NEO_JEE_11_P1_PHY_H_Waves and Sound_S12_209.pdf
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Oct 07, 2024
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About This Presentation
These notes clear your concepts for jee mains and advance
Slide Content
Welcome to
Waves
Waves
Waves
It is a disturbance in propagation that
•Transports energy and momentum
•Does not involve the bulk movement
of matter
It is a disturbance in propagation that
•Transports energy and momentum
•Does not involve the bulk movement
of matter
Classification of Waves
Heading 1 Heading 1
Heading 1
Based on the necessity of medium
Mechanical Non -Mechanical
•Requires a material
medium to propagate
through.
•Elasticityand inertiaof the
medium are key
parameters
•Some examples are -
sound waves, water
ripples, seismic waves
•Does not necessarily
require a material
medium for
propagation.
•All kinds of
electromagnetic
waves -Light waves,
radio waves, X-rays,
etc.
•They can travel as
fast as 3×10
8
�/�.
Heading 1 Heading 1
Heading 1
Based on the necessity of medium
Mechanical Non -Mechanical
•Requires a material
medium to propagate
through.
•Elasticityand inertiaof the
medium are key
parameters
•Some examples are -
sound waves, water
ripples, seismic waves
•Does not necessarily
require a material
medium for
propagation.
•All kinds of
electromagnetic
waves -Light waves,
radio waves, X-rays,
etc.
•They can travel as
fast as 3×10
8
�/�.
Classification of Waves
•Particles of the medium vibrate in a direction perpendicular to
•the direction of propagation of the wave.
•Examples are waves on a string and waves on water.
Ԧ�
�⊥Ԧ�
�
Ԧ�
�-Wave velocity Ԧ�
�-Particle velocity
Consider a transverse wave on a string
•Particles of the medium vibrate in a direction
parallel to the direction of propagation of the
wave.
•Examples are sound waves.
•For any point of the wave,
Ԧ�
�∥Ԧ�
�
Ԧ�
�-Wave velocity
Ԧ�
�-Particle velocity
•Particles of the medium vibrate in a direction perpendicular to
•the direction of propagation of the wave.
•Examples are waves on a string and waves on water.
Ԧ�
�⊥Ԧ�
�
Ԧ�
�-Wave velocity Ԧ�
�-Particle velocity
Consider a transverse wave on a string
•Particles of the medium vibrate in a direction
parallel to the direction of propagation of the
wave.
•Examples are sound waves.
•For any point of the wave,
Ԧ�
�∥Ԧ�
�
Ԧ�
�-Wave velocity
Ԧ�
�-Particle velocity
Classification of Waves
Heading 1
1�
Heading 1
2�
Heading 1
3�
Heading 1
Based on the direction of
wave propagation
Seismic waves during earthquakesWave pulse on a string Ripple
Heading 1
1�
Heading 1
2�
Heading 1
3�
Heading 1
Based on the direction of
wave propagation
Seismic waves during earthquakesWave pulse on a string Ripple
Simple harmonic wave
•The nature of the wave is simple harmonic due to the motion of the
source(periodic up and down motion)
•Every particle of the string follows the motion of source and executes
SHM of the same amplitude and frequency in vertical direction.
�
Note: Attenuation of amplitude during the propagation is neglected
•The nature of the wave is simple harmonic due to the motion of the
source(periodic up and down motion)
•Every particle of the string follows the motion of source and executes
SHM of the same amplitude and frequency in vertical direction.
�
Note: Attenuation of amplitude during the propagation is neglected
�
Generation of a simple harmonic wave
Consider an external source executing SHM in the vertical directionwhile creating a wave pulse (up and down
motion) in a string fixed at the other end.
It creates a wave packet in the string as shown in the figure.
Equation of motion of particle at �=0is
�(�,�)=�sin(��+??????)
The displacement vs time�−�graph for a particle on
string at �=0and executing SHM in vertical direction
starting from its mean position at �=0.
Generation of a simple harmonic wave
Consider an external source executing SHM in the vertical directionwhile creating a wave pulse (up and down
motion) in a string fixed at the other end.
It creates a wave packet in the string as shown in the figure.
Equation of motion of particle at �=0is
�(�,�)=�sin(��+??????)
The displacement vs time�−�graph for a particle on
string at �=0and executing SHM in vertical direction
starting from its mean position at �=0.
Time period and frequency
Time Period
Itisthetimetakenforonecompleteoscillation.
Frequency
Itisthenumberofoscillationsperunittime.
�
�
�
�
�
�
Crest
Trough
Frequency of the wave = Frequency of the oscillation of the source.
Time Period
Itisthetimetakenforonecompleteoscillation.
Frequency
Itisthenumberofoscillationsperunittime.
�
�
�
�
�
�
Crest
Trough
Frequency of the wave = Frequency of the oscillation of the source.
Relation between velocity, wavelength and frequency
The distance travelled by the wave during
one full oscillation of the particle is the
wavelength of the wave.
�=��
�=
1
�
⇒??????=????????????
�
�Crest
Trough
�
�
�
�
The distance travelled by the wave during
one full oscillation of the particle is the
wavelength of the wave.
�=��
�=
1
�
⇒??????=????????????
�
�Crest
Trough
�
�
�
�
Wave Equation for a Transverse wave on a String
The equation of motion of the source
executing SHM is given by,
��
�
�
�
�
��=0,�=�sin��+??????
A particle �of the medium at a distance �from
the source, is also executing SHM in the y-
direction about its mean position (�,0)
The equation of motion of the source
executing SHM is given by,
��
�
�
�
�
��=0,�=�sin��+??????
A particle �of the medium at a distance �from
the source, is also executing SHM in the y-
direction about its mean position (�,0)
Wave Equation
��
�
�
�
�
•If the velocity of the wave is �,then the
time taken by the disturbance to reach
from x = 0 to �=�is
�
�
.
•The displacement of particle �at any time
t is equal to the displacement of the
source at time �−
�
�
.
The wave equation is given as,
��,�=�sin��−
�
�
+??????
��,�=��??????���−
�
�
�+??????
��
�
�
�
�
•If the velocity of the wave is �,then the
time taken by the disturbance to reach
from x = 0 to �=�is
�
�
.
•The displacement of particle �at any time
t is equal to the displacement of the
source at time �−
�
�
.
The wave equation is given as,
��,�=�sin��−
�
�
+??????
��,�=��??????���−
�
�
�+??????
Angular Wave number
It is the total number of complete wave
cycles overits wavelength
�
�Crest
Trough
�
�
�
�=
2�
�
Thewaveequationcanbewrittenas,
�=�sin��−��+??????
�=
2�
�
�
=
2��
�
�=
�
�
��,�=��??????���−
�
�
�+??????
Angular wavenumber (�)is defined asthe number of wavelengths per unit distance.
It is the total number of complete wave
cycles overits wavelength
�
�Crest
Trough
�
�
�
�=
2�
�
Thewaveequationcanbewrittenas,
�=�sin��−��+??????
�=
2�
�
�
=
2��
�
�=
�
�
��,�=��??????���−
�
�
�+??????
Angular wavenumber (�)is defined asthe number of wavelengths per unit distance.
Travelling Wave
��+
�
�
��+��or−�
��−
�
�
��−��or+�
Wavefunction representing
a travelling wave
•For a wave pulse,
Phase=�±��=constant
Wavevelocity,�=
��
��
•Velocity and amplitude of the wave
depend on themedium while the
rest depend on the source.
��+
�
�
��+��or−�
��−
�
�
��−��or+�
Wavefunction representing
a travelling wave
•For a wave pulse,
Phase=�±��=constant
Wavevelocity,�=
��
��
•Velocity and amplitude of the wave
depend on themedium while the
rest depend on the source.
A wave pulse is travelling on a string at 2�/�along positive �-direction.
Displacement �of the particle at �=0at any time �is given by �=
2
�
2
+1
. Find:
(a) the function �=(�,�)i.e. displacement of a particle at position �and time �.
(b) the shape of the pulse at �=0and �=1�.
⇒��,�=
2
�−
�
2
2
+1
��,�=��=0,�−
�
�
�
�
�=1�
2�
�=0
2 4 6�−2−4
1
2
(a) �0,�=
2
�
2
+1
,�=+2�/�Given: (b)
Displacement �of the particle at �=0at any time �is given by �=
2
�
2
+1
. Find:
(a) the function �=(�,�)i.e. displacement of a particle at position �and time �.
(b) the shape of the pulse at �=0and �=1�.
⇒��,�=
2
�−
�
2
2
+1
��,�=��=0,�−
�
�
�
�
�=1�
2�
�=0
2 4 6�−2−4
1
2
(a) �0,�=
2
�
2
+1
,�=+2�/�Given: (b)
A wave equation is given as �=2sin50��−20�+30°��, where �is in seconds and
�is in centimeters. Find out
�=2��
�=50����/�
�=
�
2�
=25��
��,�=�sin��+��+??????
�=
1
�
=
1
25
�
�is in centimeters. Find out
�=2��
�=50����/�
�=
�
2�
=25��
��,�=�sin��+��+??????
�=
1
�
=
1
25
�
Phase of a wave
�=�sin��−��+??????
��
�
�
�
�(�,�)=�sin[��−
x
v
+??????)
Phase of a wave: (��−��+??????)
•If the x-coordinate is fixed, then the expression of y(t)
converts into the equation of SHM.
•If time t is fixed, then the expression of �(�)gives the
shape of the string at that instant.
−��+??????��−
��
�
+??????
phase constant of SHM
�=�sin��−��+??????
��
�
�
�
�(�,�)=�sin[��−
x
v
+??????)
Phase of a wave: (��−��+??????)
•If the x-coordinate is fixed, then the expression of y(t)
converts into the equation of SHM.
•If time t is fixed, then the expression of �(�)gives the
shape of the string at that instant.
−��+??????��−
��
�
+??????
phase constant of SHM
Relation between phase difference and path difference
�
1=�sin(��−��
1+??????)
�
2=�sin(��−��
2+??????)
Δ??????=
2�
�
Δ�
Thedifferenceinphasebetweentheparticlesat
agiventimeinstantisgivenby
Δ??????=��−��
1+??????−(��−��
2+??????)
Δ??????=��
2−�
1=�Δ�
The path difference
between any two
consecutivecrest and
trough of a wave is
�
2
Δ??????=
2�
�
×Δ�
Δ??????=
2�
�
×
�
2
=�
The path difference between two successive
crests of a wave is �
Δ??????=
2�
�
×Δ�
Δ??????=
2�
�
�=2�
�
1=�sin(��−��
1+??????)
�
2=�sin(��−��
2+??????)
Δ??????=
2�
�
Δ�
Thedifferenceinphasebetweentheparticlesat
agiventimeinstantisgivenby
Δ??????=��−��
1+??????−(��−��
2+??????)
Δ??????=��
2−�
1=�Δ�
The path difference
between any two
consecutivecrest and
trough of a wave is
�
2
Δ??????=
2�
�
×Δ�
Δ??????=
2�
�
×
�
2
=�
The path difference between two successive
crests of a wave is �
Δ??????=
2�
�
×Δ�
Δ??????=
2�
�
�=2�
�
�
�
From the snapshot at �=0,
�=0at �=0
Standard expression for a progressive
wave,
�=�sin(��−��+??????)
Applying �=0,�=0,�=0,
sin??????=0
⇒??????=�
At �=0,slope is −��
Solution :
�
�
From the snapshot at �=0,
�=0at �=0
Standard expression for a progressive
wave,
�=�sin(��−��+??????)
Applying �=0,�=0,�=0,
sin??????=0
⇒??????=�
At �=0,slope is −��
Solution :
�
�=
��
��
�=constant
=
��
��
The velocity of the particle at P(�,�)
�=�sin��−��+??????
��
��
=�
�=��cos��−��+??????
�
�=��
2
−�
2
Relation between Wave and Particle Velocity
tan??????=
��
��
��
��
=−��cos(��−��+??????)
�
�=−�
��
��
��
��
=−�
�
�
cos(��−��+??????)
⟹Particle velocity =−(wave velocity) ×slope of �−�curve
�=
��
��
�=constant
=
��
��
The velocity of the particle at P(�,�)
�=�sin��−��+??????
��
��
=�
�=��cos��−��+??????
�
�=��
2
−�
2
Relation between Wave and Particle Velocity
tan??????=
��
��
��
��
=−��cos(��−��+??????)
�
�=−�
��
��
��
��
=−�
�
�
cos(��−��+??????)
⟹Particle velocity =−(wave velocity) ×slope of �−�curve
Figure shows a snapshot of a vibrating string at �=0. The particle �is observed
moving up with velocity 203��/�. The tangent at �makes an angle 60°with the
�-��??????�. Find
�)the direction in which wave is moving
�)equation of wave
a
The direction in which wave is moving
�
�=−�tan??????
203=−�tan60°
⇒�=−20���
−1
Thus, wave is travelling along negative �-��??????�
⇒�=�sin��+��+??????
Suppose �=�sin��±��+??????
b
And �=4��
⇒�=4sin��+��+??????
�=
2�
�
=
�
2
��
−1
�=��=10�����
−1
Also, from initial conditions ??????=
�
4
,
3�
4
Finally,�=4sin10��+
�
2
�+
�
4
From the figure, �=5.5−1.5=4��
Equation of wave
Solution :
moving up with velocity 203��/�. The tangent at �makes an angle 60°with the
�-��??????�. Find
�)the direction in which wave is moving
�)equation of wave
a
The direction in which wave is moving
�
�=−�tan??????
203=−�tan60°
⇒�=−20���
−1
Thus, wave is travelling along negative �-��??????�
⇒�=�sin��+��+??????
Suppose �=�sin��±��+??????
b
And �=4��
⇒�=4sin��+��+??????
�=
2�
�
=
�
2
��
−1
�=��=10�����
−1
Also, from initial conditions ??????=
�
4
,
3�
4
Finally,�=4sin10��+
�
2
�+
�
4
From the figure, �=5.5−1.5=4��
Equation of wave
Solution :
General partial differential equation for a wave
�
2
�
��
2
=−��
2
sin��−��+??????
��
��
=�
�=��cos��−��+??????
��
��
=−��cos��−��+??????
�
2
�
��
2
=−��
2
sin��−��+??????
�
2
�
��
2
=�
2
�
2
�
��
2
Acceleration of
particle
Square of speed
of wave
Rate of
change of
slope
•Equation of Wave
Differentiating w.r.t t
Differentiating w.r.t �
�
2
�
��
2
=−��
2
sin��−��+??????
��
��
=�
�=��cos��−��+??????
��
��
=−��cos��−��+??????
�
2
�
��
2
=−��
2
sin��−��+??????
�
2
�
��
2
=�
2
�
2
�
��
2
Acceleration of
particle
Square of speed
of wave
Rate of
change of
slope
•Equation of Wave
Differentiating w.r.t t
Differentiating w.r.t �
Velocity of transverse waves on a string
�=
�
�
=
ElasticProperty
InertialProperty
•�is the tension in the string that
depends on its elastic property
•�is the linear mass density which is
an inertial property
•The velocityof the wave is the
property of the medium
�
�
�
� �
�
�=
�
�
=
ElasticProperty
InertialProperty
•�is the tension in the string that
depends on its elastic property
•�is the linear mass density which is
an inertial property
•The velocityof the wave is the
property of the medium
�
�
�
� �
�
Time taken by the wave to travel from �to �
In the given figure, a string of mass 200�and length 2�is fixed at two ends �and
�. If the tension in the string is 100�and a wave is generated at �, then find the
following
A BWavelength of the wave when its frequency is 10��
�=
�
�
=
100
0.1
=1010��
−1
�=
1
510
�
The linear mass density of the string is given
by,
�=
�
�
=
2
1010
⇒
�=
�
�
=
0.2
2
=0.1���
−1
�=10�
�=
�
�
=
1010
10
�=1010��
−1
�=10��
Solution :
In the given figure, a string of mass 200�and length 2�is fixed at two ends �and
�. If the tension in the string is 100�and a wave is generated at �, then find the
following
A BWavelength of the wave when its frequency is 10��
�=
�
�
=
100
0.1
=1010��
−1
�=
1
510
�
The linear mass density of the string is given
by,
�=
�
�
=
2
1010
⇒
�=
�
�
=
0.2
2
=0.1���
−1
�=10�
�=
�
�
=
1010
10
�=1010��
−1
�=10��
Solution :
In the given figure, mass of the string is �and its length is �. If a wave is generated
at �, find the time taken by the wave to travel from �to �.
�
�
�
�
��
�=
�
�
��
න
0
�
��
�
=�න
0
�
��
�=
�
�
�=2
�
�
�=
�
�
��
��
=
�
�
=
���/�
�/�
=��
Solution :
at �, find the time taken by the wave to travel from �to �.
�
�
�
�
��
�=
�
�
��
න
0
�
��
�
=�න
0
�
��
�=
�
�
�=2
�
�
�=
�
�
��
��
=
�
�
=
���/�
�/�
=��
Solution :
KE and PE per unit length of an element of the string
The Kinetic energy ��associated with a string element of mass ��is given by:
��
��
=
��
��
=
�
2
�
2
�
2
cos
2
(��−��+??????)
��
��
=
1
2
��.�
2
……(1)
To find �, we differentiate wave eq. w.r.t �??????��, keeping �
constant :
��=
1
2
���−��
2
cos
2
(��−��)
�=
��
��
=−��cos��−��……(2)
Using this relation and putting ��=�dx, we rewrite above equation as:
�=�sin��−��=−��??????�(��−��)
�
�
��
� ��
��
�
��
Where �is transverse speed of oscillating string element.
Potential energy per unit length
��
��
=
�
2
�
2
�
2
cos
2
(��−��+??????)
The Kinetic energy ��associated with a string element of mass ��is given by:
��
��
=
��
��
=
�
2
�
2
�
2
cos
2
(��−��+??????)
��
��
=
1
2
��.�
2
……(1)
To find �, we differentiate wave eq. w.r.t �??????��, keeping �
constant :
��=
1
2
���−��
2
cos
2
(��−��)
�=
��
��
=−��cos��−��……(2)
Using this relation and putting ��=�dx, we rewrite above equation as:
�=�sin��−��=−��??????�(��−��)
�
�
��
� ��
��
�
��
Where �is transverse speed of oscillating string element.
Potential energy per unit length
��
��
=
�
2
�
2
�
2
cos
2
(��−��+??????)
Total Energy per unit length
Total Energy =Total Kinetic Energy +Total Potential Energy
��
��
=�
2
�
2
�cos
2
(��−��+∅)
��=��+��
��
��
=
��
��
+
��
��
��
��
=2
�
2
�
2
�
2
cos
2
��−��+??????
��
��
=
�
2
�
2
�
2
cos
2
(��−��+??????)
��
��
=
�
2
�
2
�
2
cos
2
(��−��+??????)
Total Energy =Total Kinetic Energy +Total Potential Energy
��
��
=�
2
�
2
�cos
2
(��−��+∅)
��=��+��
��
��
=
��
��
+
��
��
��
��
=2
�
2
�
2
�
2
cos
2
��−��+??????
��
��
=
�
2
�
2
�
2
cos
2
(��−��+??????)
��
��
=
�
2
�
2
�
2
cos
2
(��−��+??????)
Power Transmitted by a Wave
Instantaneous Power
�=
��
��
=
��
��
∙
��
��
�=4�
2
�
2
�
2
��cos
2
(��−��+??????)
��
��
=�
2
�
2
�cos
2
(��−��+??????)
��
��
=�
�=
��
��
=
��
��
∙
��
��
Average Power
�
���=
���
��
�=4�
2
�
2
�
2
��cos
2
(��−��+??????)
�
���=2�
2
�
2
�
2
��
�
���=4�
2
�
2
�
2
��
0
2??????/??????
cos
2
(��−��+??????)��
0
2??????/??????
��
Average value is equal to
1
2
Instantaneous Power
�=
��
��
=
��
��
∙
��
��
�=4�
2
�
2
�
2
��cos
2
(��−��+??????)
��
��
=�
2
�
2
�cos
2
(��−��+??????)
��
��
=�
�=
��
��
=
��
��
∙
��
��
Average Power
�
���=
���
��
�=4�
2
�
2
�
2
��cos
2
(��−��+??????)
�
���=2�
2
�
2
�
2
��
�
���=4�
2
�
2
�
2
��
0
2??????/??????
cos
2
(��−��+??????)��
0
2??????/??????
��
Average value is equal to
1
2
�
���∝�
2
⇒
�
�
�
??????
=
�
�
�
??????
2
=
3
2
2
⇒�
�=
9
4
�
??????
⇒�
�=
9
4
×0.2�
�
�=0.45�
Solution :
���∝�
2
⇒
�
�
�
??????
=
�
�
�
??????
2
=
3
2
2
⇒�
�=
9
4
�
??????
⇒�
�=
9
4
×0.2�
�
�=0.45�
Solution :
A uniform rope of linear mass density 6�/�is suspended from the ceiling and is kept under a
tension of 60�.
(a) Find the average rate at which the energy is transmitted across a given point onthe
string. [�=1��,�=200��]
(b) Find the total energy associated with the wave in a 2�long portion of the string.
(a)
�=
�
�
=
60
6×10
−3
=100�/�
�
���=2�
2
�
2
�
2
��
=2�
2
200
2
×10
−32
×6×10
−3
×100
Given: �=6×10
−3
��/�, T=60�, �=200��, �=10
−3
�, �=2�
�
���=48�
2
��
(b) <�>=<�>�
0
<�>=48�
2
×10
−3
×
2�
100�/�
<�>=96�
2
×10
−5
�
<�>=960�
2
��
To Find: aAverage rate of Energy bTotal Energy
Solution:
<�>=48�
2
×10
−3
×
�
�
tension of 60�.
(a) Find the average rate at which the energy is transmitted across a given point onthe
string. [�=1��,�=200��]
(b) Find the total energy associated with the wave in a 2�long portion of the string.
(a)
�=
�
�
=
60
6×10
−3
=100�/�
�
���=2�
2
�
2
�
2
��
=2�
2
200
2
×10
−32
×6×10
−3
×100
Given: �=6×10
−3
��/�, T=60�, �=200��, �=10
−3
�, �=2�
�
���=48�
2
��
(b) <�>=<�>�
0
<�>=48�
2
×10
−3
×
2�
100�/�
<�>=96�
2
×10
−5
�
<�>=960�
2
��
To Find: aAverage rate of Energy bTotal Energy
Solution:
<�>=48�
2
×10
−3
×
�
�
Intensity of Wave
�=2�
2
�
2
�
2
��
�=��
2
�
�=
�
���
����
�=
2�
2
�
2
�
2
��
��
2
Where �is density of medium
�=2�
2
�
2
�
2
��
�=��
2
�
�=
�
���
����
�=
2�
2
�
2
�
2
��
��
2
Where �is density of medium
Case 1and 2considered as of the same rope with amplitudes �and 2�,
respectively. Then comment on �
1and �
2.
�=2�
2
�
2
�
2
��
Case 1
�
1=�
�
1=�
�
1=�
�
1=2�
Case 2
�
2=2�
�
2=�
�
2=�
�
2=�
�
1
�
2
=
�
1
2
�
1
2
�
2
2
�
2
2
=
�×2�
2
2�
2
�
2
=1
�=
�
�
=��=��������
�
1�
1=�
2�
2
�
1�=�
2×2�
�
1=2�
2
Solution :
respectively. Then comment on �
1and �
2.
�=2�
2
�
2
�
2
��
Case 1
�
1=�
�
1=�
�
1=�
�
1=2�
Case 2
�
2=2�
�
2=�
�
2=�
�
2=�
�
1
�
2
=
�
1
2
�
1
2
�
2
2
�
2
2
=
�×2�
2
2�
2
�
2
=1
�=
�
�
=��=��������
�
1�
1=�
2�
2
�
1�=�
2×2�
�
1=2�
2
Solution :
Principle of superposition
Ԧ�
���=Ԧ�
1+Ԧ�
2+Ԧ�
3……+Ԧ�
�
Net displacement of the particle at
any point is equal to the vector sum of
the displacements due to the
individual waves.
⇒
Ԧ�
���=Ԧ�
1+Ԧ�
2
The pulses cancel out each other for a moment
Ԧ�
���=Ԧ�
1+Ԧ�
2+Ԧ�
3……+Ԧ�
�
Net displacement of the particle at
any point is equal to the vector sum of
the displacements due to the
individual waves.
⇒
Ԧ�
���=Ԧ�
1+Ԧ�
2
The pulses cancel out each other for a moment
�
??????→Amplitudeofincident wave
Medium 1 Medium 2
Incident Wave
Reflected Wave Transmitted Wave
�
??????
�
��
�
�
1
�
2
�
1
�
1 �
2
Boundary
�
�→Amplitude of reflected wave
�
�→Amplitude of transmitted wave
�
2>�
1�
1 �
2
??????→Amplitudeofincident wave
Medium 1 Medium 2
Incident Wave
Reflected Wave Transmitted Wave
�
??????
�
��
�
�
1
�
2
�
1
�
1 �
2
Boundary
�
�→Amplitude of reflected wave
�
�→Amplitude of transmitted wave
�
2>�
1�
1 �
2
�
1<�
2
Tension (??????) is same in both strings
⇒�
1=
�
�
1
,�
2=
�
�
2
⇒�
1>�
2
Rarer Medium to Denser Medium
�
�=
2�
2
�
1+�
2
�
??????
�
�=
�
2−�
1
�
1+�
2
�
??????
Since �
2<�
1⇒�
�<0and �
�<�
??????
Since �
2<�
1⇒�
�>0and �
�<�
??????
Also, since �
�<0,
Thus, reflection from a denser mediuminduces
an additional phase shift of �.
⇒�
�=−�
�sin�
1�+��=�
�sin�
1�+��+�
1<�
2
Tension (??????) is same in both strings
⇒�
1=
�
�
1
,�
2=
�
�
2
⇒�
1>�
2
Rarer Medium to Denser Medium
�
�=
2�
2
�
1+�
2
�
??????
�
�=
�
2−�
1
�
1+�
2
�
??????
Since �
2<�
1⇒�
�<0and �
�<�
??????
Since �
2<�
1⇒�
�>0and �
�<�
??????
Also, since �
�<0,
Thus, reflection from a denser mediuminduces
an additional phase shift of �.
⇒�
�=−�
�sin�
1�+��=�
�sin�
1�+��+�
Tension (??????) is same in both strings
�
1>�
2
Denser Medium to Rarer Medium
⇒�
1=
�
�
1
,�
2=
�
�
2
⇒�
1<�
2
Since �
1<�
2⇒�
�>0and �
�<�
??????
Since �
1<�
2⇒�
�>0and�
�>�
??????
Thus, there exists noadditional phase
shift of �when reflection occurs from a
rarer medium.
�
�=
2�
2
�
1+�
2
�
??????
�
�=
�
2−�
1
�
1+�
2
�
??????
�
1>�
2
Denser Medium to Rarer Medium
⇒�
1=
�
�
1
,�
2=
�
�
2
⇒�
1<�
2
Since �
1<�
2⇒�
�>0and �
�<�
??????
Since �
1<�
2⇒�
�>0and�
�>�
??????
Thus, there exists noadditional phase
shift of �when reflection occurs from a
rarer medium.
�
�=
2�
2
�
1+�
2
�
??????
�
�=
�
2−�
1
�
1+�
2
�
??????
Transmitted and Reflected Amplitudes
??????
??????=
??????
�−??????
�
??????
�+??????
�
??????
??????
Reflected
from
??????
??????=
�??????
�
??????
�+??????
�
??????
??????
Inversion of
reflected wave
Phase
Change in
reflected
wave
Denser
medium
Rarer
medium
Yes
No
−��
+��
+��
+��
�
0
|�
�|<�
??????
�
�<�
??????
�
�<�
??????
�
�>�
??????
??????
??????=
??????
�−??????
�
??????
�+??????
�
??????
??????
Reflected
from
??????
??????=
�??????
�
??????
�+??????
�
??????
??????
Inversion of
reflected wave
Phase
Change in
reflected
wave
Denser
medium
Rarer
medium
Yes
No
−��
+��
+��
+��
�
0
|�
�|<�
??????
�
�<�
??????
�
�<�
??????
�
�>�
??????
Twostringswith�
1=�and�
2=4�arejoinedtogether.Awavewithamplitude
�
??????travelsalongthestringwith�
1=�asshowntowardsthejunction.Findout
theamplitudeofreflectedandtransmittedwave.
�
�=
2�
2
�
1+�
2
�
??????=
2�
??????
3
Given:
Amplitude of wave =�
??????.
To Find:
�
�=?&�
�=?
Solution:
�
�=
�
2−�
1
�
1+�
2
�
??????=
−�
??????
3
1=�and�
2=4�arejoinedtogether.Awavewithamplitude
�
??????travelsalongthestringwith�
1=�asshowntowardsthejunction.Findout
theamplitudeofreflectedandtransmittedwave.
�
�=
2�
2
�
1+�
2
�
??????=
2�
??????
3
Given:
Amplitude of wave =�
??????.
To Find:
�
�=?&�
�=?
Solution:
�
�=
�
2−�
1
�
1+�
2
�
??????=
−�
??????
3
Reflection from a fixed end,
�
�=−�
??????
The wave shows a phase change of�without
changing its amplitude after reflection from a
fixed end.
Fixed
end
Fixed
end
�
�=−�
??????
The wave shows a phase change of�without
changing its amplitude after reflection from a
fixed end.
Fixed
end
Fixed
end
For free end,
�
�=�
??????
The wave pulseafter reflection
from a free end is not inverted.
�
�
�
�
1
�
1
�
??????
�
??????
Reflection of a Wave from a Free End
The shape of the resultant pulse during reflection can
be found by superimposingan imaginary pulse,
identical to the incident pulse but travelling in the
opposite direction, on the incident pulse.
�
�=�
??????
The wave pulseafter reflection
from a free end is not inverted.
�
�
�
�
1
�
1
�
??????
�
??????
Reflection of a Wave from a Free End
The shape of the resultant pulse during reflection can
be found by superimposingan imaginary pulse,
identical to the incident pulse but travelling in the
opposite direction, on the incident pulse.
A pulse shown here is reflected from the rigid wall �and
then from free end �. The shape of the string after these
two reflections will be
•The initial shape of the wave
pulse.
•The shape of the wave pulse
after the first reflection.
•The shape of the wave pulse
after the second reflection.
Solution :
then from free end �. The shape of the string after these
two reflections will be
•The initial shape of the wave
pulse.
•The shape of the wave pulse
after the first reflection.
•The shape of the wave pulse
after the second reflection.
Solution :
Standing Waves
�
�
1=�sin��−��
�
�
2=�sin��+��
Due to the superposition of these waves, standingwaves are formed.
Consider a string fixed at both ends. Two identical waves are traveling in
opposite directions such that the frequency, wavelength and amplitude of
the waves are the same.
�
�
1=�sin��−��
�
�
2=�sin��+��
Due to the superposition of these waves, standingwaves are formed.
Consider a string fixed at both ends. Two identical waves are traveling in
opposite directions such that the frequency, wavelength and amplitude of
the waves are the same.
�=2�sin(��)cos(��)
�
���=�
1+�
2
⇒�
���=�sin��−��+�sin��+��
⇒�
���=2�sin
��−��+��+��
2
cos
��−��−��+��
2
⇒�
���=2�sin��cos��∵cos−??????=cos??????
�
1=�sin��−��; �
2=�sin��+��
∵sin�+sin�=2sin
�+�
2
cos
�−�
2
Standing Waves
•Each particle executes SHM.
•Amplitude of particles executing SHMdepends upon the
location of the particles.
�=��cos(��)
��=2�sin��⇒Resultant Amplitude
�=��cos��⇒Equation of SHM
�
���=�
1+�
2
⇒�
���=�sin��−��+�sin��+��
⇒�
���=2�sin
��−��+��+��
2
cos
��−��−��+��
2
⇒�
���=2�sin��cos��∵cos−??????=cos??????
�
1=�sin��−��; �
2=�sin��+��
∵sin�+sin�=2sin
�+�
2
cos
�−�
2
Standing Waves
•Each particle executes SHM.
•Amplitude of particles executing SHMdepends upon the
location of the particles.
