Question 51: Waves

Question

Two loudspeakers M and N are located 20m apart and emit sound at frequencies 118 Hz and 121 Hz, respectively. A car in initially at a point P, 1800 m away from the midpoint Q of the line MN and moves towards Q constantly at 60 km/h along the perpendicular bisector of MN. It crosses Q and eventually reaches a point R, 1800 m away from Q.Let v(t) represent the beat frequency measured by a person sitting in the car at time t. Let vP, vQ and vR be the beat frequencies measured at locations P, Q and R respectively. The speed of sound in air is 330 ms$$-$$1. Which of the following statement(s) is (are) true regarding the sound heard by the person?

Accepted Answer

Consider the time $t\left(\leq t_Qight)$ when car is at $S_1$ between $P$ and $Q$. The distance travelled by the car in time $t$ is $\mathrm{PS}_1=u t$. At this instant, the lines $\mathrm{S}_1{N}$ and $\mathrm{S}_1 \mathrm{M}$ both make angle $	heta$ with the velocity vector $\vec{u}$. The component of observer (person sitting in the car) velocity towards the sources $N$ and $M$ is $u_o=u \cos 	heta$. The sources $N$ and $M$ are at rest i.e., $u_s=0$. Apply Doppler's effect equation to get frequencies of the sources $N$ and $M$ heard by the observer as
$$
\begin{aligned}

u_N^{\prime} & =\frac{v+u_o}{v-u_s} 
u_N=\frac{v+u \cos 	heta}{v} 
u_N \\
& =\left(1+\frac{u}{v} \frac{D-u t}{\sqrt{d^2+(D-u t)^2}}ight) 
u_N, \\

u_M^{\prime} & =\frac{v+u_o}{v-u_s} 
u_M=\frac{v+u \cos 	heta}{v} 
u_M \\
& =\left(1+\frac{u}{v} \frac{D-u t}{\sqrt{d^2+(D-u t)^2}}ight) 
u_M .
\end{aligned}
$$
The beat frequency heard by the observer at time $t(\leq$ $\left.t_Qight)$ is
$$
\begin{aligned}

u(t) & =
u_N^{\prime}-
u_M^{\prime} \\
& =\left(1+\frac{u}{v} \frac{D-u t}{\sqrt{d^2+(D-u t)^2}}ight)\left(
u_N-
u_Might) ........(1)
\end{aligned}
$$
Now, consider the time $t\left(\geq t_Qight)$ when car is at $S_2$ between $Q$ and $R$. The distance travelled by the car in time $t$ is $\mathrm{PS}_2=u t$. At this instant, the lines $\mathrm{S}_2 \mathrm{~N}$ and $\mathrm{S}_2 \mathrm{M}$ both make angle $\left(180^{\circ}-	hetaight)$ with the velocity vector $\vec{u}$. The component of observer velocity towards the sources $N$ and $M$ is $u_o=-u \cos 	heta$. Apply Doppler's effect equation to get
$$
\begin{aligned}

u_N^{\prime} & =\frac{v+u_o}{v-u_s} 
u_N=\frac{v-u \cos 	heta}{v} 
u_N \\
& =\left(1-\frac{u}{v} \frac{u t-D}{\sqrt{d^2+(u t-D)^2}}ight) 
u_N, \\

u_M^{\prime} & =\frac{v+u_o}{v-u_s} 
u_M=\frac{v-u \cos 	heta}{v} 
u_M \\
& =\left(1-\frac{u}{v} \frac{u t-D}{\sqrt{d^2+(u t-D)^2}}ight) 
u_M .
\end{aligned}
$$
The beat frequency heard by the observer at time $t(\geq$ $\left.t_Qight)$ is
$$
\begin{aligned}

u(t) & =
u_N^{\prime}-
u_M^{\prime} \\
& =\left(1-\frac{u}{v} \frac{u t-D}{\sqrt{d^2+(u t-D)^2}}ight)\left(
u_N-
u_Might) ..........(2)
\end{aligned}
$$
Substitute $t=0$ and $t=t_Q={D \over u}$ in equation (1) to get $
u_P$ and $
u_Q$ and substitute $t=t_R=2 {D \over u}$ in equation (2) to get $
u_R$ i.e.,
$$
\begin{aligned}
& 
u_P=
u(t=0)=\left(1+\frac{u}{v} \frac{D}{\sqrt{d^2+D^2}}ight)\left(
u_N-
u_Might) \\
& 
u_Q=
u(t=D / u)=\left(
u_N-
u_Might) \\
& 
u_R=
u(t=2 D / u)=\left(1-\frac{u}{v} \frac{D}{\sqrt{d^2+D^2}}ight)\left(
u_N-
u_Might)
\end{aligned}
$$
$$ 	herefore $$ $$

u_P+
u_R=2 
u_Q
$$
Differentiate equations (1) w.r.t. time t to get rate of change of beat frequency
$$
\frac{\mathrm{d} 
u(t)}{\mathrm{d} t}=-\left(
u_N-
u_Might) \frac{u^2}{v} \frac{d^2}{\left(d^2+(D-u t)^2ight)^{3 / 2}}
$$ ...........(3)
From equation (3), the slope is negative and its magnitude is maximum when $t={D \over u}=t_Q$ (denominator is minimum). Thus the rate of change of beat frequency is maximum when car passes through $Q$.

Answer

The plot below represents schematically the variation of beat frequency with time

Answer

The rate of change in beat frequency is maximum when the car passes through Q

Answer

vP + vR = 2vQ

Answer

The plot below represents schematically the variations of beat frequency with time

Waves

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