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Displacement Wave Equation

s(x, t) = s0 sin(±ωt ± kx + ϕ)

2π
ω= ω = 2πf
T

2π ω
k= v=
λ k
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ds
vP = −Vw
dx
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Variation of excess pressure in gas due

to propagating longitudinal wave

ds
ΔP = −β
dx
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Pressure Equation of Longitudinal Progressive Wave

∆P = ∆P0cos(ωt – kx + ϕ)

+ve
Compression
Real pressure = Patm + ∆P
Rarefaction
– ve
Pmax = P0 + ∆P0 = P0 + βks0

Pmin = P0 − ∆P0 = P0 − βks0

Density Wave Equation

π
Δρ = Δρ0 sin ωt − kx +
2
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ds S Rarefaction Compression
ΔP = −β
dx
S
x
x
P
𝐏𝐚𝐭𝐦 + ∆𝐩max

𝐏𝐚𝐭𝐦 − ∆𝐩max

Key Point
π
ΔP = ΔP0 sin ωt − kx + ϕ +
2
s = s0 sin ωt − kx + ϕ
π
Pressure wave differs in phase by from the displacement wave.
2
The pressure maxima occurs where the displacement is zero and displacement
maxima occurs where the pressure is at normal level.
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Velocity of Sound Wave

For Liquid & Gases:

β β = Bulk Modulus of Elasticity of Fluid or Gas

v= ρ
ρ = Density of fluid or Gas

For Solid:

Y Y = Young’s Modulus of Elasticity of Solids

v=
ρ ρ = Density of solid

The velocity of sound at a given temperature is given by:

γRT
v=
M
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Laplace Correction

γp γRT T = Kelvin
Vw = =
ρ M M = Molecular mass
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Intensity & Power Transmission in Longitudinal Waves

▪ Intensity of Sound Waves

ds ds
I = −β General Formula
dx dt

ΔP02 1 β
I= <I>= (ωs0 )2
2ρV 2 Vw

1 1
▪ For Point Source I∝ and s0 ∝
r2 r

1 1
▪ For Line Source I∝ and s0 ∝
r r

▪ For Plane Wave Source I = constant and

s0 = constant
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Characteristics of Sound Waves

▪ Pitch
▪ Loudness
▪ Quality of Wave form
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Loudness of Sound

I
L = 10log dB
I0

i. e. I0 = 10−12 W/m2
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Reflection of Sound Wave

At the closed end of any tube/pipe, there is no phase change in the

pressure wave of sound during reflection.

At the open end of any tube/pipe, there is a phase change of  in the

pressure wave of sound during reflection.
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Conclusion
Pressure wave
Closed end Open end
1. No phase change Phase change of 

2. Reflected pressure wave is Reflected pressure wave gets

similar in shape with incident inverted
pressure wave

Displacement wave
Closed end Open end
1. Phase change of  No phase change

2. Reflected displacement wave Reflected displacement wave

gets inverted is similar in shape
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Normal Modes of Vibration of a Closed Organ Pipe

No. of No. of
Modes of Vibration Frequency Harmonic Overtone
Nodes Antinodes

First 1 1

Third First 2 2

n n+1
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Normal Modes of Vibration of an Open Organ Pipe

No. of No. of
Modes of Vibration Frequency Harmonic Overtone
Nodes Antinodes

1 2
First
2 1

2 3
Second First
3 2
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For Open Organ Pipe:

v
▪ Fundamental Frequency f0 =
2l

nv
▪ Frequency of nth Harmonic f= = nf0
2l

For Closed Organ Pipe:

v
▪ Fundamental Frequency f0 =
4l

▪ Frequency of nth Overtone f = (2n + 1)f0

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Interference of Sound Waves

Constructive Interference Destructive Interference

2 2 2 2
Pmax = P01 + P02 + 2P01 P02 cos0o Pmin = P01 + P02 + 2P01 P02 cos𝜋

Pmax = P01 + P02 Pmin = P01 − P02

2 2
Imax = I1 + I2 Imin = I1 − I2

∆ϕ = 0,2π, 4π……2nπ ∆ϕ = π, 3π, 5π……(2n + 1)π

Same phase 𝐜𝐨𝐬∆𝛟 = 𝟏 Opposite phase 𝐜𝐨𝐬∆𝛟 = −𝟏

λ 3λ 5λ 2n+1 λ
∆x = 0, λ, 2λ, 3λ……nλ ∆x = , , ……
2 2 2 2

Here n = 0, 1, 2, 3…….. Here n = 0, 1, 2, 3……..

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Q. A loudspeaker L is placed in the hall with two doors D1 and D2 open to the
playground, as shown below. The distance between D1 and D2 is 74.25m.
The loudspeaker L is equidistant from D1 and D2 . Monotonic sound waves
are emitted from the loudspeaker, and it is found that at both points P & Q,
sound intensities are minimum. Distance of point P from D1 is 6 m. The line
joining D1 , P and Q is perpendicular to the line joining D1 and D2 . No other
minimum intensity locations can be found between PQ and beyond Q along
the PQ line. The wavelength in meter of the sound wave generated by the
loudspeaker is.........
Sol. D2
L
Hall
D1
P Q
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Sol.

Ans. 3
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Q. Three coherent sonic sources emitting sound of single wavelength ℓ are

−λ 11 λ 11
placed on the x-axis at points ,0 , 0,0 , ,0 . The intensity
6 6
5λ
reaching a point 0, from each source has the same value I0 . Then, the
6
resultant intensity at this point due to the interference of the three waves
will be:
(A) 6I0 (B) 7I0 (C) 9I0 (D) 5I0
Sol.
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Sol.

Ans. (B)
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Q. The three identical loud speakers in figure play a 170 Hz frequency tone in
a room where speed of sound is 340 m/s. At point exactly in front of source
S2 the amplitude of the wave from each speaker is a. What is the ratio of
the resultant intensity at P and the intensity due to a single speaker?
Sol. S1

3m
S2
P

3m

S3

4m
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Sol.

Ans. 1
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Two sound waves of slightly different frequencies have amplitude ratio

Q.
11
. What is the difference of sound levels in decibels of maximum and
9
minimum intensities heard at a point?
(A) 100 (B) 10 (C) 16 (D) 20

Sol.
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Sol.

Ans. (D)
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The air column in a pipe closed at one end is made to vibrate in its second
Q.
overtone by a tuning fork of frequency 440 Hz. The speed of sound in air is
330 ms −1 . End corrections may be neglected. Let P0 denote the mean
pressure at any point in the pipe & D P0 the maximum amplitude of
pressure variation.
(i) Find the length L of the air column.
(ii) What is the amplitude of pressure variation at the middle of the
column?
(iii) What are the maximum & minimum pressures at the open end of the
pipe?
(iv) What are the maximum & minimum pressures at the closed end of the
pipe?
Sol.
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Sol.

15 ΔP0
Ans.(i) L = m, (ii) m,
16 2
(iii) Pmax = Pmin = P0 (iv) Pmax = P0 + ΔP0 , Pmin = P0 − ΔP
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Resonance Tube Experiment

Tuning Meter Scale Water

Fork Reservoir

In this
experiment, the
velocity of sound
in air is to be
found by using
Cylindrical tuning forks of
Tube known frequency.
The wavelength
of the sound will
be determined by
making use of the
resonance of an
air column.
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Mathematical Analysis of Resonance Tube

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Calculation of End Correction

In the resonance tube, the antinode is not formed exactly at the open end but
slightly outside at a distance, e. Hence, the length of the air column in the first
and second states of resonance are (ℓ1 + e) and (ℓ2 + e) then:

e e
l1
l2
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Calculation of End Correction

λ
For First resonance, ℓ1 + e = 4
3λ
For Second resonance, ℓ2 + e = 4

Subtract the above resonance length to obtain the wavelength of the wave.

3λ λ
ℓ2 − ℓ1 = −
4 4
λ = 2(ℓ2 − ℓ1 )
λ
ℓ2 − ℓ1 =
2
Put the value of λ from the 3rd equation to the 1st equation to find the
value of e.
2(ℓ2 − ℓ1 )
ℓ1 + e =
4
ℓ2 − 3ℓ1
After solving the above equation, e=
2
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In a resonance column apparatus, first resonance is obtained when the

Q.
water filling beaker (of cylindrical shape) is just empty as shown. The
water filling beaker is lowered down and it is seen that second resonance
is obtained when beaker is filled up to brim. The wavelength of sound is
α
given by 10 m Find the value of α. 2cm
Sol.
4cm
10cm
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Sol.

