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Click To See The Adjacent Topics.
A Glance At Waves.
Beat.
Diffraction.
Doppler Effect.
Application Of Doppler Effect.
Observer Stationary and Source Moving.
Source Stationary and Observer Moving.
A Glance at Waves
• The displacement of the particle at x at
n can
time t is abbreviated as y and wave eq
be written as
• y(x ,t) = f(t-x /v) represents a wave
travelling in the positive x direction with
constant speed v.
• Such a wave is called a travelling wave or
progressive wave.

• The time t and the position x must appear
in the wave eqn in the combination
(t-x /v) only.
• With respect to angular frequency and
amplitude the eqn can be written as
• y(x ,t) = a sin( kx ± ωt+ Φ) where y(x ,t) is
displacement as a function of position x
and time t, a is amplitude of the wave, ω is
angular frequency , k is angular wave
number, ( kx ± ωt+ Φ) is phase angle.
Beat

• The phenomenon of periodic variation of
intensity of sound when two sound waves
of slightly different frequencies interfere is
called Beat.
• So now let us consider two harmonic sound
waves having equal angular
frequency, amplitudes and travelling in a
medium in a same direction but having
slightly different frequencies.
• The equation of the two waves are
s1= a cos ω1t
s2= a cos ω2t
• Let ω1> ω2 . So the resultant displacement
is, by the principle of superposition.
• S = s1 + s2 = a ( cos ω1t + cos ω2t )
• =
( 2a cos ωbt ) cos ωat ,
• Where ( ωb = {ω1 - ω2}/2 ) and
( ωa = {ω1 + ω2}/2 )
• If |ω1 – ω2| << ω1 , ω2 , ωa >> ωb
• If we assume |ω1 – ω2| << ω1 , which
means ωa >> ωb , then the resultant wave
is oscillating with the average angular
frequency ωa . The amplitude is largest
when cos ωbt takes the limit +1 and -1.
• So the resultant intensity of wave
frequency which is 2ωb = ω1 – ω2 .
• Since ω = 2πν
• So,
2πνb = 2πν1 - 2πν2
• So,
νb = ν1 - ν2
• Therefore the beat frequency is
νb = |ν1 - ν2|
• The first two are the harmonic waves with
slight variation in their frequencies .
• The last one is the beat produced by the two
harmonic waves.
• For beats to be audible, the frequency |ν1 ν2| should not be very large .
• Thus, the difference of the component of the
two frequencies of the harmonic waves should
be less than 16 Hz for the beats to be heard.
Diffraction
• Bending of waves from an obstacle or an
opening is called diffraction.
• For example; If a small cardboard is placed
between a source and a listener, the sound
beyond the cardboard is not completely
stopped, rather the waves bend at the edges of
the cardboard to reach the listener.
• If a plane wave is passed through a small hole
or an opening, spherical waves are obtained on
the other side as if the hole is the source of the
sound.
• The diffraction effect are appreciable when the
dimensions of the opening or the obstacles are
comparable or smaller than the wavelength of
the wave.
• Diffraction Through A Small opening Produces
Spherical Shapes Of Waves.
• Diffraction Through A Small Obstacle Makes
Bend Shape Of Waves.
Doppler Effect
• If the source and the observer are at rest with
respect to the medium. Each compression and
rarefaction pulse, sent by the tuning fork, takes
the same time to reach the air near the ear.
• Thus, the pressure near the ear oscillates as
many times as the fork oscillates in a given
interval of time.
• The frequency observed is equal to the frequency
of the source.
• If the source or the observer or both, move
with respect to medium, the frequency
observed may be different from the
frequency of the source.

