SATHEE: Wave Motion 5 Question 14

Wave Motion 5 Question 14

14. Two loudspeakers $M$ and $N$ are located $20 m$ apart and emit sound at frequencies $118 H z$ and $121 H z$ , respectively. A car in initially at a point $P, 1800 m$ away from the mid-point $Q$ of the line $M N$ and moves towards $Q$ constantly at $60 k m / h$ along the perpendicular bisector of $M N$ . 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 $v_{P}, v_{Q}$ and $v_{R}$ be the beat frequencies measured at locations $P, Q$ and $R$ respectively. The speed of sound in air is $330 m s^{- 1}$ . Which of the following statement(s) is (are) true regarding the sound heard by the person?

(2016 Adv.)

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

(b) The rate of change in beat frequency is maximum when the car passes through $Q$

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

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Answer:

Correct Answer: 14. (b, c, d)

Solution:

Speed of car, $v = 60 k m / h = \frac{500}{3} m / s$ At a point $S$ , between $P$ and $Q$

$\begin{aligned} v_{M}^{'} & = v_{M} (\frac{C + v \cos θ}{C}); \\ v_{N}^{'} & = v_{N} (\frac{C + v \cos θ}{C}) \\ \Rightarrow Δ v & = (v_{n} - v_{M}) (1 + \frac{v \cos θ}{C}) \end{aligned}$

Similarly, between $Q$ and $R$

$\begin{aligned} Δ v & = (v_{N} - v_{M}) (1 - \frac{v \cos θ}{C}) \\ \frac{d (Δ v)}{d t} & = \pm (v_{N} - v_{M}) \frac{v}{C} \sin θ \frac{d θ}{d t} \end{aligned}$

$θ = 0^{\circ}$ at the $P$ and $R$ as they are large distance apart.

$\Rightarrow$ Slope of graph is zero.

at $Q, θ = 90^{\circ} \sin θ$ is maximum also value of $\frac{d θ}{d t}$ is maximum

as $\frac{d θ}{d t} = \frac{v}{r}$ , where $v$ is its velocity and $r$ is the length of the line joining $P$ and $S$ and $r$ is minimum at $Q$ .

$\Rightarrow$ Slope is maximum at $Q$ .

At $P, v_{P} = Δ v = (v_{N} - v_{M}) (1 + \frac{v}{C}) (θ \approx 0^{\circ})$

At $R, v_{R} = Δ v = (v_{N} - v_{M}) (1 - \frac{v}{C}) (θ \approx 0^{\circ})$

At $Q, v_{Q} = Δ v = (v_{N} - v_{M}) (θ = 90^{\circ})$

From these equations, we can see that

$v_{P} + v_{R} = 2 v_{Q}$

Wave Motion 5 Question 14

Answer:

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Wave Motion 5 Question 14

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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