A stationary source emits sound of frequency = 492 Hz. The sound is reflected by a large car approaching the source with a speed of 2 . The reflected signal is received by the source and superposed with the original. What will be the beat frequency of th

Question

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.) [2017]

Smart Achievers · Accepted Answer

(6)

$Frequency observed at car$

$ν_{1} = ν_{0} (\frac{v + v_{c}}{v})$

$Frequency of reflected sound as observed at the source$

$ν_{2} = ν_{1} (\frac{v}{v - v_{c}}) = ν_{0} (\frac{v + v_{c}}{v - v_{c}})$

$∴$ $Beat frequency = ν_{2} - ν_{0}$

$= ν_{0} [\frac{v + v_{c} - v + v_{c}}{v - v_{c}}] = ν_{0} [\frac{2 v_{c}}{v - v_{c}}]$

$= 492 [\frac{2 \times 2}{330 - 2}] = \frac{492 \times 4}{328} = 6 Hz$

Solution

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Solution

Ans. (6) Frequency observed at car ν1=ν0(v+vcv) Frequency of reflected sound as observed at the source ν2=ν1(vv-vc)=ν0(v+vcv-vc) ∴ Beat frequency=ν2-ν0 =ν0[v+vc-v+vcv-vc]=ν0[2vcv-vc] =492[2×2330-2]=492×4328=6 Hz

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