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If we take a source with a frequency of $230Hz$ moving at $30ms^{-1}$ right and the listener moving $25ms^{-1}$ right.Let the source be behind the listener at some distance so that the distance between them decreases.Take the speed of sound as $340ms^{-1}$. enter image description here

With the speaker as the frame of reference, the listener is moving $5ms^{-1}$ towards the source.

$$f_L = \left(\frac{c \pm v_L}{c \pm v_S}\right)f_S$$

$$\qquad = \left(\frac{340 + 0}{340 - 5}\right)230$$

$$\quad = 233.4328Hz$$

With the Listener as the frame of reference, the source is moving $5ms^{-1}$ towards the listener.

$$f_L = \left(\frac{340 + 5}{340 + 0}\right)230$$

$$\quad= 233.3824Hz$$

With the system as the frame of reference, the source is moving at $30ms^{-1}$ right and the listener moving $25ms^{-1}$ right.

$$f_L = \left(\frac{340 - 25}{340 - 30}\right)230$$

$$ = 233.7097Hz$$

Why are the answers all different? Mathematically it makes sense but in terms of the real world, I don't understand.

Thank you.

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  • $\begingroup$ Isn’t the listener moving away from the source? In any frame, the distance between them must always be increasing. $\endgroup$ – Superfast Jellyfish Apr 20 '20 at 14:26
  • $\begingroup$ @SuperfastJellyfish I see that there is an ambiguity in my question and will make an edit $\endgroup$ – Oliver Hope Apr 20 '20 at 14:31
  • $\begingroup$ If both are moving in the same direction, the distance won’t decrease. $\endgroup$ – Superfast Jellyfish Apr 20 '20 at 14:42
  • $\begingroup$ @SuperfastJellyfish If the velocities are different they will. Should I attach a diagram? $\endgroup$ – Oliver Hope Apr 20 '20 at 14:43
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    $\begingroup$ @SuperfastJellyfish I have added a diagram $\endgroup$ – Oliver Hope Apr 20 '20 at 15:01
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You don't have the freedom to change reference frames as you do in your question, because the speed of sound is relative to the stationary air through which it travels. If you are using $c=340$ m/s, then you need to similarly measure the velocities of the observer and the source in a reference frame in which the air isn't moving.

If you use the listener's frame of reference for example, then you would have $v_L=0$ and $v_S=5$, as you say. However, the air would be moving with constant speed $25$ m/s to the left, which means that sound would travel at $340+25=365$ m/s to the left and $340-25=315$ m/s to the right. Since the waves that the listener hears are moving to the right, you would have

$$f_L = f_0\left(\frac{315}{315-5}\right)$$

which matches the calculation performed in the frame in which the air is stationary.

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  • $\begingroup$ I see, that makes sense thank you for explaining.😊 I was worried it was going to be more complex😆 $\endgroup$ – Oliver Hope Apr 20 '20 at 15:12
  • $\begingroup$ I believe it should be 315 - 5 as the source is moving towards the listener. $\endgroup$ – Oliver Hope Apr 20 '20 at 17:18
  • $\begingroup$ @OliverHope Yes, you're right. Fixed. $\endgroup$ – J. Murray Apr 20 '20 at 17:55

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