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TRANSCRIPT
XI PHYSICS
[WAVES AND SOUND] CHAPTER NO. 8
M. Affan Khan LECTURER – PHYSICS, AKHSS, K [email protected] https://promotephysics.wordpress.com
OSCILLATORY MOTION “A motion in which an object moves to and fro or up and down about a fixed point is called oscillatory or vibratory motion.” SIMPLE HARMONIC MOTION (S.H.M) “An oscillatory motion in which acceleration is directly proportional to the displacement and is always directed toward the mean position is called simple harmonic motion.” Mathematically,
𝐚 ∝ −𝐱 Here negative shows that acceleration is always directed towards its mean position. CONDITIONS For simple harmonic motion following conditions must be satisfied. The system must have property of inertia. A restoring force is required. Motion must be oscillatory. Acceleration must be directly proportional to the displacement. CHARACTERISTICS A simple harmonic motion has following characteristics. Oscillatory motion. Time period Frequency Amplitude MOTION OF MASS SPRING SYSTEM Consider an object of mass m placed on a horizontal frictionless surface. Let this mass is attached with a spring of spring constant k as shown in figure. If an external force is applied to the object and it covers some displacement x than the restoring force can be found by using Hook’s law,
F = - k x------- (1) Where negative shows that restoring force is opposite to the applied force. According to the Newton’s 2nd law
F = m a ----------- (2) Comparing equation (1) and (2)
ma = −kx
a = −k
mx
This is the equation of acceleration of mass spring system Since k/m is constant, therefore
a ∝ −x Result: This expression shows that motion of mass spring system is simple harmonic. TIME PERIOD: Acceleration of mass spring system is given by,
a = −x√k
m
Since mass spring system is a simple harmonic oscillator therefore its acceleration can also be written as,
a = −ω2x Comparing above two accelerations we have,
−ω2x = −x√k
m
ω = √k
m
But also, ω =2π
T
2π
T= √
k
m
T = 2π√m
k
The above relation shows that time period of mass spring system is directly proportional to the square root of mass and inversely proportional to the square root of spring constant. FREQUENCY: Number of vibration or oscillation or rotation per second is called frequency. OR The reciprocal of time period is called frequency UNIT: S.I unit of frequency is per second or hertz.
FREQUENCY OF MASS SPRING SYSTEM: As we know that time period of spring mass system is,
T = 2π√m
k
But, f =1
T
Therefore,
𝐟 =𝟏
𝟐𝛑√
𝐤
𝐦
It is clear from this formula that time period of simple pendulum is independent of mass of
the bob and it depends upon length of the string.
RELATION BETWEEN UNIFORM CIRCULAR MOTION
AND SIMPLE HARMONIC MOTION:
Let us consider a particle of mass ‘m’ moving around a
vertical circle of radius ‘x0’ with constant angular
velocity ′ω′. If ‘θ′ is the angular displacement swept
during time ‘t’ then θ = ωt.
The projection ‘θ’ of particle ‘P’ on the diameter AB of
the circle as ‘P’ moves along circular path. It is also observed that the projection ‘Q’ speeds
up when it moves towards the center ‘O’ and slows down when it moves away from the
center.
Thus the instantaneous acceleration of projection ‘Q’ is directed towards the center and it
has vibratory motion.
Now, to show that motion of Q is simple harmonic, we calculate its acceleration.
The motion of ‘Q’ is associated with the motion of ‘P’ hence the acceleration of ‘Q’ must be a
component of the acceleration of the motion of particle P.
The acceleration of the particle P is centripetal acceleration i.e. directed towards the centre
of the circle along the line PO and it given by,
ac =v2
r
Or
ac = −vp
2
x0
ac = −x0ω
2
x0
ac = −x0ω2
The acceleration of projection Q is equal to the component of acceleration of particle P
along x-axis i.e.
ax = accosθ − − − (1)
In right triangle POQ.
cosθ =OQ
OP
cosθ =x
x0
Putting the values of centripetal acceleration and cosθ, we get,
ax = −x0ω
2x
x0
ax = −ω2x
ax ∝ −x
Hence it is proved that the motion of projection Q is simple harmonic.
i) Instantaneous Displacement
At some instant of time t, the angle between OP and the x-axis is θ.
