An observer moves towards a stationary source of sound with a velocity one-fifth of the velocity of sound. The precentage change in the apparent frequency is ..... $\%$
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When an observer moves towards a stationary source, the apparent frequency heard by the observer is
$\mathrm{v}^{\prime}=\mathrm{v}_{0}\left(\frac{\mathrm{v}+\mathrm{v}_{0}}{\mathrm{v}}\right)$
where
$\mathrm{v}_{0}=$ frequency fo the source
$\mathrm{v}_{0}=$ velocity of the observer
$\mathrm{v}=$ velocity of the sound
$\therefore $ $\mathrm{v}^{\prime}=\mathrm{v}_{0}\left(\frac{\mathrm{v}+\frac{\mathrm{v}}{5}}{\mathrm{v}}\right)=\frac{6 \mathrm{v}_{0}}{5}$
Percentage change in apparent frequency
$=\frac{v^{\prime}-v_{0}}{v_{0}} \times 100$
$=\frac{\left(\frac{6 v_{0}}{5}-v_{0}\right)}{v_{0}} \times 100=\frac{1}{5} \times 100=20 \%$
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