Ans. 8
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Doppler Effect of Sound or Acoustic Doppler Effect

The apparent change in frequency or pitch due to relative motion of source
and observer along the line of sight is called Doppler Effect.
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Beats

When two sound waves of Nearly Equal Frequencies (but not exactly
equal) travel in the same direction, at a given point due to their
superposition, their intensity alternately increases and decreases
periodically. This periodic change in sound intensity w.r.t. time at a given
position is known as the Beat Phenomenon.

Beat frequency = f2 − f1
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Two electric trains run at the same speed of v = 90 km/hr along a

Q.
straight track one after the other with an interval of ℓ = 2.0 km between
them. At the instant when they are located symmetrically relative to point
A at a distance of b = 1.0 km from the track (figure) both trains give a
brief sound signal of the same frequency of n = 50 Hz. What will the
number of beats heard per second at point A when the vibrations
produced by the signals arrive at it? The speed of sound is v = 350 m/s.
Round off to nearest integer. A

Sol.

ℓ
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Sol.

Ans. 5
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As shown above there are two circles having radius R and 2R respectively.
Q.
Sound source is moving in inner circle in anticlockwise direction. Observer
is moving on outer circle in clockwise direction with angular velocity ω1
and ω2 respectively. If frequency of source is f then, frequency observed
by observer is (Given: ω1 = 10 rad/sec, ω2 = 10 rad/s, R = 10 m and
velocity of sound is 330 m/s)
23 47
(A) 𝑓 (B) 𝑓
43 23 2R
Observer
43
(C) 𝑓 (D) 𝑓 R 60°
23
Sol. Source
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Sol.

Ans. (C)
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A man while sitting on an oscillating swing whistles at a frequency

Q.
π
1000 Hz. The swing is oscillating with an amplitude ± 6 radian about the
mean position in a vertical plane. Man is sitting over the swing which is at
a distance of 2 m from the point of suspension of the swing. Another man
standing just in front of the mean position of the swing hears the whistle.
The maximum frequency heard by the man standing in front of the swing
is: (Speed of sound in air = 332 m/s)
Sol.
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Sol.

Ans. 1007
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Two trains move in the same direction on two close tracks. The train A
Q.
sounds a horn of single frequency of 1 kHz. At a certain instant of time,
when both trains move at the same speed of 36 km/hr, the two straight
tracks deviate at an angle of 60° with each other. Velocity of sound in air
is 300 m/s. Which of the following statement(s) is/are true?
(A) The apparent frequency heard by a passenger on the second train B
just after he passes the bend will be 1.05 kHz.
(B) The apparent frequency heard by a passenger on the first train A, just
after the second train B passes the bend will be 1 kHz.
(C) The apparent frequency heard by a passenger on the second train B
just after he passes the bend will be 983 Hz.
(D) If the first train A, that produces the sound, were to pass the bend
while the other train B goes on the straight track then the apparent
frequency heard by a passenger on the latter train B will be less than
1 kHz.
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1 uA = 36 km/h uA
60°

2 uB = 36 km/h
uB
Sol.
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Sol.

Ans. (BCD)
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A boat is travelling in a river with a speed of 10 m/s along the stream

Q.
flowing with a speed 2 m/s. From this boat, a sound transmitter is
lowered into the river through a rigid support. The wavelength of the
sound emitted from the transmitter inside the water is 14.45 mm. Assume
that the attenuation of sound in water and air is negligible.
(a) What will be the frequency detected by a receiver kept inside the river
downstream?
(b) The transmitter and the receiver are now pulled up into air. The air is
blowing with a speed 5 m/sec in the direction opposite the river stream.
Determine the frequency of the sound detected by the receiver.
(Temperature of the air and water = 20°C ; Density of river water =
103 Kg/m3 ; Bulk modulus of the water = 2.088 × 109 Pa; Gas constant
R = 8.31 J/mol − K; Mean molecular mass of air = 28.8 × 10–3 kg/mol;
CP/CV for air = 1.4)
Sol.
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Sol.

Ans. (a) 100696 Hz (b) 103038 Hz

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What is Wave?

A wave is a disturbance that transfers energy and momentum

from one place to another without the transport of matter.
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Classification
of
Waves

Based on the Based on the

necessity of propagation
the medium of energy

Based Based on the

on vibration of
dimensions particle
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Equation of Progressive Transverse Wave

y = A sin(ωt ± kx + ϕ)

2π
ω= ω = 2πf
T

2π ω
k= v=
λ k

If the signs of both locations are different, then the wave is propagating on
the + axis of propagation.