In right angled triangle POQ
cosθ =OQ
OP
cosθ =x
x0
x = x0cosθ
If θ = ωt + ϕ, where ϕ is the initial phase angle at time t = 0.
So we may write
x = x0cos (ωt + ϕ)
This expression represents instantaneous displacement of projection Q and its value lies
between +x0 and −x0.
Where x0 is called amplitude of SHM of projection Q.
ii) Instantaneous Velocity
As the motion of Q depends upon the motion of P then the speed of projection Q is equal to
the horizontal component of the speed of the particle P, i.e.
sinθ =RP
PS
sinθ =Vx
Vp
Vx = Vpsinθ
Vx = x0ωsinθ
From trigonometry,
cos2 θ + sin2 θ = 1
sinθ = √1 − cos2 θ
sinθ = √1 − (x
x0)2
sinθ =√x0
2 − x2
x0
Now, we can substitute value of sine in above equation,
Vx = x0ω√x0
2 − x2
x0
Vx = ω√x02 − x2
As ‘x’ is instantaneous displacement then ‘Vx’ will be instantaneous velocity of projection Q,
i.e. at any time ‘t’.
ENERGY OF A PARTICLE EXECUTING SHM
Let us consider a body of mass ‘m’ connected with one end of a spring whose other end is
connected with rigid wall and it can execute SHM on a frictionless surface as shown in fig.
Now we find expression for instantaneous kinetic and potential energies:
i) Kinetic Energy
When displacement of body from its mean position
is ‘x’ then instantaneous velocity will be,
V = ω√x02 − x2
Where x0 is the amplitude of SHM
We have ω = √k
m
V = √k
m√x0
2 − x2
As we know that K. E =1
2mv2
K. E =1
2m [√
k
m√x0
2 − x2]
2
K. E =1
2m ×
k
m× (x0
2 − x2)
K. E =1
2k(x0
2 − x2)
When x = ±x0
Then K. E =1
2(x0
2 − x02)
K. E = 0 i.e minimum
When x = 0
Then K. E =1
2kx0
2 i.e maximum
ii) Potential Energy
According to Hooke’s law, the magnitude of restoring force acting on a body
attached with spring at displacement x from its mean position is
F = kx
At mean position, F = 0
Therefore the average force on the mass ‘m’ during displacement ‘x’ is
F =0 + kx
2=
1
2kx
Now,
W = F . x
W = Fxcosθ
W = (1
2kx) xcos00
W =1
2kx2
Here W = P. E
Therefore,
P. E =1
2kx2
Total Energy
At any instant of SHM the body has partly P.E. and partly K.E.
Therefore, total energy at any instant will be
E = P. E + K. E
E =1
2kx2 +
1
2k(x0
2 − x2)
E =1
2kx2 +
1
2kx0
2 −1
2kx2
E =1
2kx0
2
Here amplitude factor defines the maximization which clearly states that the total energy at
every point is constant and there will be just a conversion of K.E to P.E or P.E to K.E.
SIMPLE PENDULUM A system which consist of a point mass (spherical bob) suspended by a light, flexible and inextensible string tied to a fixed rigid and frictionless support is called simple pendulum. When the bob is displaced from its mean position and then released, it begins to oscillate periodically which is one of the condition of SHM. To check whether its motion is simple harmonic or not, we find acceleration of the motion of the bob. Let us consider a bob of a mass ‘m’ and weight ‘W’ is suspended by means of inextensible light string of length ‘l’. If the amplitude of oscillation is very small, then its motion will be SHM. When the displacement of bob is ‘x’ from its mean position, then its weight ‘W’ can be resolved into its components as shown in fig.
cosθ =W∥
W=> W∥ = Wcosθ
sinθ =W⊥
W=> W⊥ = Wsinθ
The components of weight W∥ along the length of string is just balanced by the tension in the string i.e.
