JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Waves & Sound

A stationary source emits sound of frequency

f_{0} = 492 Hz

. The sound is reflected by a large car approaching the source with a speed of

2 ms^{- 1}

. The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz? (Given that the speed of sound in air is

330 ms^{- 1}

and the car reflects the sound at the frequency it has received).

Answer: 6

Solution

The car acts as a moving observer approaching the source, receiving frequency

f_{1} = f_{0} (1 + \frac{v _{c a r}}{v _{so u n d}}) = 492 (1 + \frac{2}{330})

. The car then acts as a moving source approaching the original source, emitting frequency

f_{2} = \frac{f _{1}}{1 - \frac{v _{c a r}}{v _{so u n d}}} = \frac{f _{0} ( 1 + \frac{2}{330} )}{1 - \frac{2}{330}}

. For small velocities,

f_{2} \approx f_{0} (1 + \frac{2 v _{c a r}}{v _{so u n d}}) = 492 (1 + \frac{4}{330}) = 492 \times \frac{334}{330}

. The beat frequency is

f_{B} = f_{2} - f_{0} = 492 \times \frac{4}{330} = \frac{1968}{330} \approx 5.96 \approx 6

Hz.

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Integer answer

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JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Waves & Sound

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