Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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A Small Source of Sound Vibrating at Frequency 500 Hz is Rotated in a Circle of Radius 100/π Cm at a Constant Angular Speed of 5.0 Revolutions per Second - Physics

Sum

A small source of sound vibrating at frequency 500 Hz is rotated in a circle of radius 100/π cm at a constant angular speed of 5.0 revolutions per second. A listener situation situates himself in the plane of the circle. Find the minimum and the maximum frequency of the sound observed. Speed of sound in air = 332 m s−1.

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Solution

Given:
Speed of sound in air v = 332 ms−1
Radius of the circle r = \[\frac{100}{\pi}\] cm =\[\frac{1}{\pi}\]  m
Frequency of sound of the source \[f_0\]= 500 Hz
Angular speed \[\omega\]= 5 rev/s
Linear speed of the source is given by:

\[v = \omega r\]

⇒ \[v = 5 \times \frac{1}{\pi} = \frac{5}{\pi} = 1 . 59  \text { m/s }\]

∴ velocity of source \[v_s\]= 1.59 m/s
Let X be the position where the observer will listen at a maximum and Y be the position where he will listen at the minimum frequency.

Apparent frequency \[\left( f_1 \right)\]at X is given by :

\[f_1  = \left( \frac{v}{v - v_s} \right) f_0\]

On substituting the values in the above equation, we get:

\[f_1  = \left( \frac{332}{332 - 1 . 59} \right) \times 500 \approx 515  \text { Hz }\]

Apparent frequency \[\left( f_2 \right)\] at Y is given by:

\[f_2  = \left( \frac{v}{v + v_s} \right) f_0\]

On substituting the values in the above equation, we get:

\[f_2  = \left( \frac{332}{332 + 1 . 59} \right) \times 500 \approx 485 \text{ Hz }\]
Concept: Speed of Wave Motion
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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 16 Sound Waves
Q 72 | Page 356
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