An observer moves away from a stationary source of sound with a velocity one-fifth of the velocity of sound. Find the percentage decrease in the apparent frequency. (Take, sound in air = $320\ \mathrm{ms}^{-1}$)

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Aicharya · Accepted Answer

Sol. Given, $v_o = \dfrac{v}{5}$ ⇒ $v_o = \dfrac{320}{5} = 64\ \mathrm{ms}^{-1}$ When observer moves away from the stationary source, then $f' = \left(\dfrac{v - v_o}{v}ight) f$ ⇒ $f' = \left(\dfrac{320 - 64}{320}ight) f$ ⇒ $f' = \left(\dfrac{256}{320}ight) f$ ⇒ $\dfrac{f'}{f} = \dfrac{256}{320}$ Hence, percentage decrease in frequency, $\left(\dfrac{f' - f}{f}ight) = \left(\dfrac{320 - 256}{320} 	imes 100ight)\ %$ $f = \left(\dfrac{64}{320} 	imes 100ight)\ % = 20\%$ Answer: $20\%$

An observer moves away from a stationary source of sound with a velocity one-fifth of the velocity of sound.…

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