Two loudspeakers M and N are located 20 m ap | vedclass

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(A) Let $v_0$ be the speed of the car and $v$ be the speed of sound. The observed frequencies from $M$ and $N$ by the car moving towards $Q$ are $f_M

Vedclass · Accepted Answer

Vedclass · Answer

(A) Let $v_0$ be the speed of the car and $v$ be the speed of sound. The observed frequencies from $M$ and $N$ by the car moving towards $Q$ are $f_M = f_M^0 \left( \frac{v + v_0 \cos 	heta}{v} ight)$ and $f_N = f_N^0 \left( \frac{v + v_0 \cos 	heta}{v} ight)$,where $	heta$ is the angle between the velocity vector of the car and the line connecting the car to the speakers.The beat frequency $v(t) = |f_N - f_M| = (121 - 118) \left( \frac{v + v_0 \cos 	heta}{v} ight) = 3 \left( 1 + \frac{v_0}{v} \cos 	heta ight)$.At point $P$,$	heta$ is acute,so $\cos 	heta > 0$,thus $v_P > 3$. At point $Q$,$	heta = 90^{\circ}$,so $\cos 	heta = 0$,thus $v_Q = 3$. At point $R$,$	heta$ is obtuse,so $\cos 	heta Since $v_P = 3(1 + k)$ and $v_R = 3(1 - k)$ where $k = \frac{v_0}{v} \cos 	heta_P$,we have $v_P + v_R = 6 = 2v_Q$. Thus,$(A)$ is true.The rate of change of beat frequency is $\frac{dv}{dt} = 3 \frac{v_0}{v} (-\sin 	heta) \frac{d	heta}{dt}$. This is maximum when $\sin 	heta$ is maximum,which occurs at $Q$ where $	heta = 90^{\circ}$. Thus,$(B)$ is true.The variation of $\cos 	heta$ with time as the car moves from $P$ to $R$ follows a sigmoid-like curve,which is represented by the left plot. Thus,$(C)$ is true.

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