
Example
𝑏 7.85 mm A wave travelling along a string is described by,
𝑐 2.09 𝑠, 0.48 Hz 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–
1, and 3.0 rad s–1). Calculate (a) the amplitude, (b) the wavelength, and

(c) the period and frequency of the wave.

Solution
 Example
a 𝜆 = 20 cm
b 10= 𝜆/2, −1.5 mm 𝐀 wave travelling on a string at a speed of 𝟏𝟎 𝐦/𝐬 causes each particle
of the string to oscillate with a time period of 𝟐𝟎 𝐦𝐬.
(a)What is the wavelength of the wave?
(b)If the displacement of a particle is 𝟏. 𝟓 𝐦𝐦 at a certain instant,
what will be the displacement of a particle 𝟏𝟎 𝐜𝐦 away from it at
the same instant?
Solution y

𝝀
𝟐 x
 Example
𝑓 = 50 Hz
𝜆 = 4 × 10−2 m A particle on a stretched string supporting a travelling wave, takes
𝑣 = 2 m/s
𝟓. 𝟎 𝐦𝐬 to move from its mean position to the extreme position. The
distance between two consecutive particles, which are at their mean
positions is 𝟐. 𝟎 𝐜𝐦. Find the frequency, the wavelength and the wave
speed.

Solution

𝜆
2
 Example
(a) Negative 𝑥-direction The equation of a wave travelling on a string is
(b)𝑣 = 10 m/s 𝒚 = 𝟎. 𝟏𝟎 𝒎𝒎 𝐬𝐢𝐧 𝟑. 𝟏𝟒 𝒎−𝟏 𝒙 + 𝟑𝟏𝟒 𝒔−𝟏 𝒕
𝜆 = 0.2 m (a)In which direction does the wave travel?
𝑓 = 50 Hz (b)Find the wave speed, the wavelength and the frequency of the
c 𝑦max = 𝐴 = 0.1 mm
wave
(c)What is the maximum displacement and the maximum speed of a
𝑢max = 10𝜋 mmΤs portion of the string?
Solution
Concept Speed of transverse waves in string

Velocity of a transverse wave propagating along a string is given by

𝑻 Note: We have assumed tension 𝑻 constant in the

𝒗= string which is valid for small amplitude.
𝝁

m
Example 2%

The percentage increase in the speed of transverse waves produced in

a stretched string if the tension is increased by 4%, will be __________%.
[JEE-Main-2021]

Solution
Example
A tuning fork A of unknown frequency produces 5 beats/s with a fork
of known frequency 340 Hz. When fork A is filed, the beat frequency
decreases to 2 beats/s. What is the frequency of fork A?
[JEE-Main-2021]
(1) 342 Hz (2) 345 Hz (3) 335 Hz (4) 338 Hz

Solution
Example
A sound wave of frequency 245 Hz travels with the speed of 300 ms–1
along the positive x-axis. Each point of the wave moves to and fro
through a total distance of 6 cm. What will be the mathematical
expression of this travelling wave?
[JEE-Main-2021]
(1) Y(x,t) = 0.03 [sin 5.1 x – (0.2 × 103)t]
(2) Y(x,t) = 0.06 [sin 5.1 x – (1.5 × 103)t]
(3) Y(x,t) = 0.06 [sin 0.8 x – (0.5 × 103)t]
(4) Y(x,t) = 0.03 [sin 5.1 x – (1.5 × 103)t]

Solution
Example
A longitudinal wave is represented by
 x
x = 10sin2  nt –  cm. The maximum particle velocity will be four
 λ
times the wave velocity if the determined value of wavelength is equal
to: [JEE-Main-2022]
5π
(1) 2 𝝅 (2) 5𝝅 (3) 𝝅 (4)
2
Solution
Example
In the wave equation
2π
y = 0.5sin (400t – x)m the velocity of the wave will be :
λ [JEE-Main-2022]
(1) 200 m/s (2) 200 2 m/s (3) 400 m/s (4) 400 2 m/s

Solution
 Example
−1
𝐴𝑛𝑠. 93 m s
A steel wire 0.72 m long has a mass of 5.0 × 10–3 kg. If the wire is
under a tension of 60 N, what is the speed of transverse waves on the
wire ? NCERT

Solution

m
 Example
𝑣𝐶𝐷 = 20 m/s Two blocks each having a mass of 𝟑. 𝟐 𝐤𝐠 are connected by a wire 𝑪𝑫 and the
system is suspended from the ceiling by another wire 𝑨𝑩. The linear mass
𝑣𝐴𝐵 = 80 m/s
density of the wire 𝑨𝑩 is 𝟏𝟎 𝐠/𝐦 and that of 𝑪𝑫 is 𝟖 𝐠/𝐦. Find the speed of a
transverse wave pulse produced in 𝑨𝑩 and in 𝑪𝑫.

