Suppose a sound source S is moving towards a stationary wall and that a listener L1 is moving with the sound source. Describe in a few lines why the reflected sound heard by the listener L1 is Doppler-shifted by about twice the amount that a second listener L2 would hear standing by the wall.(adsbygoogle = window.adsbygoogle || []).push({});

The sound waves in front of the moving source are compressed causing an apparent increase in frequency to [itex]f_{L_2} = \frac{v}{v - v_S}f_S[/itex], where v is the speed of sound. The sound reflected from the wall has the same frequency [itex]f_{L_2}[/itex]. L1 is approaching the wall with speed [itex]v_S[/itex], so the relative speed of the wavefronts of sound is [itex]v + v_S[/itex]. The frequency observed by L1 is [itex]f_{L_1} = \frac{v + v_S}{v}f_{L_2} = \frac{v + v_S}{v - v_S}f_{S}[/itex] which is approximately equal to [itex]2f_{L_2}[/itex] for [itex]v_S[/itex] close to [itex]v[/itex]. Does this answer make sense, or is it unnecessarily complicated?

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# Homework Help: Doppler effect: Moving source with reflection

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