If the signs of both locations are the same, then the wave is propagating on
the – axis of propagation.
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Q. For a transverse wave travelling along a straight line, the distance

between two peaks (crests) is 5 m, while the distance between one crest
and one trough is 1.5 m. The possible wavelengths (in m) of the waves are
1 1 1
(1) 1, 2, 3, . . . (2) 2 , 4 , 6 , . . .
1 1 1
(3) 1, 3, 5, . . . (4) 1 , 3 , 5 , . . .
Sol.

Ans. 4
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Phase Difference (∆𝛟)

y
…. SHM Eq.
A
Wave Eq. x
O x1
x2 B
…. SHM Eq.
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Particle Velocity (𝐕𝐩 )

𝜕y
vp = −v
𝜕x

V
A
A
VA

Particle A follows the previous particle’s motion.

Hence, the velocity of A is downwards.
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Differential Equation of Wave

𝜕2y 1 𝜕2y
=
𝜕x 2 v 2 𝜕t 2

Wave Velocity

T T
v= = (w.r.t string)
μ ρA

If 𝐓 ∝ 𝒙 then the wave moves with constant acceleration.

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Q. Find the time taken by the wave to reach the top.

Sol.

Rod
𝑙
M

ℓ
Ans. t = 2
g
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Q. The linear density of a wire under tension T, varies linearly from 𝜇1 to 𝜇2 .

Calculate the time that a pulse needs to pass from 1 end 2 another. The
length of the wire is 𝑙.
Sol.
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3/2 3/2
2𝑙 𝜇2 − 𝜇1
Ans. 𝑡 =
3 𝑇 𝜇2 − 𝜇1
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Q. A block M hangs vertically at the bottom end of a uniform rope of

constant mass per unit length. The top end of the rope is attached to a
fixed rigid support at O. A transverse wave pulse (Pulse 1) of wavelength
λ0 is produced at point O on the rope. The pulse takes time TOA to reach
point A. If the wave pulse of wavelength λ0 is produced at point A (Pulse
2) without disturbing the position of M, it takes time TAO to reach point
O. Which of the following options is/are correct: [JEE Advanced 2017]
(A) The time TAO = TOA
(B) The velocities of the two pulses (Pulse 1 and O Pulse 1
Pulse 2) are the same at the midpoint of the
rope
(C) The wavelength of Pulse 1 becomes longer
when it reaches point A
A Pulse 2
(D) The velocity of any pulse along the rope is
M
independent of its frequency and wavelength
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Sol.

Ans. A,D
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Q. A nonuniform rope of length  hangs from a ceiling. Mass

per unit length of rope (m) changes as μ = μ0ey, where y is
the distance along the string from its lowest point. Then the
graph between the square of the velocity of the wave and y 𝑦
will be best represented as:

𝑣𝑦2 𝑣𝑦2 𝑣𝑦2 𝑣𝑦2

(A) (B) (C) (D)

𝑦 𝑦 𝑦 𝑦

Sol.
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Ans. A
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Mechanical Energy of a Wave

The kinetic energy of the particles executing SHM.

d K 1
= μA2 ω2 cos2 (ωt − kx + ϕ)
d𝑥 2

Potential energy due to the elongation of the string.

dU 1 2 2
= μA ω cos2 ωt − kx + ϕ
d𝑥 2

The total mechanical energy of the wave per unit length is given by,

dE dK dU
= + = μ2 A2 ω2 cos2 ωt − kx + ϕ
dx dx dx

dE
= μ2 A2 ω2 cos 2 ωt − kx + ϕ
dx
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Important Points y

B
2
dE 2 2 2
𝜕y x
=μ A k O dx
dx 𝜕x
dx

K.E. = maximum
At B: VP = maximum ⟹
P.E. = maximum

K.E. = minimum
At A: Particles are momentarily at rest ⟹
P.E. = minimum
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Power Transmitted Along a String

dE
= μ2 A2 ω2 v cos 2 ωt − kx + ϕ
dt

Pavg = 2π2 A2 f 2 μv

In Different Expressions for Power

1 TA2 ω2
1. <P> =
2 v
1
2. <P> = 2 μvA2 ω2 ∵ T = μv 2

3. <P> = 2π2 μvA2 f 2 ∵ ω = 2πf

T
4. <P> = 2π2 TμA2 f 2 ∵v= μ
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Intensity of Wave

I = 2π2 f 2 A2 ρv

If the medium is the same, then ρ and v, both are constant.

