Question 341: Waves

Question

An observer moves towards a stationary source of sound with a velocity of one-fifth of the velocity of sound. The percentage increase in the apparent frequency is

Accepted Answer

To determine the percentage increase in the apparent frequency when an observer moves towards a stationary sound source, we use the Doppler effect formula. The formula for the observed frequency ($f'$) is given by:
$$ f' = f \left( \frac{v + v_o}{v} ight) $$
Where:
$ f $ is the frequency of the source,
$ v $ is the velocity of sound in the medium,
$ v_o $ is the velocity of the observer.
Given that the observer moves with a velocity that is one-fifth of the velocity of sound, we have:
$$ v_o = \frac{v}{5} $$
Substitute this value into the formula:
$$ f' = f \left( \frac{v + \frac{v}{5}}{v} ight) $$
Simplify the expression:
$$ f' = f \left( \frac{5v + v}{5v} ight) = f \left( \frac{6v}{5v} ight) = f \left( \frac{6}{5} ight) $$
The apparent frequency $ f' $ is $ \frac{6}{5} 	imes f $, which means it increases by a factor of $\frac{1}{5}$ of the original frequency:
The percentage increase in frequency is:
$$ \left( \frac{f' - f}{f} ight) 	imes 100\% = \left( \frac{\frac{6}{5}f - f}{f} ight) 	imes 100\% $$
Simplify the expression:
$$ = \left( \frac{6f - 5f}{5f} ight) 	imes 100\% = \left( \frac{f}{5f} ight) 	imes 100\% = \frac{1}{5} 	imes 100\% = 20\% $$
The percentage increase in the apparent frequency is
20%
.
Option C
: $20\%$ is the correct answer.

Answer

$5 \%$

Answer

$10 \%$

Answer

$20 \%$

Answer

$25 \%$

Waves

An observer moves towards a stationary source of sound with a velocity of one-fifth of the velocity of sound. The percentage increase in the apparent frequency is

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