Doppler effect -- find the frequency

[jangchen]

Homework Statement

An avant-garde composer wants to use the Doppler effect in his new opera. As the soprano sings, he wants a large bat to fly toward her from the back of the stage. The bat will be outfitted with a microphone to pick up the singer's voice and a loudspeaker to rebroadcast the sound toward the audience. The composer wants the sound the audience hears from the bat to be, in musical terms, one half-step higher in frequency than the note they are hearing from the singer. Two notes a half-step apart have a frequency ratio of 21/12 = 1.059. With what speed must the bat fly toward the singer?

Relevant Equations

f' = f×(c+v/c) , f''=f'×(c/c-v)

I don't know what is wrong

i think

when bat gets sound
f' = f×(c+v/c)

when audience hears
f''=f'×(c/c-v)

f''=f×1.059

but it is wrong TT

 

[TSny]

Two notes a half-step apart have a frequency ratio of 21/12 = 1.059.

21/12 does not equal 1.059. But I think the value of 1.059 is correct for the frequency ratio for a half-step. Did you mean to type 2^1/12 instead of 21/12?

when bat gets sound
f' = f×(c+v/c)

when audience hears
f''=f'×(c/c-v)

f''=f×1.059

Looks good so far, except your parentheses are not placed correctly.
You should have f' = f×(c+v)/c and similarly for f''.
You haven't shown your work for getting the answer for v.

 

[collinsmark]

In addition to what @TSny said, Keep in mind that this situation is for a moving observer (not moving source). Choose your formula accordingly. Once you've applied the Doppler effect for a moving observer you can stop there.

Since the loudspeaker is not moving toward the audience, there is no need to apply the Doppler effect a second time for a moving source.

Scratch that. I re-read the problem and the loudspeaker is attached to the bat. I missed that the first time. So yes, you need to apply the Doppler effect twice. Once for a moving observer, and again for a moving source.

[Edit: Btw, this sounds like a great opera.]

 

[TSny]

[Edit: Btw, this sounds like a great opera.]

 

[jangchen]

21/12 does not equal 1.059. But I think the value of 1.059 is correct for the frequency ratio for a half-step. Did you mean to type 2^1/12 instead of 21/12?

Oh not 21/12 but 2^1/12
The answer is 19m/s...
I tried what you taught.. but i got 9.97xxx m/s..

 

[haruspex]

Oh not 21/12 but 2^1/12
The answer is 19m/s...
I tried what you taught.. but i got 9.97xxx m/s..

Seems to me your answer is about right.

 

[TSny]

Oh not 21/12 but 2^1/12
The answer is 19m/s...
I tried what you taught.. but i got 9.97xxx m/s..

Maybe @collinsmark 's original interpretation is what was intended. If you assume the loudspeaker is fixed on the stage instead of being carried by the bat (!), you will get an answer of about 20 m/s. (Not sure what you're using for a value of the speed of sound.)

 

[haruspex]

If you assume the loudspeaker is fixed on the stage instead of being carried by the bat

But it quite clearly states

The bat will be outfitted with ... and a loudspeaker

(And we are told it is a large bat.)

 

[TSny]

But it quite clearly states

(And we are told it is a large bat.)

Yes, I agree.

 

[collinsmark]

I really want to see this opera.