I ∝ A2 f 2

If the medium and frequency are both constant, then ρ , v and f are
constant.
I ∝ A2
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Q. Consider two cases of the same rope having amplitudes A and 2A,
respectively, as shown in the figure. Comment on their intensities.
Sol.
Y

A 2A
X
O

Ans. 1
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−2
Q. A string with linear mass density λ = 5.00 × 10 kg/m is under a tension
of 80.0 N. How much power must be supplied to the string to generate
sinusoidal waves at a frequency of 60 Hz and amplitude of 6.00 cm?
Sol.

Ans. 512W
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Superposition of Wave

yresultant = y1 + y2 + y3 + ⋯

This is called the superposition phenomenon.

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Coherent Source
Those sources, due to which, at a particular point, there is no change in
phase difference with respect to time, are known as coherent sources. It
highly implies that ω is the same.

Incoherent Source
Those sources, due to which, at a particular point where the phase
difference continuously changes with respect to time, are known as
incoherent sources. It highly implies that ω is different.
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Mathematical Analysis
Let the two waves be:

yresultant = y1 + y2

A= A21 + A22 + 2A1 A2 cosΔϕ

A2 sinΔϕ α Δϕ
tanθ =
A1 + A2 cosΔϕ
A1
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Resultant Intensity

I = I1 + I2 + 2 I1 I2 cos ∆ϕ

If I1 = I2 = I

Inet = 2I(1 + cos∆ϕ)

∆ϕ
Inet = 4Icos2
2

π∆x
Inet = 4Icos2
λ
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Constructive Interference Destructive Interference

+ +

= =
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Constructive Interference Destructive Interference

Amax = A21 + A22 + 2A1 A2 cos 0° Amin = A21 + A22 + 2A1 A2 cos π

Amax = A1 + A2 Amin = A1 − A2

2 2
Imax = I1 + I2 Imin = I1 − I2

∆ϕ = 0,2π, 4π……2nπ ∆ϕ = π, 3π, 5π……(2n − 1)π

Same phase 𝐜𝐨𝐬∆𝛟 = 𝟏 Opposite phase 𝐜𝐨𝐬∆𝛟 = −𝟏

λ 3λ 5λ 2n−1 λ
∆x = 0, λ, 2λ, 3λ……nλ ∆x = , , ……
2 2 2 2

Here, n = 0, 1, 2, 3…….. Here, n = 1, 2, 3……..

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Amplitudes of Reflected and Transmitted Waves

2v2
At = A
v1 + v2 i

v2 − v1
Ar = A
v1 + v2 i
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Q. Two strings of linear mass densities μ and 4μ are attached, as shown in

the figure. A wave with amplitude Ai is incident on the light string and
strikes the other. Find the amplitude of the reflected and transmitted
waves. v1

Sol.
Ai

μ1 = μ μ2 = 4μ

𝟐 𝟏
Ans. 𝑨𝒕 = 𝑨𝒊 , 𝑨𝒓 = 𝑨𝒊
𝟑 𝟑
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Reflection of Waves

Reflection from Rigid End

Reflection from Free End

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Reflection from Rigid End

When a transverse wave reflects from a rigid end, then the crest
will reflect in the form of a trough, and the trough will reflect in
the form of a crest.

The phase difference between the incident and reflected

wave is 180°.
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Reflection from Free End

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Reflection from Free End

When a transverse wave reflects from the free end, then the crest
will reflect in the form of a crest, and the trough will reflect in the
form of a trough.
Phase difference between incident and reflected wave is 𝟎°.
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Standing Wave

If two sinusoidal waves of the same amplitude and wavelength travel in

opposite directions along a stretched string, then interference with each
other produces a standing wave.

λ λ λ
2 2 2
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Analysis of Standing Wave

y = 2A sin kx + ϕ1 sin(ωt + ϕ2 )

Comparing with an equation of SHM, 𝐲 = 𝐀 𝐱 𝐬𝐢𝐧(𝛚𝐭)

Amplitude Function of time

Conclusion

As this equation satisfies the wave equation, it represents a wave.

