Two sources A and B are producing notes of fr | vedclass

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(B) Given:Frequency of sources $f_A = f_B = 680 \,Hz$.Speed of sound $v_s = 340 \,ms^{-1}$.Velocity of the listener is $v$.Beat frequency $n = 10 \,Hz

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$2.5$

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(B) Given:Frequency of sources $f_A = f_B = 680 \,Hz$.Speed of sound $v_s = 340 \,ms^{-1}$.Velocity of the listener is $v$.Beat frequency $n = 10 \,Hz$.As the listener moves from $A$ to $B$, they are moving towards $B$ and away from $A$.The apparent frequency from $A$ is $f'_A = f_A \left( \frac{v_s - v}{v_s} \right)$.The apparent frequency from $B$ is $f'_B = f_B \left( \frac{v_s + v}{v_s} \right)$.The beat frequency is $n = f'_B - f'_A$.$10 = 680 \left( \frac{340 + v}{340} \right) - 680 \left( \frac{340 - v}{340} \right)$.$10 = \frac{680}{340} (340 + v - 340 + v)$.$10 = 2 (2v)$.$10 = 4v$.$v = \frac{10}{4} = 2.5 \,ms^{-1}$.

Two sources $A$ and $B$ are producing notes of frequency $680 \,Hz$. $A$ listener moves from $A$ to $B$ with a constant velocity $v$. If the speed of sound in air is $340 \,ms^{-1}$, the value of $v$ so that he hears $10$ beats per second is: (in $\,ms^{-1}$)

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