�=��cos(��)
��=2�sin��⇒Resultant Amplitude
�=��cos��⇒Equation of SHM
Locations of Maximum Amplitude -Antinodes
�
���=±2�
�=2�sin(��)cos(��)
�=
�
4
,
3�
4
,
5�
4
,
7�
4
,……
⇒sin(��)=±1
⇒��=
�
2
,
3�
2
,
5�
2
,……
⇒
2�
�
�=
�
2
,
3�
2
,
5�
2
,……
�
���=±2�
�=2�sin(��)cos(��)
�=
�
4
,
3�
4
,
5�
4
,
7�
4
,……
⇒sin(��)=±1
⇒��=
�
2
,
3�
2
,
5�
2
,……
⇒
2�
�
�=
�
2
,
3�
2
,
5�
2
,……
Locations of Minimum Amplitude -Nodes
�=2�sin(��)cos(��)
�
m??????�=0
�=0
−2�
2�
�
4
3�
4
�
2
�
�=0,
�
2
,�,
3�
2
,……
⇒sin��=0
⇒��=0,�,2�,3�,……
⇒
2�
�
�=0,�,2�,3�,……
�=2�sin(��)cos(��)
�
m??????�=0
�=0
−2�
2�
�
4
3�
4
�
2
�
�=0,
�
2
,�,
3�
2
,……
⇒sin��=0
⇒��=0,�,2�,3�,……
⇒
2�
�
�=0,�,2�,3�,……
Nodes and Antinodes
�=2�sin(��)cos(��)
Distancebetween2successiveNodes=
�
2
−2�
2�
�
�??????�=0(Nodes)
�
���=2�(Antiodes)
Distancebetween2successiveAntinodes=
�
2
�=2�sin(��)cos(��)
Distancebetween2successiveNodes=
�
2
−2�
2�
�
�??????�=0(Nodes)
�
���=2�(Antiodes)
Distancebetween2successiveAntinodes=
�
2
Modes of vibration in a string fixed at both ends
Fundamental Mode or First
Harmonic
Second Harmonic or First
Overtone
Third Harmonic or Second
Overtone
�
�ℎ
Harmonic or �−1
�ℎ
Overtone
Fundamental Mode or First
Harmonic
Second Harmonic or First
Overtone
Third Harmonic or Second
Overtone
�
�ℎ
Harmonic or �−1
�ℎ
Overtone
�=2�sin(��)cos(��)
�=0,cos�×0=1
�=
�
4
,cos
2�
�
×
�
4
=0
�=
�
2
,cos
2�
�
×
�
2
=−1
�=
3�
4
,cos
2�
�
×
3�
4
=0
�=�,cos
2�
�
�=1
Standing Waves
�=
3�
4
�=
�
4
−2�
2�
�=
�
2
�=�
−2�
2�
�=0
−2�
2�
�
4
3�
4
�
2
�
�=0,cos�×0=1
�=
�
4
,cos
2�
�
×
�
4
=0
�=
�
2
,cos
2�
�
×
�
2
=−1
�=
3�
4
,cos
2�
�
×
3�
4
=0
�=�,cos
2�
�
�=1
Standing Waves
�=
3�
4
�=
�
4
−2�
2�
�=
�
2
�=�
−2�
2�
�=0
−2�
2�
�
4
3�
4
�
2
�
Standing Waves Vs Travelling Waves
Travelling waves Standing waves
All particles of the medium oscillate with
the same frequencyand amplitude.
All particles except nodes oscillate with
the same frequency but different
amplitudes.
The phase difference between two
particles can be any value
between 0and 2�.
The phase difference between any two
particles is either 0or �.
These waves transmitenergy from one
point of the medium to the other.
These waves donottransmitenergy.
There is noinstantwhen all the particles
are at the meanpositionstogether.
All the particles cross their mean
positionstogether.
Travelling waves Standing waves
All particles of the medium oscillate with
the same frequencyand amplitude.
All particles except nodes oscillate with
the same frequency but different
amplitudes.
The phase difference between two
particles can be any value
between 0and 2�.
The phase difference between any two
particles is either 0or �.
These waves transmitenergy from one
point of the medium to the other.
These waves donottransmitenergy.
There is noinstantwhen all the particles
are at the meanpositionstogether.
All the particles cross their mean
positionstogether.
Three one-dimensional mechanical waves in an elastic medium are given as :
�
1=3�sin(��−��),�
2=�sin(��−��+�)and �
3=2�sin(��+��)
are superimposed with each other. The maximum displacement amplitude of the
medium particle would be :
Given :�
1=3�sin��−��;�
2=�sin(��−��+�);�
3=2�sin(��+��)
Here, �
2=�sin(��−��+�)=−�sin(��−��)
Now,�
1+�
2=3�sin��−��−�sin(��−��)
⇒�
1+�
2=2�sin��−��
As,y
1+y
2=2�sin(��−��)and�
3=2�sin(��+��)
Therefore, their combination (�
1+�
2)+�
3will produce a
standing wave whose amplitude is given by :
�
���=22�
�
���=4�
Solution :
�
1=3�sin(��−��),�
2=�sin(��−��+�)and �
3=2�sin(��+��)
are superimposed with each other. The maximum displacement amplitude of the
medium particle would be :
Given :�
1=3�sin��−��;�
2=�sin(��−��+�);�
3=2�sin(��+��)
Here, �
2=�sin(��−��+�)=−�sin(��−��)
Now,�
1+�
2=3�sin��−��−�sin(��−��)
⇒�
1+�
2=2�sin��−��
As,y
1+y
2=2�sin(��−��)and�
3=2�sin(��+��)
Therefore, their combination (�
1+�
2)+�
3will produce a
standing wave whose amplitude is given by :
�
���=22�
�
���=4�
Solution :
For a standing wave having equation�=2�sin��cos��, where �is in ��,
find the minimumdistance between two points which have an
amplitude �in magnitude.
Given : �=2�sin��cos��
�
??????=
�
12
,
5�
12
,
7�
12
,
11�
12
,…
At �=0
�=2�sin��=±�
⇒sin��=±
1
2
⇒sin
2�
�
�=
1
2
orsin
2�
�
�=−
1
2
2�
�
�
??????=
�
6
,�−
�
6
,�+
�
6
,2�−
�
6
,…
Where, ??????=1,2,3,4,…
These points have an amplitude of �
and the points 2and 3(and any such
pairs) have minimum distance.
The minimum distance is given by :
�
3−�
2=
7�
12
−
5�
12
=
�
6
�
3−�
2=
�
6
Solution :
find the minimumdistance between two points which have an
amplitude �in magnitude.
Given : �=2�sin��cos��
�
??????=
�
12
,
5�
12
,
7�
12
,
11�
12
,…
At �=0
�=2�sin��=±�
⇒sin��=±
1
2
⇒sin
2�
�
�=
1
2
orsin
2�
�
�=−
1
2
2�
�
�
??????=
�
6
,�−
�
6
,�+
�
6
,2�−
�
6
,…
Where, ??????=1,2,3,4,…
These points have an amplitude of �
and the points 2and 3(and any such
pairs) have minimum distance.
The minimum distance is given by :
�
3−�
2=
7�
12
−
5�
12
=
�
6
�
3−�
2=
�
6
Solution :
Normal Modes of a String Fixed at Both Ends
Fundamental Mode or First Harmonic
The minimum frequency at which it
can oscillate.
�
1=
�
�
1
=
�
2�
=
1
2�
�
�
�
1=
�
�
1
=
�
2�
=
1
2�
�
�
�
1:fundamentalfrequencyorfirstharmonic.
The distance between consecutive nodes
is λ/2
Second Harmonic or First Overtone
�
2=
�
�
2
=
2�
2�
=
2
2�
�
�
If we increase the frequency the
string may vibrate in two loops.
�
2=
�
�
2
=
2�
2�
=
2
2�
�
�
�
2∶secondharmonicorfirstovertone.
�
�ℎ
Harmonic or �−1
�ℎ
Overtone
Integral multiples of the
fundamental frequency are called
its harmonics.
�
�=
��
2�
=
�
2�
�
�
•In general,
�
�ℎ
harmonic=�−1
�ℎ
overtone
Or
�
�ℎ
overtone=�+1
�ℎ
harmonic
Fundamental Mode or First Harmonic
The minimum frequency at which it
can oscillate.
�
1=
�
�
1
=
�
2�
=
1
2�
�
�
�
1=
�
�
1
=
�
2�
=
1
2�
�
�
�
1:fundamentalfrequencyorfirstharmonic.
The distance between consecutive nodes
is λ/2
Second Harmonic or First Overtone
�
2=
�
�
2
=
2�
2�
=
2
2�
�
�
If we increase the frequency the
string may vibrate in two loops.
�
2=
�
�
2
=
2�
2�
=
2
2�
�
�
�
2∶secondharmonicorfirstovertone.
�
�ℎ
Harmonic or �−1
�ℎ
Overtone
Integral multiples of the
fundamental frequency are called
its harmonics.
�
�=
��
2�
=
�
2�
�
�
•In general,
�
�ℎ
harmonic=�−1
�ℎ
overtone
Or
�
�ℎ
overtone=�+1
�ℎ
harmonic
In normal modes of vibration of a string tied at both ends, the differencein
frequencies of fifth harmonic and second overtone is 54��. Calculate the
fundamental frequency of vibration?
�
5−�
3=54��
�
1=
�
2�
,�
3=
3�
2�
=3�
1,�
5=
5�
2�
=5�
1
5�
1−3�
1=54��
2�
1=54��
�
1=27��
5
�ℎ
harmonic
�
2
��
overtone
�
Solution :
frequencies of fifth harmonic and second overtone is 54��. Calculate the
fundamental frequency of vibration?
�
5−�
3=54��
�
1=
�
2�
,�
3=
3�
2�
=3�
1,�
5=
5�
2�
=5�
1
5�
1−3�
1=54��
2�
1=54��
�
1=27��
5
�ℎ
harmonic
�
2
��
overtone
�
Solution :
A uniform horizontal rod of length 40��and mass 1.2��is supported by two identical wires as
shown in figure. Where should a mass of 4.8��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 right into its first
overtone? Take (�=10�/�
2
)
Since vibrations in both wires are caused by the same source, �
1=�
2.
⇒1×
�
1
2�
1
=2×
�
2
2�
2
⇒
�
1
�
1
=2
�
2
�
2
Applying translational equilibrium condition,
⇒�
1=4�
2 1
�
1+�
2=��+4��
2
Solving equations and ,1 2�
1=4��,�
2=��
Applying rotational equilibrium condition,
ΣԦ??????
�=0
⇒4���+��
�
2
−(�
2)(�)=0
⇒4���+��
�
2
=��� ⇒4�=
�
2
=
40
2
=20 ⇒�=5��from the left end
shown in figure. Where should a mass of 4.8��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 right into its first
overtone? Take (�=10�/�
2
)
Since vibrations in both wires are caused by the same source, �
1=�
2.
⇒1×
�
1
2�
1
=2×
�
2
2�
2
⇒
�
1
�
1
=2
�
2
�
2
Applying translational equilibrium condition,
⇒�
1=4�
2 1
�
1+�
2=��+4��
2
Solving equations and ,1 2�
1=4��,�
2=��
Applying rotational equilibrium condition,
ΣԦ??????
�=0
⇒4���+��
�
2
−(�
2)(�)=0
⇒4���+��
�
2
=��� ⇒4�=
�
2
=
40
2
=20 ⇒�=5��from the left end
Sonometer
Bridges
Steel Wire
Load
Mass hanger
•A sonometer is practically used to verify the laws of transverse vibrations of a string.
•A string is attached from one end of the box and a load is hung at the other.
•There are two movable bridges in the box that create the nodes.
•The distance between the two bridges can be adjusted to change the vibrating length
of the wire.
Bridges
Steel Wire
Load
Mass hanger
•A sonometer is practically used to verify the laws of transverse vibrations of a string.
•A string is attached from one end of the box and a load is hung at the other.
•There are two movable bridges in the box that create the nodes.
•The distance between the two bridges can be adjusted to change the vibrating length
of the wire.
