Two sources emit sound with the same frequency. One source is at rest and the other...

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.

Solutions

Expert Solution

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!

Any feedback would be greatly appreciated .

genius_generous answered 1 year ago

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