T = W∥ T = Wcosθ
Hence the net force acting on the bob is W⊥, which is responsible for the oscillation of the bob i.e.
F = −W⊥ F = −Wsinθ
Here –ve sign indicates that the force is directed towards mean position. According to Newton’s second law of motion, F = ma Therefore,
ma = −mgsinθ a = −gθ
As we know that S = rθ Here,
x = lθ => θ =x
l
We may write now,
a = −g(x
l)
As the string is inextensible then g
l= constant
a = −(constant)x
a ∝ −x
This result shows that the motion of simple pendulum is simple harmonic.
TIME PERIOD
As we know that the acceleration of SHM of projection of circular motion of a particle is,
a = −ω2x
By comparing of acceleration for Simple Pendulum with the above equation, we get
ω2 = (g
l)
=> ω = (g
l)
12
As we know that
T =2π
ω
By putting the value of ‘ω’ we get
T = 2π√l
g
WAVE MOTION
The mechanism by which energy is transferred from one place to another is called wave
motion.
Types of Waves
We can categorize the types of waves in 3 categories.
a) W.r.t medium
b) W.r.t propagation
c) W.r.t mode of vibration
a) Types of Waves w.r.t Medium
i) Mechanical Waves
ii) Electromagnetic Waves
Mechanical waves are those which require a medium for their propagation, whereas,
electromagnetic waves are those which do not require any medium for their propagation.
For example sound waves are mechanical and light waves are electromagnetic in nature.
b) Types of Waves w.r.t mode of vibration
i) Longitudinal Waves
ii) Transverse Waves
i) Longitudinal Waves
Such waves in which particles of the disturbed medium undergo displacement in a
direction parallel to the direction of wave motion are called longitudinal waves. In the
presence of longitudinal waves in a medium, the particles of the medium vibrate parallel to
the direction of propagation i.e. these particles
move back and forth about their mean position.
These waves are produced in substances which
are elastic and compressible like gases and wire
spring. Also sound waves are longitudinal waves
as disturbance corresponds to a series of high
and low pressure regions that travel through air
or through any other material medium with
certain velocity. Similarly a longitudinal pulse
can be produced in a spring as shown in fig.
ii) Transverse Waves
Such waves in which particles of the medium
vibrate in a direction perpendicular to the propagation of waves are called transverse
waves. E.g. the fig. shows that the wave pulse travels along the rope, each segment of
the rope which is disturbed moves in a direction perpendicular to the wave motion and
there is no motion in any part of rope in the direction of propagation of wave. Hence
this pulse is transverse wave. Electromagnetic waves such as light x – rays, radio and
television transmission waves are also transverse.
c) Types of waves w.r.t propagation
i) Travelling waves
ii) Standing waves
i) Travelling Waves
A travelling wave is that in which the displacement of wave depends upon both time and
space. In case of travelling wave through a medium, every particles or segment of the
medium vibrate simple harmonically with a frequency equal to the frequency of vibration
of the source that drives the wave into the medium.
A travelling wave can be produced in a string (medium) by flipping it at on end as shown in
fig. above for transverse waves.
The shape of the disturbed part of the string containing the wave pulse can be described at
a given time mathematically by a wave function f(x) such that
y = f(x)
Here y is the vertical displacement of a particle from its equilibrium position and x is the
horizontal displacement of the particle from the point where the displacement y is zero.