If �and �are constant
If �and �are constant
If �and �are constant
Laws of Transverse Vibrations of a String
�∝
1
�
(a) Law of Length
�∝�(b) Law of Tension
�∝
1
�
(c) Law of Mass
�
1
�
2
=
�
1
�
2
�
1
�
2
=
�
2
�
1
�
1
�
2
=
�
2
�
1
�=�
�
2�
=
�
2�
�
�
⇒�=
�
2�
�
�
If �and �are constant
If �and �are constant
Laws of Transverse Vibrations of a String
�∝
1
�
(a) Law of Length
�∝�(b) Law of Tension
�∝
1
�
(c) Law of Mass
�
1
�
2
=
�
1
�
2
�
1
�
2
=
�
2
�
1
�
1
�
2
=
�
2
�
1
�=�
�
2�
=
�
2�
�
�
⇒�=
�
2�
�
�
First Harmonic or Fundamental Frequency
�
1=
�
4�
=
1
4�
�
�
Free end -Antinode
Fixed end -Node
�=
�
1
4
�
1=4��
1=
�
�
1
�
1=
�
4�
=
1
4�
�
�
Normal Modes of a String Fixed at One End
First Overtone or Third Harmonic
�=
3�
3
4
�
3=
4�
3
�
3=
�
�
3
�
2=
3�
4�
=
3
4�
�
�
�
2=
3�
4�
=
3
4�
�
�
�
�ℎ
Overtone or 2�+1
�ℎ
Harmonic
�
�=
(2�+1)
4�
�
�
•In general,
�
�ℎ
overtone=2�+1
�ℎ
harmonic
•Only odd harmonics are present.
�
1∶�
3∶�
5∶………=1∶2∶3∶………
�
1=
�
4�
=
1
4�
�
�
Free end -Antinode
Fixed end -Node
�=
�
1
4
�
1=4��
1=
�
�
1
�
1=
�
4�
=
1
4�
�
�
Normal Modes of a String Fixed at One End
First Overtone or Third Harmonic
�=
3�
3
4
�
3=
4�
3
�
3=
�
�
3
�
2=
3�
4�
=
3
4�
�
�
�
2=
3�
4�
=
3
4�
�
�
�
�ℎ
Overtone or 2�+1
�ℎ
Harmonic
�
�=
(2�+1)
4�
�
�
•In general,
�
�ℎ
overtone=2�+1
�ℎ
harmonic
•Only odd harmonics are present.
�
1∶�
3∶�
5∶………=1∶2∶3∶………
A 2�long rope, having a mass of 80�,is fixed at one end and is tied to a light
string at the other end. The tension in the string is 256�.
a)Find the frequencies of the fundamental and the first two overtones.
b)Find the wavelength in the fundamental and the first two overtones.
2�
string at the other end. The tension in the string is 256�.
a)Find the frequencies of the fundamental and the first two overtones.
b)Find the wavelength in the fundamental and the first two overtones.
2�
Linear mass density, �=
�
�
=
80
2×1000
=
1
25
��/�
a)Find the frequencies of the fundamental and the first two overtones.
Wave velocity, v=
??????
�
=
256
(1/25)
=80�/�
Fundamental frequency, �
0=
??????
4�
=
80
4×2
⇒�
0=10��
First overtone, �
1=3�
0⇒�
1=30��
Second overtone, �
2=5�
0⇒�
2=50��
Given: �=2�,�=80�,�=256�
2�
��Fixed end free end
Given: �=2�,�=80�,�=256�
b) Find the wavelength in the fundamental and the first two overtones.
In the fundamental frequency, �=
�
4
⇒�=8�
In the first overtone, �=
3�
1
4
⇒�
1=
8
3
�=2.67�
In the second overtone, �=
5�
2
4
⇒�
2=
8
5
�=1.6�
�
�
=
80
2×1000
=
1
25
��/�
a)Find the frequencies of the fundamental and the first two overtones.
Wave velocity, v=
??????
�
=
256
(1/25)
=80�/�
Fundamental frequency, �
0=
??????
4�
=
80
4×2
⇒�
0=10��
First overtone, �
1=3�
0⇒�
1=30��
Second overtone, �
2=5�
0⇒�
2=50��
Given: �=2�,�=80�,�=256�
2�
��Fixed end free end
Given: �=2�,�=80�,�=256�
b) Find the wavelength in the fundamental and the first two overtones.
In the fundamental frequency, �=
�
4
⇒�=8�
In the first overtone, �=
3�
1
4
⇒�
1=
8
3
�=2.67�
In the second overtone, �=
5�
2
4
⇒�
2=
8
5
�=1.6�
Sound Wave
Sound is a mechanical, longitudinal, 3D wave.
•Mechanical wave
•Longitudinal wave
Ԧ�
�∥Ԧ�
�Ԧ�
�×Ԧ�
�=0
Sound is a mechanical, longitudinal, 3D wave.
•Mechanical wave
•Longitudinal wave
Ԧ�
�∥Ԧ�
�Ԧ�
�×Ԧ�
�=0
Sound Wave
�
�
Displacement Equation
�=0 �=�
�=�
0sin(��−��)
For sound waves, the oscillation is parallel to the direction the wave travels.
For a wave traveling in the +�direction, the displacement of the particles in the
medium in the x-direction can be described by:
�
Displacement Equation
�=0 �=�
�=�
0sin(��−��)
For sound waves, the oscillation is parallel to the direction the wave travels.
For a wave traveling in the +�direction, the displacement of the particles in the
medium in the x-direction can be described by:
Bulk Modulus
�
�
�
�
�
�
�
�
�
�
�
�
�
??????
�
out
�=−
∆�
∆�/�
The reciprocalof the bulk modulus is called compressibility.
It is defined as the ratio of the volumetric stress to the volume strain.
�=
Volumetricstress
Volumestrain
�
�
�
�
�
�
�
�
�
�
�
�
�
??????
�
out
�=−
∆�
∆�/�
The reciprocalof the bulk modulus is called compressibility.
It is defined as the ratio of the volumetric stress to the volume strain.
�=
Volumetricstress
Volumestrain
�=
−∆�
∆�/�
⇒∆�=−�
∆�
�
∆�=�
�−�
??????=���+��−���=���
�
??????=�=���
∆�=−�
∆�
�
=−�
���
���
=−�
��
��
Excess pressure =−�×slope of �−�curve
��
��
=�
0−�cos��−��=−�
0�cos��−��
∆�=���
0cos��−��
Excess pressure amplitude
��+��
�
�=�
0sin��−��
�+���
Pressure Wave
−∆�
∆�/�
⇒∆�=−�
∆�
�
∆�=�
�−�
??????=���+��−���=���
�
??????=�=���
∆�=−�
∆�
�
=−�
���
���
=−�
��
��
Excess pressure =−�×slope of �−�curve
��
��
=�
0−�cos��−��=−�
0�cos��−��
∆�=���
0cos��−��
Excess pressure amplitude
��+��
�
�=�
0sin��−��
�+���
Pressure Wave
Pressure Wave and Displacement Wave
�=�
0sin(��−��)
??????�=���
0sin��−��+
�
2
⇒??????�=���
0cos��−��
There’s always a phase
difference of
??????
2
between
pressurewave and
displacement wave.
�
���=�
0+∆�
0
Excess pressure amplitude=���
0=∆�
0
�
�??????�=�
0−∆�
0
Pressure oscillates between �
���and �
�??????�.
�=�
0sin(��−��)
??????�=���
0sin��−��+
�
2
⇒??????�=���
0cos��−��
There’s always a phase
difference of
??????
2
between
pressurewave and
displacement wave.
�
���=�
0+∆�
0
Excess pressure amplitude=���
0=∆�
0
�
�??????�=�
0−∆�
0
Pressure oscillates between �
���and �
�??????�.
A sound wave of wavelength 40��travels in air. If the difference between the
maximum and minimum pressures at a given point is 1.0×10
−3
�/�
2
, find the
amplitude of vibration of the particles of the medium. The bulk modulus of air is
1.4×10
5
�/�
2
.
�
���=�
0+∆�
0 �
�??????�=�
0−∆�
0
�
���−�
�??????�=2∆�
0=1.0×10
−3
�/�
2
∆�
0=0.5×10
−3
�/�
2
Excess pressure amplitude=∆�
0=���
0
�
0=
∆�
0
��
=
∆�
0�
2��
=
0.5×10
−3
�/�
2
×40×10
−2
�
2×3.14×1.4×10
5
�/�
2
�
0=2.2×10
−10
�
�
0=2.2×10
−10
�
�=40���=1.4×10
5
�/�
2
�
���−�
�??????�=1.0×10
−3
�/�
2
Solution :
maximum and minimum pressures at a given point is 1.0×10
−3
�/�
2
, find the
amplitude of vibration of the particles of the medium. The bulk modulus of air is
1.4×10
5
�/�
2
.
�
���=�
0+∆�
0 �
�??????�=�
0−∆�
0
�
���−�
�??????�=2∆�
0=1.0×10
−3
�/�
2
∆�
0=0.5×10
−3
�/�
2
Excess pressure amplitude=∆�
0=���
0
�
0=
∆�
0
��
=
∆�
0�
2��
=
0.5×10
−3
�/�
2
×40×10
−2
�
2×3.14×1.4×10
5
�/�
2
�
0=2.2×10
−10
�
�
0=2.2×10
−10
�
�=40���=1.4×10
5
�/�
2
�
���−�
�??????�=1.0×10
−3
�/�
2
Solution :
�=
�����������
����??????��
=
�
�
In a longitudinal wave, the constituents of the medium oscillate forward and
backward along the direction of propagation of the wave.
Hence,
Whenalongitudinalwavetravelsthroughgases,thecompression
andrarefactionsbringaboutchangesinvolumeoccupiedbythe
gasandhencethegasundergoesvolumetric/bulkstrain
�:bulk modulus
�:mass density
(�)=
∆�
∆�/�
Bulk modulus
Speed of longitudinal waves in gases
�����������
����??????��
=
�
�
In a longitudinal wave, the constituents of the medium oscillate forward and
backward along the direction of propagation of the wave.
Hence,
Whenalongitudinalwavetravelsthroughgases,thecompression
andrarefactionsbringaboutchangesinvolumeoccupiedbythe
gasandhencethegasundergoesvolumetric/bulkstrain
�:bulk modulus
�:mass density
(�)=
∆�
∆�/�
Bulk modulus
Speed of longitudinal waves in gases
A
lateral expansion is negligible
B
�
??????=
??????
�
Thus, the speed of longitudinal waves in a solid rod is given by
For a linear medium, like a solid rod, the lateral expansion of the bar is
negligible and we may consider it to be only under longitudinal strain.
??????:Young modulus of the material of the rod.
�:Density of the material of the rod.
Speedof longitudinal waves in a rod
lateral expansion is negligible
B
�
??????=
??????
�
Thus, the speed of longitudinal waves in a solid rod is given by
For a linear medium, like a solid rod, the lateral expansion of the bar is
negligible and we may consider it to be only under longitudinal strain.
??????:Young modulus of the material of the rod.
�:Density of the material of the rod.
Speedof longitudinal waves in a rod
•Assumption: Newton assumed that the sound
propagation is an isothermal process .
