EK Physics #758

[drillers]

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Which of the following will result in the greatest increase in the observed frequency of the sound?

a. the source moves toward the observer at 30 m/s
b. the observer moves toward the source at 30 m/s
c. no way
d. A and B will result in exactly the same frequency change

I know words like always, exactly, never, etc. can raise red flags on the mcat, but I surely thought that with the same velocity, A and B result in the same frequency change.

I mean if a train goes 30 m/s toward me or if I go 30 m/s toward a train, isn't the observed frequency the same?

EK says A is the correct answer and I do not understand their reasoning for it at all.

 

[FeinMS]

I agree with D too..

 

[milski]

EK is actually correct. One way to think about is what will happen if 30 m/s was (almost) the speed of sound. When the transmitter moves with that speed, by the time she transmits the next wave, she will have caught up with the previous and the transmission frequency will be very, very high, almost infinite.

Now, let's consider the receiver moving with the same speed towards the transmitter. To cover the distance between two waves, he'll need only half the time that he'd need if he was stationary and that will result in only doubling the frequency.

The difference comes from the fact that the receiver deals with waves which already have some fixed space between each peak. That is not true for the transmitter - she catches up with the previous wave.

The brute force way is just to use the formula f=(c+v_r)/(c-v_t) f_orig

 

[FeinMS]

What the.. lol that's hard.. I think I got it though. OP, if you want a mathematical proof, here it is.

If receiver moves toward the source at a speed of "A meters per second," F=(V+A)/V*f, where F is shifted frequency and f is original frequency (didn't want to right fo bc it's ugly.) If the source moves toward the receiver, F=V/(V-A)*f. Dividing the former by the latter gives, ((V+A)/V)/(V/(V-A))=(V^2-A^2)/V^2=1-(A/V)^2. Since A>0, V>0, 1-(A/V)^2<1. This means (V+A)/V < V/(V-A), so the second F is bigger than the first F. No way you would be doing this on MCAT unless you can finish PS with extra 10 minutes left, but I still don't understand how conceptually it works.

 

[milski]

Yes, doing the math during the exam would be hard or at least time consuming. Conceptually I would at least remember that the sender and the receiver are not equivalent, in a sense that same change in speed for them does not result in the same change of frequency.

And easy way to do the math is to take the numbers to some extreme, like A=V. Should be easy to see that one becomes infinity and the other only doubles.

 

[drillers]

Yes, doing the math during the exam would be hard or at least time consuming. Conceptually I would at least remember that the sender and the receiver are not equivalent, in a sense that same change in speed for them does not result in the same change of frequency.

And easy way to do the math is to take the numbers to some extreme, like A=V. Should be easy to see that one becomes infinity and the other only doubles.

Thanks for explaining the correct answer even if it sort of boggles my noodle. I will keep my fingers crossed something like this would not be on the MCAT 🙂