The amplitude of the wave As = 2Asin kx is not constant but varies

periodically with position (and not with time).
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Important Points

Nodes Nodes Nodes Nodes Nodes

𝛌/2

𝛌/4 𝛌/2

Antinodes Antinodes Antinodes Antinodes

Distance between two successive nodes or successive antinodes is λ/2.

Distance successive node to antinode or antinode to the tinode to node
is λ/4.
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Note

(a) Particles of the same loop are in phase, and particles from adjacent
loops are in opposite phase.

(b) The phase difference between any two particles in a standing wave is
either zero or π.
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Important Points
All particles reach their equilibrium position simultaneously.

All particles reach their extreme position simultaneously, but their

amplitude is position-dependent.

Either side particles of a node are in the opposite phase.

On either side, particles of an antinode are in the same phase.

In 1 complete vibration, the string appears straight twice.

Since the loop pattern does not transfer in stationary waves, the speed
of the stationary waves is zero.
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Normal Modes of Vibration of a String

When both ends of the string are fixed

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No. of No. of PHYSICS
Modes of Vibration Frequency Harmonic Overtone
Nodes Antinodes

λ = 2l
v First 2 1
f=
2l

λ=l
v Second First 3 2
f=
l

2l
λ=
3
3v
Third Second 4 3
f=
2l

2l
λ=
n
nv n (n − 1)th n+1 n
f=
2l
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Normal Modes of Vibration of a String

When one end of a string is free

AN

N
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No. of No. of PHYSICS
Modes of Vibration Frequency Harmonic Overtone
Nodes Antinodes

λ = 4l
v First 1 1
f=
4l

4l
λ=
3
Third First 2 2
3v
f=
4l

4l
λ=
5
5v
Fifth Second 3 3
f=
4l

4l
λ=
2n + 1
(2n + 1)v (2n + 1)th n n+1 n+1
f=
4l
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Sonometer

The relation between frequency, length of a string, the tension in the

string and linear mass density of the string can be found
experimentally with this instrument known as a sonometer.

The portion of the wire between the movable bridges forms the
"string" fixed at both ends.

By sliding these bridges, the length of the wire may be changed.

The tension of the experimental wire may be changed by changing

the weights on the hanger.

One can remove the experimental wire itself and put another wire in
its place, thereby changing the mass per unit length.
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Sonometer
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Q. A wire having a linear mass density of 9.0 × 10 kg/m is stretched

−4

between two rigid supports with a tension of 900 N. The wire resonates
at a frequency of 500 Hz. The next higher frequency at which the same
wire resonates is 550 Hz. The length of the wire is ______m.
Sol.
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Sol.

Ans. 10
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Q. A wire of length L and mass per unit length 6.0×10–3 kgm–1 is put under
tension of 540 N. The two consecutive frequencies at which it resonates
are: 420 Hz and 490 Hz. Then L in meters is:
(1) 8.1 m (2) 5.1 m (3) 1.1 m (4) 2.1 m
Sol.

Ans. 4
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Q. Two uniform strings of mass per unit length μ and 4μ, and length L and
2L, respectively, are joined at point O, and tied at two fixed ends P and
Q, as shown in the figure. The strings are under a uniform tension T. If
1 T
we define the frequency v0 = 2L , which of the following statements is
μ
correct? [JEE Advanced 2024]

m O 4μ
P Q
L 2L

(A) With a node at O, the minimum frequency of vibration of the

composite string is v0 .
(B) With an antinode at O, the minimum frequency of vibration of the
composite string is 2v0 .
(C) When the composite string vibrates at the minimum frequency with a
node at O, it has 6 nodes, including the end nodes.
(D) No vibrational mode with an antinode at O is possible for the
composite string.
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Sol.

Ans. A,C,D
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Q. A string of length 1 m and mass 2 × 10 kg is under tension T. When

−5

the string vibrates, two successive harmonics are found to occur at

frequencies 750 Hz and 1000 Hz. The value of tension T is _______
Newton. [JEE Advanced 2023]

Sol.
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Ans. 5
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Q. Four harmonic waves of equal frequencies and equal intensities I0 have

phase angles 0, π/3, 2π/3 and π. When they are superposed, the
intensity of the resulting wave is nI0 . The value of n is.
Sol. [JEE Advanced 2015]
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Ans. 3
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Q. One end of a taut string of length 3m along the x–axis is fixed at x = 0.