Solution
A

B
𝟑. 𝟐 𝐤𝐠
C

D
𝟑. 𝟐 𝐤𝐠
 Example
𝑣 = 𝑔𝑥
A heavy but uniform rope of length 𝑳 is suspended from a ceiling.
(a) Write the velocity of a transverse wave travelling on the string as a
𝐿 function of
𝑡=2 the distance from the lower end.
𝑔
(b) If the rope is given a sudden sideways jerk at the bottom, how long will it
take for the pulse to reach the ceiling?

Solution

P
L
x

A
 Example
𝑨𝒏𝒔. 𝟎. 𝟓 𝒔
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?
Solution NCERT

L
Concept Energy, Power and Intensity in Wave Motion

Power 𝑷𝒂𝒗
𝑰= = 𝐼 = 2𝜋 2 𝑓 2 𝐴2 𝜌𝑣
area of cross−section 𝑺
𝑃av = 2𝜋 2 𝑓 2 𝐴2 𝜇𝑣
 Example
𝑃 = 4.9 × 10−2 𝑊 A transverse wave of amplitude 𝟎. 𝟓𝟎 𝐦𝐦 and frequency 𝟏𝟎𝟎 𝐇𝐳 is
produced on a wire stretched to a tension of 𝟏𝟎𝟎 𝐍. If the wave speed
is 𝟏𝟎𝟎 𝐦/𝐬, what average power is the source transmitting to the
wire?

Solution
Concept Relation between path difference phase difference and time difference

Δ𝜙 Δ𝑥 Δ𝑡
= =
2𝜋 𝜆 𝑇
Concept Superposition of Waves
𝑆1
𝑥
𝑆1 : 𝑦1 = 𝐴1 sin(𝑘𝑥 − 𝜔𝑡) (i)
𝑃

𝑆2 : 𝑦2 = 𝐴2 sin[𝑘𝑥 − 𝜔𝑡 + 𝜙] (ii)
𝑆2

𝑦 = 𝑦1 + 𝑦2 = 𝐴 sin(𝑘𝑥 − 𝜔𝑡 + 𝜃)

𝐴= 𝐴12 + 𝐴22 + 2𝐴1 𝐴2 cos 𝜙

𝐴2 sin 𝜙
tan 𝜃 =
𝐴1 + 𝐴2 cos 𝜙

Intensity ∝ (amplitude) 2
Concept Interference of Waves
Condition of Maxima or Constructive Interference
cos 𝜙 = 1
𝜙 = 0, 2𝜋, … , 2𝑛𝜋
Δ𝑥 = 0, 𝜆, … , 𝑛𝜆
𝑛 = 0,1,2, …

𝐴max = 𝐴1 + 𝐴2 ,
2
𝐼max ∝ 𝐴1 + 𝐴2
Condition of Minima or Destructive Interference

cos 𝜙 = −1 𝜙 = cos(2𝑛 + 1)𝜋

𝜆 3𝜆 𝜆
Δ𝑥 = , , ⋯ , 2𝑛 + 1
2 2 2
𝐴min = 𝐴1 − 𝐴2 ,
𝐼min ∝ 𝐴1 − 𝐴2 2
Regarding Intensity

𝐼 = 𝐼1 + 𝐼2 + 2𝐼1 𝐼2 cos 𝜙

2 2
𝐼max 𝐴1 + 𝐴2 𝐼1 + 𝐼2
= =
𝐼min 𝐴1 − 𝐴2 𝐼1 − 𝐼2
Concept Stationary Waves formation

𝑦1 = 𝐴 sin(𝜔𝑡 − 𝑘𝑥)

𝑦2 = 𝐴 sin(𝜔𝑡 + 𝑘𝑥)

𝑦 = 2𝐴 cos 𝑘𝑥 sin 𝜔𝑡 = 𝐴𝑠 sin 𝜔𝑡 𝐴𝑠 = 2𝐴 cos 𝑘𝑥

𝜆/2 𝜆/2
𝜆/4
2𝐴
𝑁 𝑁 𝑁
𝑂 𝐴𝑁 𝐴𝑁 𝐴𝑁 𝐴𝑁 𝑡 = 0

𝜆 3𝜆 5𝜆
𝜆 𝜆
4 4 4 3𝜆
2
2
 Example
𝜙 = 3𝜋
Two waves, each having a frequency of 𝟏𝟎𝟎 𝐇𝐳 and a wavelength of 𝟐. 𝟎 𝐜𝐦,
𝜙 = 4𝜋 are travelling in the same direction on a string. What is the phase difference
𝐴𝑅 = 0 𝐴𝑅 = 4 mm between the wave
(a) If the second wave was produced 𝟎. 𝟎𝟏𝟓 𝐬 later than the first one at the
same place?
(b)If the two waves were produced at the same instant but the first one was
produced a distance 𝟒. 𝟎 𝐜𝐦 behind the
(c) If each of the waves has an amplitude of 𝟐. 𝟎 𝐦𝐦, what second one? would
be the amplitudes of the resultant waves in part (a) and (b)?
Solution
 Concept Vibrations of Strings

