# Doppler effect with frequency ratio

• wolfsrain159
In summary: It gives me 20.217. But that only arranges that the frequency picked up by the bat is a half step higher. Note Doc Al's second question.In summary, 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 2^(1/12) = 1.059.
wolfsrain159

## 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 2^(1/12) = 1.059.

## The Attempt at a Solution

I used f' = (c/c + Vs)f and f' = (c/c-Vs) because I was unsure which one to use. I plugged in c = 340 m/s and just divided f'/f and had that equal the frequency ratio. I kept getting answers like 19 m/s and 20 m/s but that isn't right

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wolfsrain159 said:
I used f' = (c/c + Vs)f and f' = (c/c=Vs)
You mean f'= c/(c-Vs).
One way to remember which to use is to consider the boundary case where the source is moving at speed c. The sound waves move with the source, piling up as an unlimited number of wavefronts at its leading edge. So when the source arrives at the receiver, all waves are heard in an instant, i.e. an infinite frequency. So it must be the c/(c-...) form.
I agree with 19m/s, or nearly. Do you know what the 'book' answer is? Could it be a matter of significant digits?

Edit: ouch - I got myself confused. I explained how to figure out which to use, then used the wrong one.
The correct conclusion is that it should be the "+" form. (Thanks, Doc Al.). But your equation for that case is inverted.

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haruspex said:
You mean f'= c/(c-Vs).
One way to remember which to use is to consider the boundary case where the source is moving at speed c. The sound waves move with the source, piling up as an unlimited number of wavefronts at its leading edge. So when the source arrives at the receiver, all waves are heard in an instant, i.e. an infinite frequency. So it must be the c/(c-...) form.
I agree with 19m/s, or nearly. Do you know what the 'book' answer is? Could it be a matter of significant digits?
Thank you for correcting that, I made a typo. This problem is from my online homework and the only instructions it says is to enter the answer with the proper units, which for velocity is m/s. Other then that I have no idea what I'm doing wrong. Using the second form c/(c-...) I got 19.098 m/s but I had already entered 19 m/s and 19.1 m/s and it said both were incorrect.

(1) What frequency does the bat observe and retransmit?
(2) What frequency does the audience observe emitted by the bat?

wolfsrain159 said:
Using the second form c/(c-...) I
Please see my edit at post #2.

haruspex said:
Please see my edit at post #2.
Using the other form c/(c+...) I got -19.1, which I'm assuming is negative for the direction. However, usually the online program will let you know if your sign is incorrect and it didn't say anything about it. I just don't want to use all my tries if it is still "incorrect"

wolfsrain159 said:
Using the other form c/(c+...) I got -19.1, which I'm assuming is negative for the direction. However, usually the online program will let you know if your sign is incorrect and it didn't say anything about it. I just don't want to use all my tries if it is still "incorrect"
Did you read the last sentence of my updated post #2? It was a second edit, so you might have missed it.

haruspex said:
Did you read the last sentence of my updated post #2? It was a second edit, so you might have missed it.
so would I use (c+Vs)/c = f' ?

wolfsrain159 said:
so would I use (c+Vs)/c = f' ?
Yes.

haruspex said:
Yes.
I got 20.247 using that equation and 20 m/s was marked incorrect. Would I inverse f' as well? Like use 1/1.0595 instead?

wolfsrain159 said:
I got 20.247 using that equation and 20 m/s was marked incorrect. Would I inverse f' as well? Like use 1/1.0595 instead?
It gives me 20.217. But that only arranges that the frequency picked up by the bat is a half step higher. Note Doc Al's second question.

## 1. What is the Doppler effect?

The Doppler effect is a phenomenon in which the frequency of a wave appears to change when the source of the wave is in motion relative to the observer.

## 2. How does the Doppler effect work with frequency ratio?

The frequency ratio is the ratio of the frequency of the source to the frequency observed by the observer. In the Doppler effect, this ratio will change depending on the relative motion of the source and observer. If the source is moving towards the observer, the observed frequency will be higher and the ratio will be greater than 1. If the source is moving away from the observer, the observed frequency will be lower and the ratio will be less than 1.

## 3. What factors affect the Doppler effect with frequency ratio?

The Doppler effect with frequency ratio is affected by the relative motion of the source and observer, as well as the speed of the source and the speed of sound in the medium through which the wave is traveling.

## 4. How is the Doppler effect with frequency ratio used in real life?

The Doppler effect with frequency ratio is used in various fields such as astronomy, meteorology, and medical imaging. For example, it is used to study the movement of stars and galaxies, to track weather patterns, and to measure blood flow in medical imaging.

## 5. Can the Doppler effect with frequency ratio be observed with all types of waves?

Yes, the Doppler effect can be observed with all types of waves, including sound waves, light waves, and electromagnetic waves. However, the magnitude of the effect may vary depending on the speed of the wave and the speed of the source and observer.

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