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JEE Main 2020 · Physics · Waves and Sound · medium

A source emitting a note of frequency 1155 Hz moves at 10 m s^-1 directly towards a large flat wall. A stationary listener stands on the line joining…

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A source emitting a note of frequency 1155 Hz1155\ \mathrm{Hz} moves at 10 m s−110\ \mathrm{m\,s^{-1}} directly towards a large flat wall. A stationary listener stands on the line joining source and wall, on the side of the source away from the wall. Taking the speed of sound as 340 m s−1340\ \mathrm{m\,s^{-1}}, the beat frequency heard by the listener (between the direct sound and the sound reflected from the wall) is
  1. A. 34 Hz34\ \mathrm{Hz}
  2. B. 68 Hz68\ \mathrm{Hz}
  3. C. 100 Hz100\ \mathrm{Hz}
  4. D. 136 Hz136\ \mathrm{Hz}
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Answer: (B) 68 Hz68\ \mathrm{Hz}

The listener is behind the source, so the direct sound comes from a receding source: f1=f vv+vs=1155⋅340350f_1 = f\,\dfrac{v}{v+v_s} = 1155\cdot\dfrac{340}{350}. The wall receives and reflects the frequency emitted forward by an approaching source, f2=f vv−vs=1155⋅340330f_2 = f\,\dfrac{v}{v-v_s} = 1155\cdot\dfrac{340}{330}; the stationary listener hears this frequency. Beats =f2−f1=1155×340×(1330−1350)=68 Hz= f_2 - f_1 = 1155\times340\times\left(\dfrac1{330}-\dfrac1{350}\right) = 68\ \mathrm{Hz}.
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