If we take a source with a frequency of 230Hz moving at 30ms^−1 right and the listener moving 25ms^−1 right.Let the source be behind the listener at some distance so that the distance between them decreases.Take the speed of sound as 340ms^−1.

spoofing44

spoofing44

Open question

2022-08-21

If we take a source with a frequency of 230 H z moving at 30 m s 1 right and the listener moving 25 m s 1 right.Let the source be behind the listener at some distance so that the distance between them decreases.Take the speed of sound as 340 m s 1 .
With the speaker as the frame of reference, the listener is moving 5 m s 1 towards the source.
f L = ( c ± v L c ± v S ) f S
= ( 340 + 0 340 5 ) 230
= 233.4328 H z
With the Listener as the frame of reference, the source is moving 5 m s 1 towards the listener.
f L = ( 340 + 5 340 + 0 ) 230
= 233.3824 H z
With the system as the frame of reference, the source is moving at 30 m s 1 right and the listener moving 25 m s 1 right.
f L = ( 340 25 340 30 ) 230
= 233.7097 H z
Why are the answers all different?

Answer & Explanation

Cristofer Briggs

Cristofer Briggs

Beginner2022-08-22Added 8 answers

You don't have the freedom to change reference frames as you do in your question, because the speed of sound is relative to the stationary air through which it travels. If you are using c = 340 m/s, then you need to similarly measure the velocities of the observer and the source in a reference frame in which the air isn't moving.
If you use the listener's frame of reference for example, then you would have v L = 0 and v S = 5, as you say. However, the air would be moving with constant speed 25 m/s to the left, which means that sound would travel at 340 + 25 = 365 m/s to the left and 340 25 = 315 m/s to the right. Since the waves that the listener hears are moving to the right, you would have
f L = f 0 ( 315 315 5 )
which matches the calculation performed in the frame in which the air is stationary.

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