JEE Advanced 2023 — Paper 2 — PHY-IV solution

JEE Advanced2023Paper 2PHY-IV

PARAGRAPH-I

$S_{1}$ and $S_{2}$ are two identical sound sources of frequency $656$ Hz. The source $S_{1}$ is located at $O$ and $S_{2}$ moves anti-clockwise with a uniform speed $42$ ms $^{- 1}$ on a circular path around $O$ , as shown in the figure. There are three points $P$ , $Q$ and $R$ on this path such that $P$ and $R$ are diametrically opposite while $Q$ is equidistant from them. A sound detector is placed at point $P$ . The source $S_{1}$ can move along direction $O P$ .

[Given: The speed of sound in air is $324$ ms $^{- 1}$ ]

Q1. When only $S_{2}$ is emitting sound and it is $Q$ , the frequency of sound measured by the detector in Hz is __________.

Your answer

Solution

Let the speed of sound be $c = 324$ m s $^{- 1}$ , source frequency $f = 656$ Hz, and source speed $v_{s} = 42$ m s $^{- 1}$ .

At $Q$ , the velocity of $S_{2}$ is tangential to the circle. With $P$ at the bottom and $Q$ on the side (anti-clockwise motion), the velocity at $Q$ points away from $P$ along a line making $4 5^{\circ}$ with the line $QP$ .

The component of source velocity along the line $S_{2} \to P$ (i.e. the velocity of recession from the detector) is

$v_{s, ∥} = v_{s} cos 4 5^{\circ} = 42 \cdot \frac{1}{2} = 4 m s^{- 1}$ .

Since the source is receding from the stationary observer, the Doppler-shifted frequency is

$f^{'} = f \cdot \frac{c}{c + v _{s, ∥}} = 656 \cdot \frac{324}{324 + 4} = 656 \cdot \frac{324}{328} = 648 Hz$ .

← Back to full paper

Practice more PHY-IV

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →