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a)f<f ' b)f=f ' c)f=2f ' d)f=(1/2)f ' e) f>f'

I chose a because as the cars moves closer the higher the frequency right? My teacher said its e and it was through the scantron score checker. Please explain how I am wrong? Thanks

-vu2

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- Thread starter VU2
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- #1

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a)f<f ' b)f=f ' c)f=2f ' d)f=(1/2)f ' e) f>f'

I chose a because as the cars moves closer the higher the frequency right? My teacher said its e and it was through the scantron score checker. Please explain how I am wrong? Thanks

-vu2

- #2

Curious3141

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a)f<f ' b)f=f ' c)f=2f ' d)f=(1/2)f ' e) f>f'

I chose a because as the cars moves closer the higher the frequency right? My teacher said its e and it was through the scantron score checker. Please explain how I am wrong? Thanks

-vu2

In the classical Doppler effect, what are the formulae relating apparent to true frequency when a) the source is moving toward a stationary receiver, and b) the receiver is moving toward a stationary source?

The forms of the equations are qualitatively different. Write the formulae down, and it should become clearer.

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Please write out and post those two equations for the specific context of this question.Its f'=f(v(+/-)vD)/(v(+/-)vS). As car 1 moves toward car 2 it would be a plus sign on the numerator.

Not quite sure what you mean. Signs don't increase or decrease.And as car 2 moves toward car 1, the sign will decrease in the denominator.

That makes it sound as though the measurement is taken while both are stationary. Both measurements are taken with the same closing speed, v.In the beginning, car 1 moves toward car 2 , then stops, and records the frequency, f.

It's nothing to do with how far apart they are. It all depends on the two speeds relative to the air.Now car 2 moves toward car 1, which will now have a higher frequency,f '. Or are both situations separate from the two? Because I'm assuming the two cars are getting closer and closer.

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Let's be accurate here. fSorry, I'm still having trouble grasping the concept. So,t he source is moving toward a stationary receiver (1) f=fo(v/v-vs), and the the receiver is moving toward a stationary source, (2) f'=f(v+vs/v).

f

f

where v is the velocity of sound in air.

In this case, you have v

Substitute that into the equations and look at what that gives you for the ratio of f

There's no such concept as a stronger frequency. Do you mean higher frequency?As they are moving closer, wouldn't the the observer hear a stronger frequency?

If they're moving closer at a steady speed, the frequency will be higher than when both stationary, but it's by a fixed amount. It won't get progressively higher as they get closer. It will get louder.

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Yes, sorry i meant higher frequency. So if I am correct, what you explained in the equations and situation shows that they both have same frequency because the ratios of fs and fd are the same right?

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Ratios of fYes, sorry i meant higher frequency. So if I am correct, what you explained in the equations and situation shows that they both have same frequency because the ratios of fs and fd are the same right?

Put v

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The equations are different. Why shouldn't they produce different answers?

What do you get for f

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So fs/fd=va^2/(va^2-v^2)

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Right. Is that more or less than 1?So fs/fd=va^2/(va^2-v^2)

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You have to consider the role of the medium.

It might help to take an extreme example. Suppose a source is travelling at almost the speed of sound. The wavefronts ahead of it become very tightly bunched. One wavefront has not got very far ahead before the next one is emitted. The frequency increases without limit.

Now suppose instead that the receiver is moving towards a static source at nearly the speed of sound. As it passes through one wavefront, the distance to the next is still the same wavelength as emitted at the source. With the receiver and wavefront each travelling at nearly the speed of sound towards each other, the time taken to cross successive wavefronts is halved. The frequency is therefore doubled at most.

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Yes, but I think you meant "the source is movingI think I understand what you meant. In general, because the the source is moving with the medium, the air, it will have a higher frequency than when the source is static and receiver moving at the same velocity?

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