Hence x, y, are the coordinates of the position of particles. Since the wave pulse is moving
along the string, its position is changing continuously with time. Hence the location of wave
pulse depends on time also. Thus the shape of the wave pulse with its location can be given
by the function f(x,t) such that
y = f(x, t)
This is called wave function. For a travelling wave, moving along +x-axis, the wave function
will be,
y = f(x′) = f(x − vt)
Sound waves are travelling waves. Similarly, the light waves are also travelling wave.
ii) Standing waves
The waves formed by superposition or overlapping of two travelling waves of same
amplitude and frequency moving in opposite direction in the same medium are called
standing waves or stationary waves. When a string is tightly stretched between two fixed
supports and then flicked upward, the crest extends over the whole distance between the
supports. So each end of the wave suffers a phase change i.e. crest on reflection becomes a
trough and the trough becomes a crest on reflection at the other end as shown in fig.
These waves are stationary or standing in the sense that these are found in the form of
loops in which vibration occur within limited space.
The point where displacement is maximum called antinode denoted by A, and that where
displacement is minimum (zero) is called a node denoted by N. The distance between two
consecutive nodes or any two consecutive antinodes is equal half of the wavelength.
SUPERPOSITION PRINCIPLE
Whenever two or more waves travel in the same space then these waves interfere with
each other, thus form a resultant wave. The net wave displacement caused by resultant
wave is found equal to the algebraic sum of the individual wave displacements of all given
waves. This is known as superposition principle and it may be written as,
Y = y1 + y2 + y3 + ⋯+ yn
Let us consider two sinusoidal waves with the same amplitude, frequency and wavelength
travelling in opposite direction i.e.
y1 = A0sin (kx − ωt) and y2 = A0sin (kx + ωt)
Where,
A0 = Amplitude of wave
k =2π
λ i. e. angular wave number
ω = 2πν =2π
T i. e. angular frequency
x = space or position coordinate
t = time coordinate
According to superposition principle the resultant wave displacement will be,
y = y1 + y2
By putting the values,
y = A0 sin (kx − ωt) + A0 sin (kx + ωt)
y = A0[sin(kx − ωt) + sin(kx + ωt)]
y = A0[2sin (kx − ωt + kx + ωt
2) cos (
kx − ωt − kx + ωt
2)]
y = 2A0sinkxcosωt
This equation represents the wave function of standing waves having angular frequency ω
and resultant amplitude equal to 2A0sinkx.
It means that the amplitude of SHM of a given particle of medium or string depends on
sinkx in case of standing waves.
Points of Maximum Amplitudes or Intensity
The resultant amplitude of standing wave is 2Aosinkx.
Therefore, amplitude will be maximum only when, sinkx = ±1
kx = sin−1(±1) 2π
λ= 900, 2700, 4500. 6300, ………
2π
λx =
π
2,3π
2,5π
2,7π
2,………
x =λ
2π(π
2,3π
2,5π
2,7π
2…………)
x =λ
4,3λ
4,5λ
4,7λ
4,………
These are the points of maximum amplitudes in standing waves and these are known as
‘Antinodes’. The distance between two consecutive antinodes is equal to λ
2
i.e. 3λ
4−
λ
4=
λ
2
or 5λ
4−
3λ
4=
λ
2
Points of minimum amplitudes or intensity
The resultant amplitude of standing wave is , 2A0sinkx
Therefore, amplitude will be minimum only when, sinkx = 0
kx = sin−1 0
kx = 00, 1800, 3600, 5400, ……… 2π
λx = 0, π, 2π, 3π, 4π………
x =λ
2π(0, π, 2π, 3π, 4π………)
x = 0,λ
2, λ,
3λ
2, 2λ
5λ
2,………
These are the points of minimum amplitudes in standing waves, and these are known as
‘Nodes’. The distance between any two consecutive nodes is equal to λ
2
i.e λ
2− 0 =
λ
2
or
λ −λ
2=
λ
2
FUNDAMENTAL FREQUENCY AND HARMONICS
Standing waves can be set up by a continuous super position of waves incident and
reflected from the extreme points of the medium. Let us consider a stretched string of
length ‘L’ which can vibrate in different modes and each mode has its own frequency as
shown in fig.
