# Frequency Problem

Car 1 moves toward car 2 with speed v. An observer in car 2 measures the frequency of the sound emitted by car 1's horn to be f. Now, car 1 remains stationary while car 2 moves toward car 1 with speed v. The observer in car 2 now measures the frequency of car 1's horn to be f '. The relationship between the observed frequencies is...

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

## Answers and Replies

Curious3141
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Car 1 moves toward car 2 with speed v. An observer in car 2 measures the frequency of the sound emitted by car 1's horn to be f. Now, car 1 remains stationary while car 2 moves toward car 1 with speed v. The observer in car 2 now measures the frequency of car 1's horn to be f '. The relationship between the observed frequencies is...

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.

Its f'=f(v(+/-)vD)/(v(+/-)vS). As car 1 moves toward car 2 it would be a plus sign on the numerator. And as car 2 moves toward car 1, the sign will decrease in the denominator. In the beginning, car 1 moves toward car 2 , then stops, and records the frequency, f. 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.

haruspex
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Its f'=f(v(+/-)vD)/(v(+/-)vS). As car 1 moves toward car 2 it would be a plus sign on the numerator.
Please write out and post those two equations for the specific context of this question.
And as car 2 moves toward car 1, the sign will decrease in the denominator.
Not quite sure what you mean. Signs don't increase or decrease.
In the beginning, car 1 moves toward car 2 , then stops, and records the frequency, f.
That makes it sound as though the measurement is taken while both are stationary. Both measurements are taken with the same closing speed, v.
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.
It's nothing to do with how far apart they are. It all depends on the two speeds relative to the air.

Sorry, 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). As they are moving closer, wouldn't the the observer hear a stronger frequency? So I solve for the first equation, setting the values vs and vd to 2 m/s and the fo to 1 hz, arbitrary numbers. I solve for the first equation, set that to the second equation then it would give a higher frequency right?

haruspex
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Sorry, 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).
Let's be accurate here. fo, the frequency at origin, is the same in both cases:
fs (frequency heard when source moves) = fova/(va-vs)
fd (frequency heard when receiver moves) = fo(va+vd)/va
where v is the velocity of sound in air.
In this case, you have vs=vd=v.
Substitute that into the equations and look at what that gives you for the ratio of fs to fd.
As they are moving closer, wouldn't the the observer hear a stronger frequency?
There's no such concept as a stronger frequency. Do you mean higher 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.

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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haruspex
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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?
Ratios of fs and fd to what? fo? No they're not.
Put vs=vd=v in the equations and write out what that gives you for fs/fd. Please post your working.

Okay, so fs=fo(va)/(va-v) and fd=fo(va+v)/va and if fo and v are constants, lets say f=1 hz and v=2. I see now, so fs is larger. How can fs be larger though? Don't fs and fd have the same fo and v in the first place? This is a conceptual problem by the way.

haruspex
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Okay, so fs=fo(va)/(va-v) and fd=fo(va+v)/va and if fo and v are constants, lets say f=1 hz and v=2. I see now, so fs is larger. How can fs be larger though? Don't fs and fd have the same fo and v in the first place? This is a conceptual problem by the way.
The equations are different. Why shouldn't they produce different answers?
What do you get for fs/fd (for the third time of asking)?

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

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

It is more than one. But I' still don't understand why they produce different numbers, unfortunately.

haruspex
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It is more than one. But I' still don't understand why they produce different numbers, unfortunately.
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.

I 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? Btw, thanks for being patient with me.

haruspex