⇒∆�=0
•Speed of sound wave is ,
�=
�
??????���ℎ�����
�
⇒�=
�
�
=
��
�
For air medium at 273.15�
�=1.01×10
5
��
�=1.29��/�
3
Newton’s Prediction
∴�=
�
�
=
1.01×10
5
��
1.29��/�
3
≈280��
−1
�
���=331�/�
�
���≠�(A discrepancy close to50��
−1
)But
propagation is an isothermal process .
⇒∆�=0
•Speed of sound wave is ,
�=
�
??????���ℎ�����
�
⇒�=
�
�
=
��
�
For air medium at 273.15�
�=1.01×10
5
��
�=1.29��/�
3
Newton’s Prediction
∴�=
�
�
=
1.01×10
5
��
1.29��/�
3
≈280��
−1
�
���=331�/�
�
���≠�(A discrepancy close to50��
−1
)But
Laplace’s correction
•Formation of compressions and rarefactions is an adiabatic process.
⇒∆�=0
•Speed of sound wave is ,
�=
�
��??????����??????�
�
⇒�=
��
�
=
���
�
For air medium at 273.15�
�=1.01×10
5
��
�=1.29��/�
3
�≈1.4
∴�=
��
�
=
1.4×1.01×10
5
��
1.29��/�
3
≈331.3�/�
�
���=331�/�
�
���≈�(Excellent agreement with the experimental value)
•Formation of compressions and rarefactions is an adiabatic process.
⇒∆�=0
•Speed of sound wave is ,
�=
�
��??????����??????�
�
⇒�=
��
�
=
���
�
For air medium at 273.15�
�=1.01×10
5
��
�=1.29��/�
3
�≈1.4
∴�=
��
�
=
1.4×1.01×10
5
��
1.29��/�
3
≈331.3�/�
�
���=331�/�
�
���≈�(Excellent agreement with the experimental value)
•Effect of temperature : As temperature increases, speed of sound increases.
�
�=
���
�
�∝�
⇒
�=
��
�
=
���
�
•Effect of Pressure : Speed of sound does not vary on varying the pressure or
density of the medium if the temperature is kept constant.
Factors affecting the speedof sound
•Effect of Humidity :As humidityincreases, speedof sound increases.
�
�=
��
�
�
ℎ��??????��??????�>�
����??????�•At same pressure, �
ℎ��??????��??????�<�
����??????�⇒
�
�=
���
�
�∝�
⇒
�=
��
�
=
���
�
•Effect of Pressure : Speed of sound does not vary on varying the pressure or
density of the medium if the temperature is kept constant.
Factors affecting the speedof sound
•Effect of Humidity :As humidityincreases, speedof sound increases.
�
�=
��
�
�
ℎ��??????��??????�>�
����??????�•At same pressure, �
ℎ��??????��??????�<�
����??????�⇒
The value of �for oxygen as well as for hydrogen is 1.40. If the speed of sound in
oxygen is 470�/�, what will be the speed in hydrogen at the same temperature
and pressure?
�
�=
���
�
(�)
�2
=
���
�
02
(�)
??????2
=
���
�
??????2
Hence,
(�)
�2
(�)
??????2
=
�
??????2
�
�2
⇒
(�)
�2
(�)
??????2
=
2
32
=
1
4
⇒(�)
??????2
=4×(�)
�2
⇒(�)
??????2
=4×470 ⇒(�)
??????2
=1880��
−1
Solution :
oxygen is 470�/�, what will be the speed in hydrogen at the same temperature
and pressure?
�
�=
���
�
(�)
�2
=
���
�
02
(�)
??????2
=
���
�
??????2
Hence,
(�)
�2
(�)
??????2
=
�
??????2
�
�2
⇒
(�)
�2
(�)
??????2
=
2
32
=
1
4
⇒(�)
??????2
=4×(�)
�2
⇒(�)
??????2
=4×470 ⇒(�)
??????2
=1880��
−1
Solution :
Intensity of a sound wave
��
Power transmitted by the wave across the
cross-section �
For sound:
�=2�
2
�
2
�
2
��
�=2�
2
�
�
2
�
2
��
�=
∆�
�
2
�
2�
��
Power transmitted by the wave across the
cross-section �
For sound:
�=2�
2
�
2
�
2
��
�=2�
2
�
�
2
�
2
��
�=
∆�
�
2
�
2�
Introduction to Beats
The phenomenon of periodic variation of intensity of sound when two sound waves
of slightly different frequencies superpose, is called beats.
�
1+�
2
Beats
WavesIn phase Waves out of phase
In-coherent waves
�
1
�
2
The phenomenon of periodic variation of intensity of sound when two sound waves
of slightly different frequencies superpose, is called beats.
�
1+�
2
Beats
WavesIn phase Waves out of phase
In-coherent waves
�
1
�
2
Theory of beats
�
2
�
1
�
1+�
2
Practically , a beat can be heard only if the number of maxima
formed in 1secare less than 10.
�
2
�
1
�
1+�
2
Practically , a beat can be heard only if the number of maxima
formed in 1secare less than 10.
Theory of beats
For a maxima to occur , ∆??????should be an even multiple of �??????.�2��.
�
2
�
1sin(�
1�+��)
�
1
�
2sin(�
2�+��)
�
1
�
2
∆??????=�
2−�
1�
�
∆??????=0
⇒�
2−�
1�
1=0
⇒�
1=0
Similarly, for the second and the third maxima,
�
2−�
1�
2=2�
⇒�
2=
2�
�
2−�
1
�
2−�
1�
3=4�
⇒�
3=
4�
�
2−�
1
(for first maxima)
,(�
1=�
2=�)
Time interval between maxima is constant and is given by ,
�
�����=
2�
�
2−�
1
=
1
�
2−�
1
Beat frequency
�
�����=
�
2−�
1
2�
=�
2−�
1
For a maxima to occur , ∆??????should be an even multiple of �??????.�2��.
�
2
�
1sin(�
1�+��)
�
1
�
2sin(�
2�+��)
�
1
�
2
∆??????=�
2−�
1�
�
∆??????=0
⇒�
2−�
1�
1=0
⇒�
1=0
Similarly, for the second and the third maxima,
�
2−�
1�
2=2�
⇒�
2=
2�
�
2−�
1
�
2−�
1�
3=4�
⇒�
3=
4�
�
2−�
1
(for first maxima)
,(�
1=�
2=�)
Time interval between maxima is constant and is given by ,
�
�����=
2�
�
2−�
1
=
1
�
2−�
1
Beat frequency
�
�����=
�
2−�
1
2�
=�
2−�
1
Beat diagram
Resultant displacement of the
superimposed wave is,
�=�
1+�
2=�sin�
1�+�sin�
2�
⇒�=�(sin�
1�+sin�
2�)
�
Minimum
intensity
Maximum
intensity
�
�
Constructive
interference
Destructive
interference
�
−�
−2�
�
2�
�=2�sin2π
�
1+�
2
2
�cos2π
�
1−�
2
2
�
Resultant displacement of the
superimposed wave is,
�=�
1+�
2=�sin�
1�+�sin�
2�
⇒�=�(sin�
1�+sin�
2�)
�
Minimum
intensity
Maximum
intensity
�
�
Constructive
interference
Destructive
interference
�
−�
−2�
�
2�
�=2�sin2π
�
1+�
2
2
�cos2π
�
1−�
2
2
�
�
A=500���
�����=5���
�=???
�
�−�
�=5
⇒�
�=�
�±5
⇒�
�=500±5
∴�
�=505��or 495��
T
Atuningfork�offrequency�=500��produces5beatspersecondwithanother
tunningfork�.Iffilingisdoneonthetunningfork�,thebeatfrequencybecomes
8.Findthefrequencyof�.
As per the question, upon filing the prongs of tuning fork�, its frequency
slightly increases.
If the frequency of tuning fork �is 495��, then upon filing, it increases by a few ��(say 497��). Therefore, the beat
frequency becomes 3��(decreases). Instead, it increases to 8��according to data.
Hence, logically, the appropriate frequency of tuning fork B should be 505��to fulfil the given criteria.
Solution :
A=500���
�����=5���
�=???
�
�−�
�=5
⇒�
�=�
�±5
⇒�
�=500±5
∴�
�=505��or 495��
T
Atuningfork�offrequency�=500��produces5beatspersecondwithanother
tunningfork�.Iffilingisdoneonthetunningfork�,thebeatfrequencybecomes
8.Findthefrequencyof�.
As per the question, upon filing the prongs of tuning fork�, its frequency
slightly increases.
If the frequency of tuning fork �is 495��, then upon filing, it increases by a few ��(say 497��). Therefore, the beat
frequency becomes 3��(decreases). Instead, it increases to 8��according to data.
Hence, logically, the appropriate frequency of tuning fork B should be 505��to fulfil the given criteria.
Solution :
�=195��
5�����/������areheardwhenaturningforkissoundedwithasonometerwire
undertension,whenthelengthofthesonometerwireiseither0.95�or1�.
Thefrequencyoftheforkwillbe
Number of beats per second heard = 5
∴
�+5
�−5
=
100
95
⇒95�+5=100(�−5)
Or, 95�+475=100�−500
Or, 5�=975
Or, �=
975
5
T
JEE Mains 2018
�
1=0.95��
2=1�
∵�=
�
2�
�
2�
1
−�=5⇒
�
2�
1
=�+5
�−
�
2�
2
=5 ⇒
�
2�
2
=�−5
……(1)
……(2)
Dividing Eq. (1)by Eq. (2), we get
Frequency of fork be �
Solution :
5�����/������areheardwhenaturningforkissoundedwithasonometerwire
undertension,whenthelengthofthesonometerwireiseither0.95�or1�.
Thefrequencyoftheforkwillbe
Number of beats per second heard = 5
∴
�+5
�−5
=
100
95
⇒95�+5=100(�−5)
Or, 95�+475=100�−500
Or, 5�=975
Or, �=
975
5
T
JEE Mains 2018
�
1=0.95��
2=1�
∵�=
�
2�
�
2�
1
−�=5⇒
�
2�
1
=�+5
�−
�
2�
2
=5 ⇒
�
2�
2
=�−5
……(1)
……(2)
Dividing Eq. (1)by Eq. (2), we get
Frequency of fork be �
Solution :
Closed Organ Pipe
•A compression pulse reflects as a compression pulse and a rarefaction pulse reflects as a rarefaction pulse
•Reflected pressure wave will have no phase change, i.e., no inversion.
At the closed end
•In the case of a closed organ pipe of length �, there will be a pressure
node (displacement antinode) at the open end
•A displacement node (pressure antinode) will be formed at the closed
end.
•A compression pulse reflects as a compression pulse and a rarefaction pulse reflects as a rarefaction pulse
•Reflected pressure wave will have no phase change, i.e., no inversion.
At the closed end
•In the case of a closed organ pipe of length �, there will be a pressure
node (displacement antinode) at the open end
•A displacement node (pressure antinode) will be formed at the closed
end.
