In: Physics
Two sources emit sound with the same frequency. One source is at rest and the other is moving at a rate of 5 m / s toward the stationary observer. Observers hear kites 6 times per second. Determine the frequency of the sound emitted by the source.
Doppler's effect states that the frequency perceived by an observer changes if there is a relative motion between the source and the observer.
f ' = ( V-Vo / V- Vs ) * f ------------- formula 1
where f ' is apparent frequency heard by the observer, f is the frequency of the source, V is speed of sound ,Vo is speed of observer and Vs is speed of source . In this formula the sign convention is that speed will be taken as positive if it is in the direction : from source to observer .
GIVEN : 1) Velocity of source 1 ( Vs1) = 5 m/s
2) Velocity of source 2 (Vs2)= 0 m/s
3) Velocity of observer ( Vo)= 0m/s
4) taking frequency of souces = F
Taking velocity of sound = 340 m/s
Observer hears kites 6 times per second which means we have the beat frequency given to us as 6 Hz.
formula of beat frequncy is modulus of ( F1 - F2 ) where F1 and F2 are the frequency of sources interfering to produce the beat.
We know that by doppler's effect observer will hear frequency F from Source 2 as there is no relative motion because Vs2 and Vo both are 0.
by doppler's effect the apparent frequency heard by observer from source 1 is F "
by appluing formula 1 : F'' = ( 340 - 0 / 340 - 5 ) * F
we get F '' = 1.0149 * F
now we know that our beat frequency is 6 Hz
applying formula of beat frequency ie. F'' - F = 6 Hz
1.0149 F - F = 6
0.0149 F = 6
F= 6 /0.0149 = 402.6 Hz
We get the frequency of sources to be F= 402.6 Hertz.
Hope it helps!
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