Physics problem...is this a mistake?

[BeHappy2013]

I found this problem on a random website. I thought pitch would increase regardless of the fact that the change in velocity is 0. The EK equation ∆f/f=v/c where f is the frequency of the source shows that there is a change in frequency even if the velocity is constant?

A person blowing a whistle is running toward another person who is standing still. As the runner approaches (at constant velocity), the stationary listener will hear a sound that:

A. continuously decreases in pitch and intensity.
B. has a fixed pitch but increases in intensity.
C. has a fixed pitch but decreases in intensity.
D. continuously increases in pitch and intensity.

ANSWER: B.
The Doppler effect tells us that if the source moves towards the detector, then the perceived frequency will be higher than that of the source. But if the source’s speed is constant, this higher pitch will remain constant as well. Finally, if the source of the sound gets closer, the intensity increases.

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[Jepstein30]

I found this problem on a random website. I thought pitch would increase regardless of the fact that the change in velocity is 0. The EK equation ∆f/f=v/c where f is the frequency of the source shows that there is a change in frequency even if the velocity is constant?

A person blowing a whistle is running toward another person who is standing still. As the runner approaches (at constant velocity), the stationary listener will hear a sound that:

A. continuously decreases in pitch and intensity.
B. has a fixed pitch but increases in intensity.
C. has a fixed pitch but decreases in intensity.
D. continuously increases in pitch and intensity.

ANSWER: B.
The Doppler effect tells us that if the source moves towards the detector, then the perceived frequency will be higher than that of the source. But if the source’s speed is constant, this higher pitch will remain constant as well. Finally, if the source of the sound gets closer, the intensity increases.

Pitch is fixed because speed is not increasing so B/C.

As for intensity, that would increase as the person approaches. Closer to source = more intense. As sound waves spread out, they have the same power but more area to cover = less intense. So closer to source = less area to cover with same power = more intense.

 

[BeHappy2013]

Thanks for your explanation! Can someone explain to me why pitch doesn't change at constant velocity? I have always been under the impression that the doppler effect says that pitch changes whenever one object moves closer or farther away.

 

[Jepstein30]

Thanks for your explanation! Can someone explain to me why pitch doesn't change at constant velocity? I have always been under the impression that the doppler effect says that pitch changes whenever one object moves closer or farther away.

Right, but as long as the velocity stays constant (i.e. no acceleration) it doesn't matter how far away the object is.

You have the equation right, all that matters is the velocity itself.. if that changes, then the shift changes. Otherwise, the shift stays constant no matter how close/far away.

 

[V5RED]

edit: nevermind...I misinterpreted their answers...

Their answer is good. The pitch is fixed at a higher pitch than if they were staying still.

 

[BeHappy2013]

Thanks! I think I get it conceptually now...if the velocity is not changing then the ∆f is the same at all times, so while the pitch is different than if the two people were both standing still, it is still not changing. Also...since this is sound, would you take the equation that EK tells you to use for light (∆f/f(s) = v/c) and swap 340 m/s for c since sound is what we're working with?

 

[Charles Darwin]

Wouldn't the frequency (and, in turn, pitch) increase as the two people got closer?

 

[Jepstein30]

Thanks! I think I get it conceptually now...if the velocity is not changing then the ∆f is the same at all times, so while the pitch is different than if the two people were both standing still, it is still not changing. Also...since this is sound, would you take the equation that EK tells you to use for light (∆f/f(s) = v/c) and swap 340 m/s for c since sound is what we're working with?

Yes, you'd substitute that for c.

Wouldn't the frequency (and, in turn, pitch) increase as the two people got closer?

Nope, just depends on relative velocity.

 

[Charles Darwin]

Nope, just depends on relative velocity.

f = f' (v + v0 / v - vs) => f increases, right?

 

[Jepstein30]

f = f' (v + v0 / v - vs) => f increases, right?

Right, so that equation only shows that velocity affects frequency.. not distance/displacement.

And its also much easier to use deltaf/f=v/c and then just know which way to increase/decrease f.

 

[ridethecliche]

If an object is moving, then the pitch (frequency) changes relative to what it would be if the object were standing still.

The question is asking about the object while it is moving. There is no change because V is constant. If V were changing, then the pitch would be changing as well.

In questions about the doppler effect, the base frequency given is if the things were stationary. This is asking for a shift relative to the pitch while moving.

Intensity does change because that's related to radius from the origin of the sound.

 

[SN2ed]

Thread closed. These types of questions belong in the study Q&A forum.