1. String fixed at both ends: 1 𝑇

𝑓1 = 𝑓1 : fundamental frequency
2𝑙 𝜇
𝑓1 : 𝑓2 : 𝑓3 : … = 1: 2: 3: …
when a string is fixed at both ends, all harmonics (odd as well even) can be
produced.
th 𝑝 𝑇
If the string is vibrating in 𝑝 harmonic 𝑓 =
2𝑙 𝜇
 Concept Vibrations of Strings

2. String fired at one end:

1 𝑇
𝑓1 = 𝑓1 : fundamental frequency
4𝑙 𝜇

𝑓1 : 𝑓2 : 𝑓3 : … = 1: 3: 5: …
From string fixed at one end only odd harmonics can be produced.
Concept Vibrations of tuning fork
Example 𝑙 = 2.1m

A wire having a linear mass density 𝟓. 𝟎 × 𝟏𝟎−𝟑 𝐤𝐠/𝐦 is stretched

between two rigid supports with a tension of 𝟒𝟓𝟎 𝐍 . The wire
resonates at a frequency of 𝟒𝟐𝟎 𝐇𝐳. The next higher frequency at
which the same wire resonates is 𝟒𝟗𝟎 𝐇𝐳. Find the length of the wire.

Solution
Example 𝑛 = 11 Hz

A steel wire of length 𝟏 𝐦, mass 𝟎. 𝟏 𝐤𝐠 and uniform cross-sectional

area 𝟏𝟎−𝟔 𝒎𝟐 is rigidly fixed at both ends. the temperature of wire is
lowered by 𝟐𝟎∘ 𝐂. If transverse waves are set up by plucking the string
in the middle, calculate the frequency of the fundamental mode of
vibration. Young's modulus of steel = 𝟐 × 𝟏𝟎𝟏𝟏 𝐍/𝒎𝟐 , coefficient of
linear expansion of steel = 𝟏. 𝟐𝟏 × 𝟏𝟎−𝟓 ∘ 𝑪 −𝟏
Solution
T T
Example 𝑛 = 70 Hz

The length of the wire shown in figure, between the pulleys is 𝟏. 𝟓 𝐦

and its mass is 𝟏𝟐. 𝟎 𝐠. Find the frequency of vibration with which the
wire vibrates in two loops having the middle point of the wire
between the pulleys at rest.
Solution

9 kg 9 kg
(a)f = 30 Hz Example
b Harmonics 3,5,7 Three resonant frequencies of a string are 90, 150 and 210 Hz
(c) Overtones 2,4.6 (a)Find the highest possible fundamental frequency of vibration of
this string.
d 𝑣 = 48 m/s (b)Which harmonics of the fundamental are the given properties?
(c)Which overtones are these frequencies?
(d)If the length of the string is 80 cm a what would be the speed of a
transverse was on this is string?
Solution (a) Highest possible fundamental frequency is given by highest
common factor: H.C.F of 90,180 and 210 is 30
 Example
𝑓 = 30 Hz The fundamental frequency of a sonometer wire increases by 𝟔 𝐇𝐳 if
Δ𝑓 = 5 Hz (decrease) its tension is increased by 𝟒𝟒%, keeping the length constant. Find the
change in the fundamental frequency of the sonometer wire when the
length of the wire is increased by 𝟐𝟎%, keeping the original tension in
the wire.
Solution
 Example
Ans. 422 Hz.
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 ?
Solution NCERT
 Example
𝐴𝑛𝑠. 2.06 × 104 N
A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should
be the tension in the wire so that speed of a transverse wave on the
wire equals the speed of sound in dry air at 20 °C = 343 m s–1.
Solution NCERT
 Example
𝐴𝑛𝑠. 8.7 𝑠
A stone dropped from the top of a tower of height 300 m splashes into
the water of a pond near the base of the tower. When is the splash
heard at the top given that the speed of sound in air is 340 m s–1? (g =
9.8 m s–2)

Solution NCERT
 Concept Sound Waves

The human ear is sensitive to waves in the frequency range from 20

to 20000 Hz, called the audible range. The sound of frequency below
20 Hz are called Infrasonic and above 20000 Hz as Ultrasonic.

