Determining Frequency Shift for Moving Sound Sources

[CAF123]

I was considering the following question;
'A fire engine traveling at 70m/s drives past a van traveling at 30m/s. The speed of sound in air is 330m/s and the fire engine has a frequency of 200Hz when stationary. At what frequency does the van driver hear the siren?'

I understand you have to use f(prime) = f((v+/-vd)/(v+/-vs)), however I find it quite difficult to decide whether to use the + or - in the calculation, when things are moving relative to one another. An explanation of when to use a + and when to use a - in each of the numerator/denominator would be most helpful.
The answer is 180Hz.
Thanks.

 

[Doc Al]

I would write the formula this way:
$$f' = f\frac{v + v_d}{v - v_s}$$
All you have to remember is that when either source or detector moves towards the other, the observed frequency must increase thus the sign for that speed must be positive (in my version of the formula).

In this problem is the source (fire engine) moving towards (+) or away (-) from the detector? Is the detector (van) moving towards or away from the source? Use that as your guide to get the proper signs.

 

[CAF123]

Thanks for your explanation.
I was just wondering if at first the van is in front of the fire engine and the fire engine is approaching it from behind (the fire engine is moving faster than the van as stated in q) then the apparent frequency detected by the van would be greater, right? Then the fire engine overtakes the van and the apparent freq decreases? How does this fit into the eqn?

Also, the answer for this question has a plus everywhere ie v+vd and v+vs.
Many thanks.

 

[Doc Al]

I was just wondering if at first the van is in front of the fire engine and the fire engine is approaching it from behind (the fire engine is moving faster than the van as stated in q) then the apparent frequency detected by the van would be greater, right? Then the fire engine overtakes the van and the apparent freq decreases? How does this fit into the eqn?

Yes, when the fire engine overtakes the van things will change. Let's get the first case settled before worrying about the second. (You'll see that the same rules will solve all cases, once you understand them.)

Also, the answer for this question has a plus everywhere ie v+vd and v+vs.

That's correct. To see if you understand what I explained earlier, answer the questions I asked in my last post.

Hint: Treat the motion of the detector (in the numerator) and the source (in the denominator) as separate effects which are combined by the formula.

 

[CAF123]

Thanks, I have the first case settled. If we consider the situation that the source is always behind the detector, then the source moves toward the detector thus we have v+vd. The detector is always moving away from the source and so this acts to decrease apparent f. v+/-vs is in the denominator and so to make this smaller we use v+vs, as needed.

Then I considered a second case. The source has overtaken the detector and is always in front of the detector. The source therefore moves away from the detector,acting to decrease f, and so we have v-vd. The detector moves towards the source, increasing f and giving v-vs on the denominator(to divide by something smaller). I think there should be an error in this second argument because this would give an apparent freq of 230Hz. Since the source moves quicker than the detector, we should expect a decrease in freq, yes?

Lastly, when answering this question how were we supposed to know whether to consider the source behind the detector or in front to get the specific freq asked for in the q?

Thanks again.

 

[Doc Al]

Thanks, I have the first case settled. If we consider the situation that the source is always behind the detector, then the source moves toward the detector thus we have v+vd. The detector is always moving away from the source and so this acts to decrease apparent f. v+/-vs is in the denominator and so to make this smaller we use v+vs, as needed.

Actually, on second thought, the given answer is for the second case where the fire engine has already passed the van, which is what the question asks about. Before the fire engine overtakes the van, the analysis should be as follows. Since the source is moving towards the detector, it tends to increase the frequency. (Thus v - 70 in the denominator.) And since the detector is moving away from the source, it acts to decrease the frequency. (Thus v - 30 in the numerator.) Crank that out to get the apparent frequency before the fire engine overtakes the van.

Then I considered a second case. The source has overtaken the detector and is always in front of the detector. The source therefore moves away from the detector,acting to decrease f, and so we have v-vd. The detector moves towards the source, increasing f and giving v-vs on the denominator(to divide by something smaller). I think there should be an error in this second argument because this would give an apparent freq of 230Hz. Since the source moves quicker than the detector, we should expect a decrease in freq, yes?

Yes, there's a mistake in your analysis. Once the fire engine has overtaken the van, the analysis should be as follows. The source is now moving away from the detector, acting to decrease f, and so we have v + 70 in the denominator. And the detector moves towards the source, so v + 30 in the numerator.

You should expect a decrease in frequency when the fire engine overtakes the van and that's just what you get.

Lastly, when answering this question how were we supposed to know whether to consider the source behind the detector or in front to get the specific freq asked for in the q?

In this case it says it "drives past", so that implies that it has passed the van. If it said "drives toward" then it would imply that it hadn't yet overtaken the van. But it always pays to ask if you are unsure (like on a test, for instance).

 

[CAF123]

I just reliased I had confused the two cases immediately after I had left the last post.
Everything is now clear for me. Thanks a lot!