• So, The apparent change in the frequency of
the wave due to the motion of the source or
the observer is called Doppler Effect.
Applications Of Doppler Effect
• The change in the frequency caused by the
moving object due to Doppler effect is used to
measure the velocities in diverse areas such as
military, medical science, astrophysics, etc.
• It is used to check the over-speeding of the
vehicles.
• It is used at airports to guide aircraft, and in
military to detect enemy aircraft. Astrophysics
use it to measure the velocities of the stars.
• Doctors use it for sonography, echocardiogram,
etc.
Observer Stationary &
Source Moving

• Now suppose the observer is at rest with respect
to the medium and the source moves towards the
observer at a speed u which less than the wave
speed v.
•
u
S

•
•

S

ut

x-ut
x
• If the frequency of the vibration of the source is
ν0 , It sends compression pulse at a regular time
interval of T=1/ ν0 .
• Suppose the separation between the source and
the observer is x when a compression pulse is
emitted at t=0, the next compression pulse will
be emitted after a time interval of T. The source
will travel a distance ut in this time and hence
this second compression will be emitted at x-ut
from the observer.
• The first pulse takes a time of x /v to reach the
observer whereas the next one takes (x-uT)/v.
• So the time interval between the consecutive
compression pulses is
• T’ = T+ (x-ut)/v – x/v = {1-(u/v)}T
• = {(v-u)/v}T
•
u
S
S
•
ut
x-ut
•
x
•
• The apparent frequency of the sound as
experienced by the observer is
• ν’ = 1/T’
•
= {u/(v-u)} ν0
• So, ν’ = {u/(v-u)} ν0
Source Stationary &
Observer Moving

• Consider the source remains stationary with
respect to the medium and the observer
approaches with the speed u.
• As the source remains stationary, compression
pulses are emitted at a time interval T. These
pulses travel down the medium with a speed v
and at any instant the separation between two
consecutive compression pulses is λ= vT
•
•
•

•

u
S

vT

S

vT’

uT’
• When the observer receives a compression
pulse, the next compression pulse is a distance
vT away from it. This second compression pulse
moves towards the observer at a speed v and the
observer moves towards it at a speed u.
• So the observer will receive this second
compression pulse a time T’

• T’ = vt/(v+u)
• The apparent frequency of the sound
experienced by the observer is then
• ν’ = 1/T
•
= {(v+u)/v} ν0
• So v’ = {(v+u)/v} ν0
• So the final equation for the frequency can be
finally written as
•
v= {(v+u0)/(v-us)} ν0
• Where, v is speed of sound in the medium, u0 is
speed of the observer with respect to the
medium, considered +ve when it moves towards
the source and vise-versa and us is the speed of
the source with respect of the medium,
considered +ve when it moves towards the
observer and vise-versa.
Beat
Beat

