The apparent frequency of a sound wave as hea | vedclass

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(C) The apparent frequency $f_{app}$ is given as $f_{app} = f + 0.1f = 1.1f$. According to the Doppler effect formula: $f_{app} = f \left[ \frac{v + v

Vedclass · Accepted Answer

$i, iv$

Vedclass · Answer

(C) The apparent frequency $f_{app}$ is given as $f_{app} = f + 0.1f = 1.1f$. According to the Doppler effect formula: $f_{app} = f \left[ \frac{v + v_o}{v - v_s} ight]$. Thus,$1.1 = \frac{v + v_o}{v - v_s}$. Case $1$: If the observer is stationary $(v_o = 0)$,then $1.1 = \frac{v}{v - v_s}$. $1.1(v - v_s) = v \Rightarrow 1.1v - 1.1v_s = v \Rightarrow 0.1v = 1.1v_s$. $v_s = \frac{0.1 	imes 330}{1.1} = 30 \, m/s$. This matches statement $(i)$. Case $2$: If the source is stationary $(v_s = 0)$,then $1.1 = \frac{v + v_o}{v}$. $1.1v = v + v_o \Rightarrow v_o = 0.1v$. $v_o = 0.1 	imes 330 = 33 \, m/s$. This matches statement $(iv)$. Therefore,the correct statements are $(i)$ and $(iv)$.

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