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Question

A source emitting sound at frequency 4000 Hz, is moving along the Y-axis with a speed of 22 m s−1. A listener is situated on the ground at the position (660 m, 0). Find the frequency of the sound received by the listener at the instant the source crosses the origin. Speed of sound in air = 330 m s−1.


Solution

Given:
Speed of sound in air v = 330 ms−1
Frequency of sound f0 = 4000 Hz
Velocity of source vs = 22 m/s
The apparent frequency heard by the listener f = ?

At t = 0, let the source be at a distance of y from the origin. Now, the time taken by the sound
to reach the listener is the same as the time taken by the sound to reach the origin.
∴​ 
y22=660+y2330 15y2=6602+y2 224y2=6602y=660224

Velocity of source along the line joining the source S and listener L:
vscosθ = 22.y660+y2=22y15y=2215

Frequency heard by the listener f is

f=vv-vs cosθ×f0f=330330-2215×4000

 f= 4017.85 ≈ 4018 Hz

Physics
HC Verma - I
Standard XI

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