1st Harmonics or Fundamental Frequency:
If the string is plucked from its midpoint and released it
begins to vibrate in one loop by forming nodes at fixed
points as shown in fig.
As we can see from the diagram the distance between
two consecutive nodes is equal to half of the
wavelength.
Then L =λ1
2 Or, λ1 = 2L
We Have V = νλ
In this case we may write with the following subscripts
ν1 =V
λ1
ν1 =V
2L
This is fundamental frequency, or the 1st harmonics of standing waves.
2nd Harmonics or 1st Overtone
If the same string is now vibrated with more frequency then there is a chance that the
string must vibrate in two segments.
Here the length of string is equal to one complete wavelength
L = λ2,
We Have V = νλ
In this case we may write with the following subscripts
ν2 =V
λ2
ν2 =V
L
We may write the above equation as
ν2 =2V
2L= 2 (
V
2L)
ν2 = 2ν1
This is called 2nd harmonics or 1st overtone.
3rd Harmonics or 2nd Overtone
The same string can also vibrate into three segments if it is plucked with more frequency
and in this case the length of string forms one and half wavelength
L =3
2λ3, Or λ3 =
2L
3
We Have V = νλ
In this case we may write with the following subscripts
ν3 =V
λ3
ν3 =V
2L/3
ν3 =3V
2L= 3 (
V
2L)
ν3 = 3ν1
This is called 3rd harmonics or 2nd overtone
By examining the equations of all harmonics we can say that if string vibrates in ‘n’ loops
then, νn = nν1
SONOMETER: It is a device used to study
the vibration of stretched string in
different number of loops. It consists of a
thin metallic wire stretched across two
bridges on the top of a hollow wooden
sounding box one meter in length. One
end of the wire is fastened to a peg at one
end of the box and other end passes over
a smooth frictionless pulley is connected
with suspended load.
By adjusting the position of bridges or by changing the postion of movable bridge C,
different modes of vibration can be achieved.
If L is the length of vibrating segment of the string, T is the tension and μ is the mass per
unit length of the wire, then the frequency produced in the string given by,
νn = nν1
vn =nV
2L
vn =n
2L√
T
μ
Where ‘n’ is an integer (indicating mode of vibration)
For n = 1
v1 =1
2L√
T
μ
As we can see here clearly in above equation
v ∝1
L (Law of Length)
v ∝ √T (Law of Tension)
v ∝1
√μ (Law of linear density)
SOUND WAVES
Sound is the form of energy and it travels in the form of longitudinal waves. As sound
waves are mechanical waves i.e. they travel through a medium then particles of the
medium vibrate along the direction of propagation of the wave motion.
Sound waves can travel through gases, solids and liquids with a speed which depends upon
the properties of the medium.
Because of longitudinal displacement of molecules of medium about their mean position, a
series of alternate high and low pressure regions called compressions and rarefactions
respectively is formed. This series ultimately reaches at ear drum which begins to vibrate
with same frequency as the sound waves have.
There are three classes of sound waves depending upon frequency,
i) Infrasonic waves
ii) Audible waves
iii) Ultrasonic waves
Such longitudinal mechanical waves which have frequency less than 20 Hz are called
infrasonic waves
Such longitudinal mechanical waves which have frequency range between 20 Hz to 20 kHz
are called audible sound waves.
Such longitudinal mechanical waves which have frequency greater than 20 kHz are called
ultrasonic waves.
SPEED OF SOUND WAVES
Newton’s Formula
The speed of sound in air was determined by Newton through his formula for speed of
longitudinal or compressional waves in a medium.
As sound waves are also longitudinal or compression waves through a medium then the
speed of sound depends upon the compressibility and the inertia of the medium. In fact the
high speed of all mechanical waves can be expressed in a general form.