SPEED OF A SOUND WAVE

𝜸𝑷 𝑪𝑷 𝜸𝑷 𝜸𝑹𝑻
𝒗= ,𝜸 = 𝒗= =
𝝆 𝑪𝑽 𝝆 𝑴𝟎

Effect of pressure, temperature and humidity on the speed of

sound in air
 Characteristic Of Sound
(a) Loudness: Loudness depends on intensity Sound level (measured in decibels,
dB) is given by
𝐼1
𝐿1 − 𝐿2 = 10 log10 (i)
𝐼2
where 𝐼 : intensity of sound
𝐼0 : reference intensity
𝐼0 : minimum intensity that is just audible at intermediate frequencies

𝐼0 = 10−12 W/m2
𝐼1
𝐿1 − 𝐿2 = 10 log10 (ii)
𝐼2
(b) Pitch: It is the sensation received by the ear due to frequency and is the
characteristic which differentiate a shrill sound from a grave sound. The pitch
depends on frequency.
(c) Quality: It is the sensation received by the ear due to 'waveform'. The
characteristic of sound by virtue of which sounds of same intensity and same
frequency can be distinguished is called quality. Quality is related to waveform and
waveform depends on overtones present, quality depends on number of overtones
i.e., harmonics present and their intensity.
 BEATS

𝑦1 = 𝐴 sin 𝜔1 𝑡 = 𝐴 sin 2 𝜋𝑓1 𝑡

𝑦2 = 𝐴 sin 𝜔2 𝑡 = 𝐴 sin 2 𝜋𝑓2 𝑡

𝑓1 + 𝑓2
Frequency of resultant wave
2

So, beat frequency i.e. number of beats per second will be

𝑓𝑏 = 𝑓1 ∼ 𝑓2
Beat frequency is equal to the difference of frequencies of two interfering waves.
 VIBRATIONS OF AIR COLUMNS

(a) Closed origin pipe

(i) 𝑓1 : 𝑓2 : 𝑓3 : … = 1: 3: 5 …

From closed pipe, only odd harmonics can be produced

(b) Open pipe

(i) 𝑓1 : 𝑓2 : 𝑓3 … = 1: 2: 3

From open pipe all frequencies (odd as well as even) can be obtained.
VIBRATIONS OF AIR COLUMNS

END CORRECTION
 Example
Frequencies between
A cylindrical metal tube has a length of 𝟓𝟎 𝐜𝐦 and is open at both
1000 and 2000 Hz
ends. Find the frequencies between 𝟏𝟎𝟎𝟎 𝐇𝐳 and 𝟐𝟎𝟎𝟎 𝐇𝐳 at which
= 1020 Hz, 1360 Hz, 1700 Hz the air column in the tube can resonate. Speed of sound in air is
𝟑𝟒𝟎 𝐦/𝐬.

Solution
 Example
𝑙max = 8.5 m
Find the greatest length of an organ pipe open at both ends that will
have its fundamental frequency in the normal hearing range (𝟐𝟎 −
𝟐𝟎𝟎𝟎𝟎 𝐇𝐳). Speed of sound in air = 𝟑𝟒𝟎 𝐦/𝐬.
Solution
 Example
1
𝑙 = m = 0.25 m
4 Two successive resonance frequencies in an open organ pipe are
𝟏𝟗𝟒𝟒 𝐇𝐳 and 𝟐𝟓𝟗𝟐 𝐇𝐳. Find the length of the tube. The speed of sound
in air is 𝟑𝟐𝟒 𝐦/𝐬.
Solution
(a) Independent of pressure. Example
Use the formula 𝜸𝑷
(b) Increases as 𝑇 𝒗= to explain why the speed of sound in air
𝝆
(c) The molecular mass of (a)is independent of pressure,
water (18) is less than that (b)increases with temperature,
of N2 (28) and O2 (32). (c)increases with humidity
Therefore. as humidity
increases, the effective Solution
molecular mass of air
decreases and hence v
increases.
 Example
𝐴𝑛𝑠. 𝑎 3.4 × 10−4 m
𝑏 1.49 × 10−3 m A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound
meets a water surface, what is the wavelength of (a) the reflected
sound, (b) the transmitted sound? Speed of sound in air is 340 m s–1
and in water 1486 m s–1.
NCERT
Solution
 Example
𝐴𝑛𝑠. 4.1 × 10−4 m
A hospital uses an ultrasonic scanner to locate tumours in a tissue.
What is the wavelength of sound in the tissue in which the speed of
sound is 1.7 km s–1 ? The operating frequency of the scanner is 4.2 MHz.
Solution NCERT