A and B are two sources generating sound wave | vedclass

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(C) Let $f_A = 500 \, Hz$ be the frequency of source $A$,$f_B$ be the frequency of source $B$,$v = 340 \, m/s$ be the speed of sound,and $v_s = 4 \, m

Vedclass · Accepted Answer

$512$

Vedclass · Answer

(C) Let $f_A = 500 \, Hz$ be the frequency of source $A$,$f_B$ be the frequency of source $B$,$v = 340 \, m/s$ be the speed of sound,and $v_s = 4 \, m/s$ be the speed of source $A$.Case $1$: When source $A$ moves towards the stationary listener at $C$,the apparent frequency $f'_A$ is given by:$f'_A = f_A \left( \frac{v}{v - v_s} \right) = 500 \left( \frac{340}{340 - 4} \right) = 500 \left( \frac{340}{336} \right) \approx 505.95 \, Hz$.Case $2$: When source $A$ moves away from the stationary listener at $C$,the apparent frequency $f''_A$ is given by:$f''_A = f_A \left( \frac{v}{v + v_s} \right) = 500 \left( \frac{340}{340 + 4} \right) = 500 \left( \frac{340}{344} \right) \approx 494.19 \, Hz$.Let $f_B$ be the frequency of source $B$. The beat frequency is $|f'_A - f_B| = 6$ and $|f''_A - f_B| = 18$.From Case $1$: $f_B = f'_A \pm 6 = 505.95 \pm 6$,so $f_B \approx 511.95 \, Hz$ or $499.95 \, Hz$.From Case $2$: $f_B = f''_A \pm 18 = 494.19 \pm 18$,so $f_B \approx 512.19 \, Hz$ or $476.19 \, Hz$.Comparing both cases,$f_B \approx 512 \, Hz$ is the common solution.

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