Modes of vibration in a closed organ pipe
�
0=
1�
�
4�
�
1=
3�
�
4�
�
2=
5�
�
4�
Fundamental Frequency
Firstovertone/3
rd
Harmonic
Secondovertone/5
th
Harmonic
For�
�ℎ
Overtone/(2�+1)Harmonic:
�
�=2�+1
�
�
4�
�=0,1,2,…
�
0=
1�
�
4�
�
1=
3�
�
4�
�
2=
5�
�
4�
Fundamental Frequency
Firstovertone/3
rd
Harmonic
Secondovertone/5
th
Harmonic
For�
�ℎ
Overtone/(2�+1)Harmonic:
�
�=2�+1
�
�
4�
�=0,1,2,…
Aclosedorganpipehaafundamentalfrequencyof1.5���.Thenumberof
overtonesthatcanbedistinctlyheardbyapersonwiththisorganpipewillbe:
(Assumethatthehighestfrequencyapersoncanhearis20,000��)
�
�=2�+1�
�
�=6
Modes of vibration in a closed organ pipe for �
�ℎ
Overtone :
�
�=2�+1
�
�
4�
�=0,1,2,…
�
�=1.5���,The maximum frequency human can hear �
�=20,000��Given:
20000=2�+1×1500
�=6.1
overtonesthatcanbedistinctlyheardbyapersonwiththisorganpipewillbe:
(Assumethatthehighestfrequencyapersoncanhearis20,000��)
�
�=2�+1�
�
�=6
Modes of vibration in a closed organ pipe for �
�ℎ
Overtone :
�
�=2�+1
�
�
4�
�=0,1,2,…
�
�=1.5���,The maximum frequency human can hear �
�=20,000��Given:
20000=2�+1×1500
�=6.1
Open Organ Pipe
•A compression pulse reflects as a rarefaction pulse and a rarefaction pulse reflects as a compression pulse
•Reflected pressure wave will have a phase change of �.
At the open end
There will be formation of pressure node or a displacement anti-node
at the open ends.
•A compression pulse reflects as a rarefaction pulse and a rarefaction pulse reflects as a compression pulse
•Reflected pressure wave will have a phase change of �.
At the open end
There will be formation of pressure node or a displacement anti-node
at the open ends.
Modes of vibration in an open organ pipe
�
0=1
�
�
2�
�
1=2
�
�
2�
Fundamental Frequency
Firstovertone/2
nd
Harmonic
Secondovertone/3
rd
Harmonic
�
2=3
�
�
2�
For�
�ℎ
Overtone/�+1
�ℎ
Harmonic:
�
�=�+1
�
�
2�
�=0,1,2,…
�
0=1
�
�
2�
�
1=2
�
�
2�
Fundamental Frequency
Firstovertone/2
nd
Harmonic
Secondovertone/3
rd
Harmonic
�
2=3
�
�
2�
For�
�ℎ
Overtone/�+1
�ℎ
Harmonic:
�
�=�+1
�
�
2�
�=0,1,2,…
The fundamental frequencyof a closed organ pipe of length 20��is equal to the
second overtone of an open organ pipe. The lengthof the open organ pipe is
Fundamental frequency of vibration of the closed organ pipe
�
�=
�
4�
1
=
�
4×20
The frequency of vibration in �
�ℎ
overtone for the open
organ pipe
�
�=(�+1)
�
2�
For the second overtone of the open organ pipe
�
2=2+1
�
2�
=
3�
2�
�
�=�
2
Given that,
�
80
=
3�
2�
�=120��
Solution :
second overtone of an open organ pipe. The lengthof the open organ pipe is
Fundamental frequency of vibration of the closed organ pipe
�
�=
�
4�
1
=
�
4×20
The frequency of vibration in �
�ℎ
overtone for the open
organ pipe
�
�=(�+1)
�
2�
For the second overtone of the open organ pipe
�
2=2+1
�
2�
=
3�
2�
�
�=�
2
Given that,
�
80
=
3�
2�
�=120��
Solution :
�=0.6��is the radius of organ pipe
�
4
=�+�
�
�=
2�+1�
�
4�+0.6�
Thus, in general,
End correction in closed pipe End correction in open pipe
�
2
=�+2�
�
�=
�+1�
�
2�+1.2�
Thus, in general,
�is the radius of organ pipe�=0.6�
�
4
=�+�
�
�=
2�+1�
�
4�+0.6�
Thus, in general,
End correction in closed pipe End correction in open pipe
�
2
=�+2�
�
�=
�+1�
�
2�+1.2�
Thus, in general,
�is the radius of organ pipe�=0.6�
Introduction to Doppler effect
The apparent change in the frequency of sound due to the relative motion
between the source and the listener is known as Doppler effect.
�
�
�
Example: A passenger standing on a railway platform can hear a loud
and high-pitched sound of the horn of a train as it approaches them.
The apparent change in the frequency of sound due to the relative motion
between the source and the listener is known as Doppler effect.
�
�
�
Example: A passenger standing on a railway platform can hear a loud
and high-pitched sound of the horn of a train as it approaches them.
�
′
=
�+�
�
�
�
0
� �
�
�
�
′
=
�−�
�
�
�
0
�
�
� �
�
�
�
′
=
�+�
�cos??????
�
�
0
� �??????
�
�
�
′
=
�−�
�cos??????
�
�
0
�
�
??????
Doppler effect: The listener alone is moving
′
=
�+�
�
�
�
0
� �
�
�
�
′
=
�−�
�
�
�
0
�
�
� �
�
�
�
′
=
�+�
�cos??????
�
�
0
� �??????
�
�
�
′
=
�−�
�cos??????
�
�
0
�
�
??????
Doppler effect: The listener alone is moving
�
′
=
�
�−�
�
�
0
� �
�
�
�
�
�
′
=
�
�+�
�
�
0
� �
�
�
�
′
=
�
�−�
�cos??????
�
0
??????
� �
�
′
=
�
�+�
�cos??????
�
0
�
�
??????
�
�
Doppler effect: Source alone is moving
′
=
�
�−�
�
�
0
� �
�
�
�
�
�
′
=
�
�+�
�
�
0
� �
�
�
�
′
=
�
�−�
�cos??????
�
0
??????
� �
�
′
=
�
�+�
�cos??????
�
0
�
�
??????
�
�
Doppler effect: Source alone is moving
Asoundsource�ismovingalongastraighttrackwithspeed�,andisemitting
soundoffrequency�
�(seefigure).Anobserverisstandingatafinitedistance,at
thepoint�,fromthetrack.Thetimevariationoffrequencyheardbythe
observerisbestrepresentedby:(�
�representstheinstantwhenthedistance
betweenthesourceandobserverisminimum)
T
A C
DB
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
��
�
�
�
�
�
�
�
soundoffrequency�
�(seefigure).Anobserverisstandingatafinitedistance,at
thepoint�,fromthetrack.Thetimevariationoffrequencyheardbythe
observerisbestrepresentedby:(�
�representstheinstantwhenthedistance
betweenthesourceandobserverisminimum)
T
A C
DB
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
��
�
�
�
�
�
�
�
�=
�
�+�
�cos??????
�
0
�=
�
�−�
�cos??????
�
0
�
��
�
??????
�cos??????
�
??????↑cos??????↓�↓
�
�
??????
�cos??????
�
??????↓cos??????↑�↓
While approaching:
While receding:
When the source is approaching the frequency
heard by the observer:
When the source is receding the frequency heard
by the observer:
�
�
�
�+�
�cos??????
�
0
�=
�
�−�
�cos??????
�
0
�
��
�
??????
�cos??????
�
??????↑cos??????↓�↓
�
�
??????
�cos??????
�
??????↓cos??????↑�↓
While approaching:
While receding:
When the source is approaching the frequency
heard by the observer:
When the source is receding the frequency heard
by the observer:
�
�
T
A C
DB
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
��
�
�
�
�
�
�
�
Asoundsource�ismovingalongastraighttrackwithspeed�,andisemitting
soundoffrequency�
�(seefigure).Anobserverisstandingatafinitedistance,at
thepoint�,fromthetrack.Thetimevariationoffrequencyheardbythe
observerisbestrepresentedby:(�
�representstheinstantwhenthedistance
betweenthesourceandobserverisminimum)
A C
DB
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
��
�
�
�
�
�
�
�
Asoundsource�ismovingalongastraighttrackwithspeed�,andisemitting
soundoffrequency�
�(seefigure).Anobserverisstandingatafinitedistance,at
thepoint�,fromthetrack.Thetimevariationoffrequencyheardbythe
observerisbestrepresentedby:(�
�representstheinstantwhenthedistance
betweenthesourceandobserverisminimum)
Doppler Effect for Moving Medium
�
��
�
�
�
�
′
=
�±�
�±�
�
�±�
�∓�
�
�
0
•If the wind blows from source
to observer,
�→�+�
�
•If the wind blows from
observer to source,
�→�−�
�
�
��
�
�
�
�
′
=
�±�
�±�
�
�±�
�∓�
�
�
0
•If the wind blows from source
to observer,
�→�+�
�
•If the wind blows from
observer to source,
�→�−�
�
An observer is at rest, and a car having a sound source �of frequency �approaches
the observer �with velocity �
�. After crossing, it recedes away with the same velocity.
Find the change in the frequency of the sound heard by the observer when the car
was moving towards the observer and receding away from the observer, if the speed
of the sound is �.
The difference in frequency
(or beats) experienced by the
observeris given by,
�
′
−�
′′
=�
��
1
�−�
�
−
1
�+�
�
��
�
�
� �
�
�
�
′
=
�
�−�
�
�
�
�
′′
=
�
�+�
�
�
�
Approaching
Receding
�
′
−�
′′
=��
�
�+�
�−�−�
�
�
2
−�
�
2
�
′
−�
′′
=
2��
��
�
�
2
−�
�
2
Solution :
the observer �with velocity �
�. After crossing, it recedes away with the same velocity.
Find the change in the frequency of the sound heard by the observer when the car
was moving towards the observer and receding away from the observer, if the speed
of the sound is �.
The difference in frequency
(or beats) experienced by the
observeris given by,
�
′
−�
′′
=�
��
1
�−�
�
−
1
�+�
�
��
�
�
� �
�
�
�
′
=
�
�−�
�
�
�
�
′′
=
�
�+�
�
�
�
Approaching
Receding
�
′
−�
′′
=��
�
�+�
�−�−�
�
�
2
−�
�
2
�
′
−�
′′
=
2��
��
�
�
2
−�
�
2
Solution :
A car moving with a speed of 50��
−1
towards the east blows the horn at
frequency �. A truck is moving at a speed of 100��
−1
towards the south as
shown in the figure. If the velocity of sound is �, find out the frequency at
which the truck driver hears the horn.
Component of the velocity of the cartowards the
truck,
Component of the velocity of the truckaway from the
car,
�
�=50cos37°=40��
−1
�
�=100cos53°=60��
−1
Theapparent frequency heard by the truck driver,
�
�=�
�
�−�
�
�−�
�
=�
�−60
�−40
�
�=�
�−60
�−40
Solution :
−1
towards the east blows the horn at
frequency �. A truck is moving at a speed of 100��
−1
towards the south as
shown in the figure. If the velocity of sound is �, find out the frequency at
which the truck driver hears the horn.
Component of the velocity of the cartowards the
truck,
Component of the velocity of the truckaway from the
car,
�
�=50cos37°=40��
−1
�
�=100cos53°=60��
−1
Theapparent frequency heard by the truck driver,
�
�=�
�
�−�
�
�−�
�
=�
�−60
�−40
�
�=�
�−60
�−40
Solution :
Coherent Sources
Coherent Sources
•Same frequency
•Same wavelength
•Constant Phase difference
•Same medium
Coherent Sources
Coherent Sources
•Same frequency
•Same wavelength
•Constant Phase difference
•Same medium
Coherent Sources
Interference of Waves
Interference of wavesis a phenomenon in which
twocoherent wavessuperpose to form a
resultantwaveof greater, lower, or the same
amplitude.
