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JEE Advanced 2023Paper 2
Exam
JEE Advanced
Year
2023
Paper
Paper 2
Subject
Physics
Chapter
Waves and Sound

JEE Advanced 2023 · Paper 2 · Physics · Waves and Sound

S _ 1 and S _ 2 are two identical sound sources of frequency 656Hz...

S1{S}_{1} and S2{S}_{2} are two identical sound sources of frequency 656Hz656Hz. The source S1{S}_{1} is located at OO and S2{S}_{2} moves anti-clockwise with a uniform speed 42ms−14\sqrt{2}m{s}^{- 1} on a circular path around OO, as shown in the figure. There are three points P,QP, Q and RR on this path such that PP and RR are diametrically opposite while QQ is equidistant from them. A sound detector is placed at point PP. The source S1{S}_{1} can move along direction OPOP.

[Given: The speed of sound in air is 324ms−1324m{s}^{- 1}]

S _ 1 and S _ 2 are two identical sound sources of frequency 656Hz . The source S _ 1 is located at O and S _ 2 moves


When only S2{S}_{2} is emitting sound and it is at QQ, the frequency of sound measured by the detector in HzHz is _____.

Correct answer

0

Step-by-step solution

S _ 1 and S _ 2 are two identical sound sources of frequency 656Hz . The source S _ 1 is located at O and S _ 2 moves

Given: Source frequency, f0=656Hz{f}_{0} = 656Hz.

Velocity of sound

=324ms−1= 324m{s}^{- 1}.

Velocity component of the source moving away from detector along like PQPQ,

Vs=42cos45o=4ms−1{V}_{s} = 4\sqrt{2}cos{45}^{o} = 4m{s}^{- 1}.

Therefore, apparent frequency, f=(vv+vs)f0=(324324+4)656f = \left(\dfrac{v}{v + {v}_{s}}\right){f}_{0} = \left(\dfrac{324}{324 + 4}\right)656

⇒f=648Hz⇒ f = 648Hz

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