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Beat

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  • 3. • • • • • • • • Click To See The Adjacent Topics. A Glance At Waves. Beat. Diffraction. Doppler Effect. Application Of Doppler Effect. Observer Stationary and Source Moving. Source Stationary and Observer Moving.
  • 4. A Glance at Waves • The displacement of the particle at x at n can time t is abbreviated as y and wave eq be written as • y(x ,t) = f(t-x /v) represents a wave travelling in the positive x direction with constant speed v.
  • 5. • Such a wave is called a travelling wave or progressive wave. • The time t and the position x must appear in the wave eqn in the combination (t-x /v) only.
  • 6. • With respect to angular frequency and amplitude the eqn can be written as • y(x ,t) = a sin( kx ± ωt+ Φ) where y(x ,t) is displacement as a function of position x and time t, a is amplitude of the wave, ω is angular frequency , k is angular wave number, ( kx ± ωt+ Φ) is phase angle.
  • 7. Beat • The phenomenon of periodic variation of intensity of sound when two sound waves of slightly different frequencies interfere is called Beat. • So now let us consider two harmonic sound waves having equal angular frequency, amplitudes and travelling in a medium in a same direction but having slightly different frequencies.
  • 8. • The equation of the two waves are s1= a cos ω1t s2= a cos ω2t • Let ω1> ω2 . So the resultant displacement is, by the principle of superposition. • S = s1 + s2 = a ( cos ω1t + cos ω2t ) • = ( 2a cos ωbt ) cos ωat , • Where ( ωb = {ω1 - ω2}/2 ) and ( ωa = {ω1 + ω2}/2 )
  • 9. • If |ω1 – ω2| << ω1 , ω2 , ωa >> ωb • If we assume |ω1 – ω2| << ω1 , which means ωa >> ωb , then the resultant wave is oscillating with the average angular frequency ωa . The amplitude is largest when cos ωbt takes the limit +1 and -1. • So the resultant intensity of wave frequency which is 2ωb = ω1 – ω2 .
  • 10. • Since ω = 2πν • So, 2πνb = 2πν1 - 2πν2 • So, νb = ν1 - ν2 • Therefore the beat frequency is νb = |ν1 - ν2|
  • 11. • The first two are the harmonic waves with slight variation in their frequencies . • The last one is the beat produced by the two harmonic waves.
  • 12. • For beats to be audible, the frequency |ν1 ν2| should not be very large . • Thus, the difference of the component of the two frequencies of the harmonic waves should be less than 16 Hz for the beats to be heard.
  • 13. Diffraction • Bending of waves from an obstacle or an opening is called diffraction. • For example; If a small cardboard is placed between a source and a listener, the sound beyond the cardboard is not completely stopped, rather the waves bend at the edges of the cardboard to reach the listener.
  • 14. • If a plane wave is passed through a small hole or an opening, spherical waves are obtained on the other side as if the hole is the source of the sound. • The diffraction effect are appreciable when the dimensions of the opening or the obstacles are comparable or smaller than the wavelength of the wave.
  • 15. • Diffraction Through A Small opening Produces Spherical Shapes Of Waves.
  • 16. • Diffraction Through A Small Obstacle Makes Bend Shape Of Waves.
  • 17. Doppler Effect • If the source and the observer are at rest with respect to the medium. Each compression and rarefaction pulse, sent by the tuning fork, takes the same time to reach the air near the ear. • Thus, the pressure near the ear oscillates as many times as the fork oscillates in a given interval of time. • The frequency observed is equal to the frequency of the source.
  • 18. • If the source or the observer or both, move with respect to medium, the frequency observed may be different from the frequency of the source. • So, The apparent change in the frequency of the wave due to the motion of the source or the observer is called Doppler Effect.
  • 19. Applications Of Doppler Effect • The change in the frequency caused by the moving object due to Doppler effect is used to measure the velocities in diverse areas such as military, medical science, astrophysics, etc. • It is used to check the over-speeding of the vehicles.
  • 20. • It is used at airports to guide aircraft, and in military to detect enemy aircraft. Astrophysics use it to measure the velocities of the stars. • Doctors use it for sonography, echocardiogram, etc.
  • 21. Observer Stationary & Source Moving • Now suppose the observer is at rest with respect to the medium and the source moves towards the observer at a speed u which less than the wave speed v. • u S • • S ut x-ut x
  • 22. • If the frequency of the vibration of the source is ν0 , It sends compression pulse at a regular time interval of T=1/ ν0 . • Suppose the separation between the source and the observer is x when a compression pulse is emitted at t=0, the next compression pulse will be emitted after a time interval of T. The source will travel a distance ut in this time and hence this second compression will be emitted at x-ut from the observer.
  • 23. • The first pulse takes a time of x /v to reach the observer whereas the next one takes (x-uT)/v. • So the time interval between the consecutive compression pulses is • T’ = T+ (x-ut)/v – x/v = {1-(u/v)}T • = {(v-u)/v}T • u S S • ut x-ut • x •
  • 24. • The apparent frequency of the sound as experienced by the observer is • ν’ = 1/T’ • = {u/(v-u)} ν0 • So, ν’ = {u/(v-u)} ν0
  • 25. Source Stationary & Observer Moving • Consider the source remains stationary with respect to the medium and the observer approaches with the speed u. • As the source remains stationary, compression pulses are emitted at a time interval T. These pulses travel down the medium with a speed v and at any instant the separation between two consecutive compression pulses is λ= vT
  • 27. • When the observer receives a compression pulse, the next compression pulse is a distance vT away from it. This second compression pulse moves towards the observer at a speed v and the observer moves towards it at a speed u. • So the observer will receive this second compression pulse a time T’ • T’ = vt/(v+u)
  • 28. • The apparent frequency of the sound experienced by the observer is then • ν’ = 1/T • = {(v+u)/v} ν0 • So v’ = {(v+u)/v} ν0
  • 29. • So the final equation for the frequency can be finally written as • v= {(v+u0)/(v-us)} ν0 • Where, v is speed of sound in the medium, u0 is speed of the observer with respect to the medium, considered +ve when it moves towards the source and vise-versa and us is the speed of the source with respect of the medium, considered +ve when it moves towards the observer and vise-versa.