A source of sound emits a sound wave of frequ | vedclass

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Question

(C) According to the Doppler effect,the apparent frequency $f'$ is given by the formula: $f' = f \left( \frac{V \pm v_o}{V \mp v_s} \right)$.Here,$V$

Vedclass · Accepted Answer

$\frac{6}{5} f$

Vedclass · Answer

(C) According to the Doppler effect,the apparent frequency $f'$ is given by the formula: $f' = f \left( \frac{V \pm v_o}{V \mp v_s} ight)$.Here,$V$ is the velocity of sound,$v_s = \frac{V}{3}$ is the velocity of the source moving towards the observer,and $v_o = \frac{V}{5}$ is the velocity of the observer moving away from the source.Since the source moves towards the observer,the denominator becomes $(V - v_s) = (V - \frac{V}{3}) = \frac{2V}{3}$.Since the observer moves away from the source,the numerator becomes $(V - v_o) = (V - \frac{V}{5}) = \frac{4V}{5}$.Substituting these values into the formula:$f' = f \left( \frac{4V/5}{2V/3} ight) = f \left( \frac{4}{5} 	imes \frac{3}{2} ight) = f \left( \frac{12}{10} ight) = \frac{6}{5} f$.Thus,the apparent frequency is $\frac{6}{5} f$.

$A$ source of sound emits a sound wave of frequency '$f$' and moves towards an observer with a velocity $\frac{V}{3}$,where $V$ is the velocity of sound. If the observer moves away from the source with a velocity $\frac{V}{5}$,the apparent frequency heard by him will be:

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