V = √(elastic property)
inertial property
Newton developed a formula for speed of sound in waves in air as,
V = √B
ρ
Where ‘B’ is bulk modulus of compressible medium i.e. air and ′ρ′ is the density which is
inertial property.
Bulk modulus is defined as the ratio of stress to the volumetric strain,
B =stress
volumetric strain
B =P
∆V/V
Newton assumed that the process of sound propagation is isothermal i.e. in this process
temperature remains constant. And in the case of isothermal process bulk modulus is only
equal to pressure.
B = P
Therefore,
V = √P
ρ
As we go through this formula, and put values of atmospheric pressure and density at STP,
we find that velocity of sound comes out to be 281 m/s, which is completely contradicting
with the experimental value of 332 m/s. Therefore, this idea of Newton was rejected.
Laplace’s Correction
Laplace realized that, the wave motion in case of sound is so rapid and the heat
conductivity is so low that there is in sufficient time for the heat produced in the
compressed regions to be conducted to the rarefied regions, therefore, the process of
propagation of sound waves is not isothermal rather it is adiabatic which is a process in
which heat does not flow into or out of the system.
Laplace, therefore, used adiabatic equation in deriving formula for speed of sound in air as
he forced that temperature of medium doesn’t remain constant. The modified formula for
the speed of sound in air is,
V = √γP
ρ
Where, ‘γ′ is the ratio of molar specific heat of gas at constant pressure to the molar specific
heat at constant volume.
i. e. γ =Cp
Cv
Effects of Temperature on Speed of Sound
The speed of sound in any medium increase with the rise in temperature. As the
translational and vibrational kinetic energy of molecules of the medium is increased with
the rise in temperature then pressure zones propagate more rapidly in the form of
alternate compressions and rarefactions after the temperature of medium air is increased.
We have the formula for speed of sound in air as,
V = √γP
ρ
According to general gas equation,
P ∨= nRT P =nRT
V
Putting in above equation
V = √γnRT
ρ ∨
V = √γnRT
M∨ ∨
V = √γnRT
M
For one mole of gas (i.e. n = 1)
V = √γRT
M
Also, if we need to find velocity of sound a particular temperature (VT), we may simple use
the following formula
VT = 332√TK
273
Where, TK is the temperature in Kelvin.
Characteristics of Musical Sound
Following are the characteristics of Musical Sound
i) Loudness
ii) Intensity Level
iii) Pitch
iv) Quality
i) Loudness
The auditory sensation produced by intensity of sound I called louness of sound.
Therefore, it depends upon both intensity of sound and the nature of humar ear.
A normal human ear is very sensitive detector sound. It can record the least
sound (10-12 Watt/m2)
ii) Intensity Level
Weber-Fechner’s Law
The loudness of sound doesn’t increase directly as the power delivered to the ear
increases but seems to vary roughly as the logarithm of the power or intensity.
This was first realized by Weber-Fechner and they stated that,
“Loudness of sound is directly proportional to the logarithm of intensity”
L ∝ log I
L = klogI
If I0 the intensity of faintest audible sound then its loudness can be written as,
L0 = klogI0
The difference of loudness of sound with the loudness of faintest audible sound
is called intensity level.
By subtracting L0 from L, we get
L − L0 = klogI − klogI0
L − L0 = k(logI − logI0)
L − L0 = klog (I
I0)
Where L − L0 = β (intensity level)
β = klog (I
I0)
Unit
In M.K.S system the unit of intensity level is “Bel”. As it is a large unit so we
measure intensity level in decibel (dB) i.e. 1 Bel = 10 dB
The unit of loudness of sound is “Sone”.
1 Sone = 40dB at 1000 Hz.
iii) Pitch
The pitch of sound reflects its class of frequency. It is the property of sound by
which we can distinguish between shrill and grave sound. The greater the
frequency, the greater will be the pitch of sound and lower the frequency, the
lower will be the pitch of sound.
For example, the frequency of women’s sound is more than the men’s sound.
