An observer moves towards a stationary source | vedclass

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(C) When an observer moves towards a stationary source of sound,the apparent frequency $n^{\prime}$ is given by the Doppler effect formula:$n^{\prime}

Vedclass · Accepted Answer

$20$

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(C) When an observer moves towards a stationary source of sound,the apparent frequency $n^{\prime}$ is given by the Doppler effect formula:$n^{\prime} = \left( \frac{v + v_{o}}{v} ight) n$Where $v$ is the velocity of sound and $v_{o}$ is the velocity of the observer.Given that $v_{o} = \frac{v}{5}$,we substitute this into the formula:$n^{\prime} = \left( \frac{v + \frac{v}{5}}{v} ight) n = \left( \frac{\frac{6v}{5}}{v} ight) n = \frac{6}{5} n = 1.2n$The increase in frequency is $\Delta n = n^{\prime} - n = 1.2n - n = 0.2n$.The percentage increase is $\left( \frac{\Delta n}{n} ight) 	imes 100 = \left( \frac{0.2n}{n} ight) 	imes 100 = 20 \%$.

An observer moves towards a stationary source of sound with a velocity of one-fifth of the velocity of sound. The percentage increase in the apparent frequency is (in $\%$)

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