A stationary source of sound A is producing s | vedclass

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(C) Let $v = 340 \, ms^{-1}$ be the speed of sound, $v_o$ be the speed of the observer, $f_A = 170 \, Hz$ be the frequency of source $A$, and $f_B = 2

Vedclass · Accepted Answer

$68$

Vedclass · Answer

(C) Let $v = 340 \, ms^{-1}$ be the speed of sound, $v_o$ be the speed of the observer, $f_A = 170 \, Hz$ be the frequency of source $A$, and $f_B = 240 \, Hz$ be the frequency of source $B$.Since the observer is moving towards $A$, the apparent frequency heard from $A$ is $f'_A = f_A \left( \frac{v + v_o}{v} \right)$.The observer is moving away from $B$ (since $B$ is moving towards $A$ and the observer is between them moving towards $A$), so the apparent frequency heard from $B$ is $f'_B = f_B \left( \frac{v - v_o}{v - v_B} \right)$, where $v_B = 20 \, ms^{-1}$.For the number of beats to be zero, $f'_A = f'_B$.$170 \left( \frac{340 + v_o}{340} \right) = 240 \left( \frac{340 - v_o}{340 - 20} \right)$.$170 \left( \frac{340 + v_o}{340} \right) = 240 \left( \frac{340 - v_o}{320} \right)$.$\frac{17}{34} (340 + v_o) = \frac{24}{32} (340 - v_o)$.$0.5 (340 + v_o) = 0.75 (340 - v_o)$.$170 + 0.5 v_o = 255 - 0.75 v_o$.$1.25 v_o = 85$.$v_o = \frac{85}{1.25} = 68 \, ms^{-1}$.

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