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Two submarines are moving towards each other. The first submarine emits a wave towards the second. The wave bounces off the second submarine and reaches the first submarine again.

How should one calculate the frequency of the reflected wave?

The main question here is what speeds should we consider for the reflection when calculating the Doppler effect. From what I understand, if the second submarine is moving when the reflection is happening, so both the observer and the source are still moving towards each other, hence:

f' = f * [ (v-o) / (v-s) ]

Where o is the speed of the observer (first sub) and s is the speed of the source (second sub).

But here's the thing, when discussing this in class, my physics teacher mentioned that the speed of the source (second sub) should be zero, because we already considered its speed when calculating the frequency of the wave before the reflection, in that case the speed of the observer.

And I really couldn't understand this.

Why shouldn't I consider anymore the speed of the source when reflecting a Doppler affected wave?

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  • $\begingroup$ Duplicate? physics.stackexchange.com/questions/154880/… $\endgroup$ – Farcher May 23 '17 at 14:39
  • $\begingroup$ The wall is stationary in that question. While that question does confirm that the reflection will be shifted -as expected- it doesn't confirm how to calculate the reflection if the reflection's source is in movement. $\endgroup$ – Johnny Bigoode May 23 '17 at 15:53
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I recommend breaking this into two smaller problems by inserting a "stationary observer" between the two subs. Denote the frequency of the signal that leaves sub 1 as $f_1$, observed at the stationary point $f_s$, observed by the (moving) sub 2 as $f_2$. Upon reflection, the frequency of the signal is unchanged (in the frame of reference of sub 2), but the signal received at the intermediate point $f_s'$ is obviously different. Now consider $f_s'$ the stationary transmitter - what is the frequency $F_1'$ that will be observed by the first sub?

Each of the steps of the above is something you know how to do. Let me step you through it: assuming velocity of sub 1 towards the stationary observer is $v_1$, velocity of sub 2 towards the observer is $v_2$, and speed of sound in water is c:

$$f_s = \frac{c}{c-v_1} f_1$$

When that signal is received by sub 2, it sees

$$f_2 = \frac{c+v_2}{c} f_s = \frac{c+v_2}{c-v_1}f_1$$

Now that observed frequency, upon reflection, becomes a source of sound moving towards SO with velocity $v_2$, so the reflected frequency seen by the stationary observer is

$$f_s' = \frac{c}{c-v_2}f_2$$

and finally, this frequency is observed by sub 1 moving towards the SO:

$$f_1' = \frac{c+v_1}{c}f_s' = \frac{(c+v_1)(c+v_2)}{(c-v_1)(c-v_2)} f_1$$

The velocity of both subs relative to the medium (water) is what matters - not their relative velocity. This is because the velocity of the sound is fixed relative to the medium. Note that the answer would be slightly different if you had light waves in vacuum, whose speed is the same regardless of the frame of reference.

If you look at this Khan academy video, around 3:30 you get the same principle explained (the reflecting object becomes a source that re-radiates at the frequency it "hears").

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  • $\begingroup$ The signal received at the intermediate point f's is not obviously different, that's the core of my question. Maybe I'm something is lost in translation, but even with the stationary observer, according to the concept my physics teacher presented, the stationary observer would see the same frequency reflected from sub2 as the frequency that sub2 would see, because, as stated in the question, you shouldn't consider sub2's velocity when calculating the Doppler effect upon reflection since sub2's speed was already in account when the wave was coming from the original source. $\endgroup$ – Johnny Bigoode May 23 '17 at 19:46
  • $\begingroup$ My point is - forget about sub 2. Look at sub 1 and the stationary observer. The SO sees a higher frequency because sub 1 comes towards him. Now emit that higher frequency (from SO) to sub 2; since sub 2 is moving towards you, the reflected frequency will be higher. Now repeat one more time, looking at SO sending the higher frequency back to sub 1 (which is moving towards you). Once more the frequency is increased. Put it all together... what do you get. $\endgroup$ – Floris May 23 '17 at 19:49
  • $\begingroup$ I'm missing the part where the wave reflects from sub2 to the SO before going back to sub1. It's clear that from SO to sub1 the frequency should be higher because sub1 is moving towards SO, but what about the reflection from sub2 to the SO? Shouldn't that frequency be higher also since sub2 is moving towards SO too? $\endgroup$ – Johnny Bigoode May 23 '17 at 19:52
  • $\begingroup$ Yes. sub 2 is moving towards SO, so you get two shifts of frequency. Once, the observed frequency from SO to S2 is shifted by $\frac{v+v2}{v}$; then, since this is now a "moving source", the frequency received by SO is shifted again by $\frac{v}{v-v2}$ so the total frequency shift that SO sees between the frequency it observed from S1 to the frequency received back from S2 is $\frac{v+v2}{v-v2}$ greater. A similar effect (but with $v_1$) is going on at the other side. You can't just take the difference in velocities between the subs as their velocity relative to the medium is what matters. $\endgroup$ – Floris May 23 '17 at 19:57
  • $\begingroup$ So the original conclusion I made was correct, right? The SO would see the reflected wave with the Doppler effect because we must consider sub2 as now a "moving source"? I'll try to study this whole answer more calmly because I'l still having issues to understand the math behind it, the whole concept still feels like a gut feeling for me somehow. $\endgroup$ – Johnny Bigoode May 23 '17 at 20:05
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The Doppler effect just depends on the relative velocity between the two objects. So we could assume the 'listener' is at rest while the 'reflector' is moving at a speed of v + v' where v' is velocity of 'listener'.

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    $\begingroup$ Not for sound. The motion of the source produces a different frequency shift than the motion of the observer. The speeds in the Doppler formulas are relative to the medium. $\endgroup$ – nasu May 23 '17 at 16:57
  • $\begingroup$ my mistake. sorry. $\endgroup$ – jmh May 23 '17 at 19:49

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