Question

A source of sound attached to the bob of a simple pendulum executes $S.H.M$. The difference between the apparent frequency of sound as received by an observer during its approach and recession at the mean position of the S.H.M. motion is $2\%$ of the natural frequency of the source. The velocity of the source at the mean position is:(velocity of sound in the air is $340m/s$)
[Assume velocity of sound source $<<$ velocity of sound in air]

• A. $1.4m/s$
• B. $3.4m/s$
• C. $1.7m/s$
• D. $2.1m/s$

Answer 1

The correct option is B $3.4m/s$

From doppler effect formula we know

apparent frequency of sound as received by an observer during its approach $f_{max}=f\frac{V}{V-V_S}$

apparent frequency of sound as received by an observer during its recession $f_{min}=f\frac{V}{V+V_S}$

where $V$ is velocity of sound and $V_S$ is velocity of source at mean position.

According to question

$\frac{f_{max}-f_{min}}{f}=2\%=1/50$

$\frac{f\frac{V}{V-V_S}-f\frac{V}{V+V_S}}{f}=\frac{1}{50}$

$\frac{2V_S}{V}=\frac{1}{50}$

$V_S=\frac{V}{100}$

$V_S=\frac{340}{100}m/s$

$V_S=3.4m/s$