The speed of the waves in the string is 100 ms −1 . The other end of the
string is vibrating in the y direction, so that stationary waves are set up
in the string. The possible waveform(s) of these stationary waves is
(are): [JEE Advanced 2014]
πx 50πt
(A) y t = A sin cos
6 3
πx 100πt
(B) y t = A sin cos
3 3
5πx 250πt
(C) y t = A sin cos
6 3
5πx
(D) y t = A sin 2
cos 250πt

Sol.
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Ans. A,C,D
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Q. If y1 = 5 (mm) sin πt is equation of oscillation of sources S1 and y2 = 5

(mm) sin (πt + π/6) be that of S2 and it takes 1 sec and ½ sec for the
transverse waves to reach point A from sources S1 and S2 respectively
then the resulting amplitude point A, is:
A
S1 S2
(A) 5 2 + 3mm (B) 5 3mm
(C) 5 mm (D) 5 2mm

Sol.
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Ans. C
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Q. A particular guitar wire is 30.0 cm long and vibrates at a frequency of 196

Hz in fundamental mode when no finger is placed on it. The next higher
notes on the scale are 220 Hz, 247 Hz, 262 Hz and 294 Hz. How far from
the end of the string must the finger be placed to play these notes?

Sol.
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Ans. 26.7 cm, 23.8 cm, 22.4cm and 20.0 cm

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Q. A uniform horizontal rod of length 40 cm and mass 1.2 kg is supported

by two identical wires as shown in the figure. Where should a mass of
4.8 kg be placed on the rod so that the same tuning fork may excite the
wire on the left into its fundamental vibrations and that on the right into
its overtone?

Sol. Take g = 10 m/s

2

40 cm
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Ans. 5 cm from the left end

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Q. Figure shows an aluminum wire of length 60 cm joined to a steel wire of

length 80 cm and stretched between two fixed supports. The tension
produced is 40 N. The cross-sectional area of the steel wire is 1.0 mm2 and
that of the aluminum wire is 3.0 mm2 . What could be the minimum
frequency of a tuning fork which can produce standing wire is 3.0 mm2.
What could be the minimum frequency of a tuning fork which can produce
standing waves in the system with the joint as a node? The density of
aluminum is 2.6 g/cm3 and that of steel is 7.8 g/cm3.
Sol. 80 cm 60 cm

Steel Aluminium
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Ans. 180 Hz
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Q. A heavy string is tied at one end to a movable support and to a light

thread at the other end, as shown in the figure. The thread goes over a
fixed pulley and supports a weight to produce tension. The lowest
frequency with which the heavy string resonates is 120 Hz. If the movable
support is pushed to the right by 10 cm so that the joint is placed on the
pulley, what will be the minimum frequency at which the heavy string can
resonate? 10 cm
Sol.
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Ans. 240 Hz
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Q. A string fixed at both ends is vibrating in the lowest mode of vibration for
which a point at a quarter of its length from one end is a point of
maximum displacement. The frequency of vibration in this mode is 100 Hz.
What will be the frequency emitted when it vibrates in the next mode such
that this point is again a point of maximum displacement?
Sol.
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Ans. 300 Hz
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Q. A straight sonometer wire is stretched by the weight of a solid hanging

block. When oscillating the complete wire, its oscillating frequency is n.
When block is immersed in water its frequency is n1 and when immersed
in a liquid its frequency becomes n2 . Find the specific gravity of the block
and that of the liquid.
Sol.
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𝑛2 − 𝑛22
Ans. 2
𝑛 − 𝑛12
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Q. A metal rod of length 100 cm is clamped at two points A and B as shown

in the figure. Distance of each clamp from the end of rod is 30 cm. If
density and Young’s modulus of elasticity of rod material and 9000 kg/
m3 and 1.44 × 1011 N/m2 respectively. Calculate the minimum and next
higher (first overtone) frequency at which stationary waves can be setup
in rod.
Sol. 100 cm

30 cm 30 cm
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Ans. 30000 Hz
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Q. If a string is horizontally stretched with a tension F in it and its one end

oscillates vertically by a source with equation y = A sin ωt where A is
small, due to these oscillations a transverse wave is setup in string. Find
the rate at which energy is supplied by oscillating source to string and
maximum rate at any instant energy is supplied. Take the linear mass
density of the string as μ.
Sol.
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Ans. FμA2 ω2
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