JEE Main 2018 · Physics · Waves and Sound · hard

A source of frequency 500 Hz moves at 10 m s^-1 directly towards a large flat wall, while a stationary observer stands on the line of motion behind…

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A source of frequency

500\ \mathrm{Hz}

moves at

10\ \mathrm{m\,s^{-1}}

directly towards a large flat wall, while a stationary observer stands on the line of motion behind the source (so the source is between the observer and the wall). The observer hears the sound coming directly from the source and also the echo from the wall. Taking the speed of sound as

340\ \mathrm{m\,s^{-1}}

, the number of beats heard per second is nearly

A. $29.4$
B. $44.1$
C. $58.8$
D. $14.7$

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Answer: (A) $29.4$

Direct sound: the source recedes from the observer, so

f_1=\dfrac{500\times340}{340+10}=485.7\ \mathrm{Hz}

. The wall is stationary and receives

f_w=\dfrac{500\times340}{340-10}=515.2\ \mathrm{Hz}

; a stationary reflector re-emits the same frequency, and the stationary observer hears

f_2=515.2\ \mathrm{Hz}

. Beat frequency

=f_2-f_1=515.2-485.7=29.4\ \mathrm{Hz}

. Using only one of the two shifts gives about half of this value,

14.7\ \mathrm{Hz}

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