Two whistles A and B produce notes of frequen | vedclass

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(B) Let the speed of sound be $v = 330 \ m/s$. The speed of the listener $v_L = 30 \ m/s$ and the speed of whistle $B$ is $v_B = 30 \ m/s$.$1$. For wh

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$4$

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(B) Let the speed of sound be $v = 330 \ m/s$. The speed of the listener $v_L = 30 \ m/s$ and the speed of whistle $B$ is $v_B = 30 \ m/s$.$1$. For whistle $B$: Since the listener and whistle $B$ are moving in the same direction with the same speed,their relative velocity is zero. Therefore,the frequency heard by the listener from $B$ remains $f_B' = 596 \ Hz$.$2$. For whistle $A$: The listener is moving away from the stationary whistle $A$ with speed $v_L = 30 \ m/s$. The observed frequency $f_A'$ is given by the Doppler effect formula: $f_A' = f_A \left( \frac{v - v_L}{v} ight)$.Substituting the values: $f_A' = 660 \left( \frac{330 - 30}{330} ight) = 660 \left( \frac{300}{330} ight) = 660 	imes \frac{10}{11} = 600 \ Hz$.$3$. The number of beats heard is the difference between the two observed frequencies: $	ext{Beats} = |f_A' - f_B'| = |600 - 596| = 4 \ Hz$.

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