Therefore, the women’s sound will be considered as shrill and the men’s sound
will be considered as grave normally.
iv) Quality of Sound (or Timbre)
The quality of sound is the property by which we can distinguish between nodes
of the same pitch and intensity when played on different instruments or sung by
different voices. Every instruments of the same kind may yield notes of different
quality. We recognize the voice of our friend over the telephone by its quality.
The difference in the sound produced by two notes of same pitch and loudness is
due to difference in their resultant waveforms. The resultant waveform of any
sound is obtained by combining the amplitudes of fundamental and the
overtones of given sound.
Difference between Musical Sound and Noise
The difference between noise and musical sound is because of the waveform of sound. If
the wave form is irrefular non-symmeteric and having random fluctuations then resulting
sound is said to be noise. Similarly, if the wave form is regular and symmetrical and having
ordered fluctuations then resulting sound produces a smooth pleasant sensation, hence
called music.
BEATS
The periodic vibration in the intensity of sound at a given point due to superposition of two
waves having slightly different frequencies is called the phenomenon of beats.
The production of beats is actually the interference of sound waves having slightly different
frequencies due to which intensity of resultant wave changes in time constructively and
destructively.
The number of beats
that one hear per
second (beat
frequency) is equal
to the difference in
frequency between
two sources. The
maximum beat
frequency that the human ear can detect is about 7 beats per second. When the beat
frequency (number of beats produced per second) is greater than seven we cannot hear
them clearly.
If we strike two tuning forks of slightly different frequencies with a rubber pad and bring
them in air then we hear only one note of changing intensity i.e. periodic rise and fall in the
intensity of sound. This is called phenomenon of beats.
Consider two travelling waves having equal amplitudes but slightly different frequencies f1
and f2 in the same direction i.e.
Y1 = A0cosω1t; Y2 = A0cosω2t
Y = Y1 + Y2
Y = A0cosω1t + A0cosω2t
Y = A0[(cosω1t + cosω2t)
Y = A0(2 cos (ω1t + ω2t
2) cos (
ω1t − ω2t
2)
Y = 2A0 cos (2πν1t − 2πv2t
2) cos (
2πv1t − 2πv2t
2)
Y = 2A0 cos (2π(v1 + v2)t
2) cos (
2π(v1 − v2)t
2)
Y = 2A0 cos (2π(v1 − v2)t
2) cos (
2π(v1 + v2)t
2)
This shows that resultant wave travels with average frequency v1+v2
2 and resultant
amplitude.
Amplitude = 2A0cos2π (v1 + v2
2)
This amplitude varies in time periodically with frequency
DOPPLER’S EFFECT
The apparent change in the pitch or frequency of sound due to relative motion of source of
sound and listener is called Doppler’s Effect as it was first observed by John Doppler, as
Austrian Physicists.
For example, when the train is approaching towards an observer at rest, the pitch of sound
increases whereas the pitch of the sound decreases when the train is moving away. This
effect can also be observed when a listener moves towards or away from source of sound.
Difference situations for the observations of Doppler’s Effect are given as,
Case 1(a): When listener moves towards source at rest.
Suppose,
V = Velocity of sound waves in air
λ = Wavelength of sound waves
v = Real frequency of sound waves
V0 = Velocity of observer
v′ = Apparent frequency of sound heard by obsever
We know that
V = vλ
Or,
v =V
λ
Now suppose listener is moving towards stationary source which emits sound waves. Then
the relative velocity of sound w.r.t the listener will be V + V0. Then for this case we may
write
Apparent frequency =Relative Velocity
Wavelength
v′ =(V + V0)
λ
v′ =V + V0
V/v
v′ =V + V0
Vv
This expression shows that the apparent frequency of sound is increased.