Interference of wavesis a phenomenon in which
twocoherent wavessuperpose to form a
resultantwaveof greater, lower, or the same
amplitude.
Resultant Amplitude
�
���=�
1
2
+�
2
2
+2�
1�
2cos∆??????
∆??????=
2�
�
Δ�+(??????
2−??????
1)
??????
2=??????
1⇒Generally,
∆??????=
2�
�
Δ�
�
2
�
1
∆??????
�
���
�
���
2
=�
1
2
+�
2
2
+2�
1�
2cos∆??????
�
���=�
1
2
+�
2
2
+2�
1�
2cos∆??????
�
1=�
1sin��−��
1+??????
1�
2=�
2sin��−��
2+??????
2
�
�
���=�
1
2
+�
2
2
+2�
1�
2cos∆??????
∆??????=
2�
�
Δ�+(??????
2−??????
1)
??????
2=??????
1⇒Generally,
∆??????=
2�
�
Δ�
�
2
�
1
∆??????
�
���
�
���
2
=�
1
2
+�
2
2
+2�
1�
2cos∆??????
�
���=�
1
2
+�
2
2
+2�
1�
2cos∆??????
�
1=�
1sin��−��
1+??????
1�
2=�
2sin��−��
2+??????
2
�
Resultant Intensity
As, �∝�
2
⇒�=��
2
Now, �
���
2
=�
1
2
+�
2
2
+2�
1�
2cos∆??????
�
���=�
1+�
2+2�
1�
2cos∆??????
⇒
�
���
�
=
�
1
�
+
�
2
�
+2×
�
1
�
�
2
�
cos∆??????
⇒�
���=�
1+�
2+2�
1�
2cos
2�
�
Δ�
�
1=�
1sin��−��
1+??????
1,�
2=�
2sin��−��
2+??????
2
�
���=�
1+�
2+2�
1�
2cos∆??????
As, �∝�
2
⇒�=��
2
Now, �
���
2
=�
1
2
+�
2
2
+2�
1�
2cos∆??????
�
���=�
1+�
2+2�
1�
2cos∆??????
⇒
�
���
�
=
�
1
�
+
�
2
�
+2×
�
1
�
�
2
�
cos∆??????
⇒�
���=�
1+�
2+2�
1�
2cos
2�
�
Δ�
�
1=�
1sin��−��
1+??????
1,�
2=�
2sin��−��
2+??????
2
�
���=�
1+�
2+2�
1�
2cos∆??????
Special Case of Resultant Intensity
If �
1=�
2=�,
�
���=�
1+�
2+2�
1�
2cos∆??????
�
���=�+�+2��cos∆??????
⇒�
���=2�1+cos∆??????
�
���=4�cos
2
∆??????
2
⇒�
���=4�cos
2
∆??????
2
⇒�
���=4�cos
2
�∆�
�
If �
1=�
2=�,
�
���=�
1+�
2+2�
1�
2cos∆??????
�
���=�+�+2��cos∆??????
⇒�
���=2�1+cos∆??????
�
���=4�cos
2
∆??????
2
⇒�
���=4�cos
2
∆??????
2
⇒�
���=4�cos
2
�∆�
�
Constructive Interference
�=�
1
2
+�
2
2
+2�
1�
2cos??????
If, cos??????=1⇒??????=2��
⇒�=�
���=�
1+�
2
Phasedifference=
2�
�
×(Pathdifference)⇒2��=
2�
�
×(∆�)
Path difference =∆�=��
Interference that produces the greatest possible amplitude is called Constructive Interference.
�
2
�
1
??????
�
�=�
1
2
+�
2
2
+2�
1�
2cos??????
If, cos??????=1⇒??????=2��
⇒�=�
���=�
1+�
2
Phasedifference=
2�
�
×(Pathdifference)⇒2��=
2�
�
×(∆�)
Path difference =∆�=��
Interference that produces the greatest possible amplitude is called Constructive Interference.
�
2
�
1
??????
�
Intensity of resultant wave
Net Intensity : �
���=�
1+�
2+2�
1�
2cos??????
For constructive Interference
cos??????=1
�
���=�
���=�
1+�
2+2�
1�
2=�
1+�
2
2
�
���=�
1+�
2
2
If �
1=�
2=�
�
���=4�
Net Intensity : �
���=�
1+�
2+2�
1�
2cos??????
For constructive Interference
cos??????=1
�
���=�
���=�
1+�
2+2�
1�
2=�
1+�
2
2
�
���=�
1+�
2
2
If �
1=�
2=�
�
���=4�
Destructive Interference
�=�
1
2
+�
2
2
+2�
1�
2cos??????
If, cos??????=−1⇒??????=(2�+1)�
⇒�=�
�??????�⇒�
1−�
2
Phasedifference=
2�
�
×(Pathdifference)⇒(2�+1)�=
2�
�
×(∆�)
Path difference =∆�=(2�+1)
�
2
Interference that produces the minimum possible amplitude is called Destructive Interference
�=�
1
2
+�
2
2
+2�
1�
2cos??????
If, cos??????=−1⇒??????=(2�+1)�
⇒�=�
�??????�⇒�
1−�
2
Phasedifference=
2�
�
×(Pathdifference)⇒(2�+1)�=
2�
�
×(∆�)
Path difference =∆�=(2�+1)
�
2
Interference that produces the minimum possible amplitude is called Destructive Interference
Intensity of resultant wave
Net Intensity : �
���=�
1+�
2+2�
1�
2cos??????
For destructive Interference
cos??????=−1
�
���=�
�??????�=�
1+�
2−2�
1�
2=�
1−�
2
2
�
�??????�=�
1−�
2
2
If �
1=�
2=�
�
�??????�=0
Net Intensity : �
���=�
1+�
2+2�
1�
2cos??????
For destructive Interference
cos??????=−1
�
���=�
�??????�=�
1+�
2−2�
1�
2=�
1−�
2
2
�
�??????�=�
1−�
2
2
If �
1=�
2=�
�
�??????�=0
Two sources �
1and �
2are separated by a distance of 3.2�. If an observer starts
moving from source �
1towards �
2, find out the number of maxima and
minimathe observer meets during his motion from �
1to �
2.
�
1 �
2
3.2�
Maxima : 3�,2�,�,0,�,2�,3�
Minima : 2.5�,1.5�,0.5�,0.5�,1.5�,2.5�
�
1 �
2
3.2�
�
3�2��0 �2�3�
2.5�1.5�0.5�0.5�1.5�2.5�
Maxima =7 Minima =6
Solution :
1and �
2are separated by a distance of 3.2�. If an observer starts
moving from source �
1towards �
2, find out the number of maxima and
minimathe observer meets during his motion from �
1to �
2.
�
1 �
2
3.2�
Maxima : 3�,2�,�,0,�,2�,3�
Minima : 2.5�,1.5�,0.5�,0.5�,1.5�,2.5�
�
1 �
2
3.2�
�
3�2��0 �2�3�
2.5�1.5�0.5�0.5�1.5�2.5�
Maxima =7 Minima =6
Solution :
Two sources �
1and �
2are separated by distance of 3.2�are symmetrically placed
about the centre of a circular path as shown. If an observer starts moving from
point �along the circular path, find the number of maximaand minimahe
meets, when he complete one revolution to reaches point �again.
Maxima : 4times∶3�,2�,�2times∶0
Minima : 4times∶2.5�,1.5�,0.5�
Maxima =14 Minima =12
�
1 �
2
3.2�
�
∆�=3.2� ∆�=3.2�
∆�=0
∆�=0
2�
0.5�
1.5�
2.5�
3�
�
Solution :
1and �
2are separated by distance of 3.2�are symmetrically placed
about the centre of a circular path as shown. If an observer starts moving from
point �along the circular path, find the number of maximaand minimahe
meets, when he complete one revolution to reaches point �again.
Maxima : 4times∶3�,2�,�2times∶0
Minima : 4times∶2.5�,1.5�,0.5�
Maxima =14 Minima =12
�
1 �
2
3.2�
�
∆�=3.2� ∆�=3.2�
∆�=0
∆�=0
2�
0.5�
1.5�
2.5�
3�
�
Solution :
T
Figure shows a tube structure in which a sound signal is sent from one end and
is received at the other end. The semicircular part has a radius of 20.0��. The
frequency of the sound source can be varied electronically between 1000��and
4000��. Find the frequencies at which maxima of intensity are detected. Take
the speed of sound in air to be 340�/�.
For constructive interference,
∆�=�−2�=��=�
�
�
Path difference between the interfering waves,
∴Frequency for maxima,
�=
��
�−2�
=
�×340
1.14×0.2
For �=1,2,3…
�
1≈1491��,�
2≈2982��,�
3≈4473��…
∆�=��−2�=�−2�
�
1≈1491��&�
2≈2982��,
Solution :
Figure shows a tube structure in which a sound signal is sent from one end and
is received at the other end. The semicircular part has a radius of 20.0��. The
frequency of the sound source can be varied electronically between 1000��and
4000��. Find the frequencies at which maxima of intensity are detected. Take
the speed of sound in air to be 340�/�.
For constructive interference,
∆�=�−2�=��=�
�
�
Path difference between the interfering waves,
∴Frequency for maxima,
�=
��
�−2�
=
�×340
1.14×0.2
For �=1,2,3…
�
1≈1491��,�
2≈2982��,�
3≈4473��…
∆�=��−2�=�−2�
�
1≈1491��&�
2≈2982��,
Solution :
Loudness of Sound
•The physiological sensation of
loudness of a sound is related to
its intensity.
•The intensity level �in decibels(��)
is given by :
�=10log
10
�
�
0
�
0=Minimum audible intensity
=10
−12
Wm
−2
�
�
�
�
•The physiological sensation of
loudness of a sound is related to
its intensity.
•The intensity level �in decibels(��)
is given by :
�=10log
10
�
�
0
�
0=Minimum audible intensity
=10
−12
Wm
−2
�
�
�
�
A small speaker delivers 2�of audio output. At what distance from the speaker
will one detect 120��intensity sound?
[Take reference intensity of sound as 10
−12
�/�
2
]
�=10log
10
�
�
0
120=10log
10
�
10
−12
Loudness of sound in decibels is given by,If the speaker is at a distance �from the observer
�=
�
4��
2
1=
2
4��
2
⇒�=
1
2�
�=0.398�≈40��
12=log
10
�
10
−12
⇒10
12
=
�
10
−12
�=1�/�
2
Solution :
will one detect 120��intensity sound?
[Take reference intensity of sound as 10
−12
�/�
2
]
�=10log
10
�
�
0
120=10log
10
�
10
−12
Loudness of sound in decibels is given by,If the speaker is at a distance �from the observer
�=
�
4��
2
1=
2
4��
2
⇒�=
1
2�
�=0.398�≈40��
12=log
10
�
10
−12
⇒10
12
=
�
10
−12
�=1�/�
2
Solution :
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