Doppler's Effect TPR

[pm1]

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The engine of a small unmanned airplane produces a known and specific sound frequency (not given). A stationary sound detection device observes that the known emitted sound frequency is 90% of the perceived sound frequency. Relative to the detection device, in which direction and at what velocity is the plane moving?

A. Towards the detector, at 34m/s
B. Away from the detector, at 34m/s
C. Towards the detector, at 64m/s
D. Away from the detector, at 64m/s

Answer: A

I was able to eliminate B and D. And then I just guessed on A thinking that 34 is 10% of 340m/s which ended up being correct but I can't understand the math on the back of the book. Could someone please work out the math behind this problem for me?

Thank you!

 

[cloak25]

found this equation on wiki. http://en.wikipedia.org/wiki/Doppler_effect

f = fo [1-(Vs-Vr)/c]
Vs-Vr = velocity of source relative to the receiver
c = speed of sound
f = 0.90fo

0.90fo = fo [1-(Vs-0)/340]
0.90 = 1 - Vs/340 = (340-Vs)/340
306 = 340 - Vs
34 = Vs

I'm not quite sure how to determine the direction though.

 

[deleted103644]

found this equation on wiki. http://en.wikipedia.org/wiki/Doppler_effect

f = fo [1-(Vs-Vr)/c]
Vs-Vr = velocity of source relative to the receiver
c = speed of sound
f = 0.90fo

0.90fo = fo [1-(Vs-0)/340]
0.90 = 1 - Vs/340 = (340-Vs)/340
306 = 340 - Vs
34 = Vs

I'm not quite sure how to determine the direction though.

Frequency increases traveling towards the detector, decreases away.

You can picture it and it makes sense - the waves are getting spaced more closely together due to the relative movement.

 

[cloak25]

so in this case, if the velocity was found to be -34, it would be away from ?

 

[MedPR]

so in this case, if the velocity was found to be -34, it would be away from ?

Not necessarily.

Directionality in doppler affects the observed frequency. If source/observer have a net movement away from each other, the observed frequency is lower than the actual frequency. Opposite if there is a net movement toward each other.

 

[pm1]

found this equation on wiki. http://en.wikipedia.org/wiki/Doppler_effect

f = fo [1-(Vs-Vr)/c]
Vs-Vr = velocity of source relative to the receiver
c = speed of sound
f = 0.90fo

0.90fo = fo [1-(Vs-0)/340]
0.90 = 1 - Vs/340 = (340-Vs)/340
306 = 340 - Vs
34 = Vs

I'm not quite sure how to determine the direction though.

thanks!

should we also memorize this version of the formula? I am not sure if I could derive from the traditional one at a reasonable amount of time.. (it took me about 5min to derive just now) 😱

 

[MedPR]

thanks!

should we also memorize this version of the formula? I am not sure if I could derive from the traditional one at a reasonable amount of time.. (it took me about 5min to derive just now) 😱

No, the way you solved it (intuitively) will be enough for the MCAT. It's also significantly faster and leaves you a smaller chance for math errors.

 

[pm1]

No, the way you solved it (intuitively) will be enough for the MCAT. It's also significantly faster and leaves you a smaller chance for math errors.

which would be by realizing that 34m/s is 10% of 340m/s thus it should give a difference in frequency of 10%?

thanks you!