Case 1(b): When listener moves away from source at rest. (Same diagram with opposite
direction of listener)
Suppose,
V = Velocity of sound waves in air
λ = Wavelength of sound waves
v = Real frequency of sound waves
V0 = Velocity of observer
v′ = Apparent frequency of sound heard by obsever
We know that
V = vλ
Or,
v =V
λ
Now suppose listener is moving away from stationary source which emits sound waves.
Then the relative velocity of sound w.r.t the listener will be V − V0. Then for this case we
may write
Apparent frequency =Relative Velocity
Wavelength
v′ =(V − V0)
λ
v′ =V − V0
V/v
v′ =V − V0
Vv
This expression shows that the apparent frequency of sound is decreased.
Case 2(a): When source is moving towards stationary observer.
Let us now consider a source of sound moving
with velocity Vs towards stationary listener. The
wave crests detected by the stationary listener
are closer together because the source is moving
in the direction of the outgoing wave resulting in
shortening of wavelength. i.e. the wavelength λ′
measured by the listener is shorter than the
original wavelength of the source (λ′ < λ)
Now Suppose,
V = Velocity of sound waves in air
λ = Wavelength of sound waves
v = Real frequency of sound waves
Vs = Velocity of source
v′ = Apparent frequency of sound heard by obsever
λ′ = Apparent wavelength recieved by observer
For this case we may write
Apparent wavelength = Orignal wvelength − Relative distance travelled by the source
λ′ = λ − S
λ′ =V
v−
Vs
v
V
v′=
V − Vs
v
v′
V=
v
V − Vs
v′ =V
V − Vsv
This result shows that the apparent frequency of sound is increased.
Case 2(b): When source is moving away from stationary observer. (Same diagram with
opposite direction of source)
Let us now consider a source of sound moving with velocity Vs away from stationary
listener. The wave crests detected by the stationary listener are farther together because
the source is moving in opposite direction of the outgoing wave resulting in increase of
wavelength. i.e. the wavelength λ′ measured by the listener is greater than the original
wavelength of the source (λ′ < λ)
Now Suppose,
V = Velocity of sound waves in air
λ = Wavelength of sound waves
v = Real frequency of sound waves
Vs = Velocity of source
v′ = Apparent frequency of sound heard by obsever
λ′ = Apparent wavelength recieved by observer
For this case we may write
Apparent wavelength = Orignal wvelength + Relative distance travelled by the source
λ′ = λ + S
λ′ =V
v+
Vs
v
V
v′=
V + Vs
v
v′
V=
v
V + Vs
v′ =V
V + Vsv
This result shows that the apparent frequency of sound is decreased.
Case 3(a): When both source and the listener are moving towards each other.
The apparent frequency heard by listener in this case will be
v′ = (V + V0
V − Vs) v
This shows that apparent frequency is increased rapidly.
Case 3(b): When both source and the listener are moving away from each other.
The apparent frequency heard by listener in this case will be
v′ = (V − V0
V + Vs) v
This shows that apparent frequency is decreased rapidly.
Applications of Doppler’s Effect:
1. Doppler’s effect has very wide range of applications in different scientific fields and in
daily life e.g. It is used in measuring the speed of automobile by traffic police
2. Ultrasound machine do their jobs on the basis of Doppler’s Effect which are used as an
alternative of x-rays
3. This effect is used in finding the motion of objects like submarine under water.
4. Doppler’s effect is also employed in light waves which is used in space and astronomical
research.
5. Radio Detection and Ranging (RADAR) is commonly used to detect the presence of any
aircraft in the airspace and it is very important in civil aviation and military purposes.
Acoustics
The branch of Physics which deals with the study of production and properties of sound is
called a acoustics. This term is also used to describe the way in which sound is recorded
and reproduced.
This subject is used in developing auditorium, seminar, halls, conference rooms, recording
rooms of radio and television stations, good acoustic conditions are required for proper
listening and recording of sound and for this purpose the loudness of each separate syllable
should be sufficiently large and echoes should be just sufficient to maintain the continuity
of sound.