According to Doppler effect, whenever there is a relative motion between a source of sound and listener, the apparent frequency of sound heard by the listener is different from the actual frequency of the sound emitted by the source.
The apparent change in the frequency of sound when the source, the listener and the medium are in relative motion is called Doppler effect.
When the distance between the source and the listener is decreased then the apparent frequency increases. The
For Example :
The frequency of a whistling engine heard by a person standing on the platform appears to increase, when the engine is approaching the platform and it appears to decrease when the engine is moving away from the platform.
Expression for Apparent frequecny in Doppler Effect
S = Source of sound
L = Listener
Vs = Velcoity of Source
VL = Velocity of Listener
V = Velocity of Sound in air
ν = Actual frequency
ν’ = Apparent frequency
SOME TEXT MISSING
ν’ = [ (V-VL)/(V-Vs)] ν
Special Cases :
1) If the source is moving away from the listener, but the listener is at rest, then VS is – ve and VL = 0.
∴ ν’ = [ (V)/(V+Vs)] ν Clearly ν ‘ < ν
2) If the source is moving towards the listener but the listener is at rest, then VS is + ve and VL=0.
∴ ν’ = [ (V)/(V-Vs)] ν Clearly ν ‘ > ν
3) If the source is at rest and the listener is moving away from the source, then Vs = 0 and VL is +ve.
∴ ν’ = [ (V-VL)/(V)] ν Clearly ν ‘ < ν
4) If the source is at rest and listener is moving towards the source, then Vs=0 and VL = -ve
∴ ν’ = [ (V+VL)/(V)] ν Clearly ν ‘ > ν
5) If the source and listener are approaching each other, then Vs = +ve & VL=-ve
∴ ν’ = [ (V+VL)/(V-VS)] ν Clearly ν ‘ > ν
6) If the source and listener are moving away from each other, then Vs = -ve and VL is +ve.
∴ ν’ = [ (V-VL)/(V+Vs)] ν Clearly ν ‘ < ν
7) If the source and listener are both in motion in the same direction and with the same velocity, then
Vs=VL =V’ (Let)
∴ ν’ = [ (V-V’)/(V-V’)] ν Clearly ν ‘ = ν
Dopler Effect Numericals
1. The apparent frequency of the whistle of an approaching train engine, changes in the ratio 9:8 as the engine crosses a stationary observer on the platform. If the speed of sound is 340 m/s. Find the speed of the train. [20 m/s]
2. A policeman on duty detects a drop of 10% in the pitch of the horn of a motor car as it crosses him. If the velocity of sound is 330 m/s. Calculate the speed of the car. [17.37 m/s]
3. Find the velocity of the source of the sound, when the frequency appears to be i) double ii) half of the original frequency to a stationary observer. Take velocity of sound = 330 m/s. [i) 165 m/s ii) 330 m/s]
When two sound waves of slightly different frequencies travelling in the same path in the same direction superpose on each other, then the intensity of the resultant sound at any point in the medium rises and falls alternately with time.
These periodical variations in the intensity of sound caused by the superposition of two sound waves of slightly different frequencies are called beats.
If the intensity of sound is maximum at t=0, one beat is said to be formed when the intensity becomes maximum again, after becoming minimum once in between. [One rise or fall of intensity
Beat Time period – The time interval between two successive beats (i.e. two successive maxima of sound or minima of sound) is called beat period.
Beat Frequency – The number of beats produced per second is called beat frequency.
Beat Frequency = Difference in frequencies of the two superposing waves i.e. ν2 – ν1
The essential condition for the formation of Beats –
For beats to be audible, the difference in the frequency of the two sound waves should not exceed 10. If the difference is more than 10, we will hear more than 10 beats per seconds but due to the persistence of hearing, our ears are not able to distinguish between two sound as separate if the interval between them is less than 1/10th of a second. Hence beats heard will not be distinct if the number of beats produced per second is more than 10.