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JEE Main 2019 · Physics · Waves and Sound · hard

A source emitting a note of frequency 680 Hz moves at 10 m s^-1 directly toward a large plane wall. A stationary observer stands on the line of…

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A source emitting a note of frequency 680 Hz680\ \mathrm{Hz} moves at 10 m s−110\ \mathrm{m\,s^{-1}} directly toward a large plane wall. A stationary observer stands on the line of motion behind the source, so that he receives sound directly from the source and also the sound reflected by the wall. The speed of sound is 340 m s−1340\ \mathrm{m\,s^{-1}}. The beat frequency heard by the observer is nearly
  1. A. 20 Hz20\ \mathrm{Hz}
  2. B. 60 Hz60\ \mathrm{Hz}
  3. C. 40 Hz40\ \mathrm{Hz}
  4. D. 80 Hz80\ \mathrm{Hz}
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Answer: (C) 40 Hz40\ \mathrm{Hz}

The observer is behind the source, which moves away from him, so the direct sound has f1=680×340340+10≈660.6 Hzf_1 = \dfrac{680\times340}{340+10}\approx 660.6\ \mathrm{Hz}. The wall is in front of the source and receives fw=680×340340−10≈700.6 Hzf_w=\dfrac{680\times340}{340-10}\approx700.6\ \mathrm{Hz}. A stationary wall reflects this same frequency back, and the stationary observer hears f2=700.6 Hzf_2=700.6\ \mathrm{Hz}. The beat frequency is f2−f1≈40 Hzf_2-f_1\approx 40\ \mathrm{Hz}. Note that the wall acts as a stationary source of the reflected wave, so no second Doppler shift occurs.
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Original question written in the JEE Main 2019 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake