Question

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

• A. Zero
• B. 5%
• C. 10%
• D. 20%

Answer 1

Answer: (D)

20%

Answer 2

When an observer moves towards a stationary source, the apparent frequency heard by the observer is

$\upsilon' = \upsilon_{0}\left( \frac{v + v_{O}}{v} \right)$

Where

$\upsilon_{O} =$Frequency of the source

$v_{O} =$Velocity of the observer

V = velocity of the sound

$\therefore\upsilon' = \upsilon_{0}\left( \frac{v + \frac{v}{5}}{v} \right) = \frac{6\upsilon_{0}}{5}$

Percentage changes in apparent frequency

$= \frac{\upsilon' - \upsilon_{0}}{\upsilon_{0}} \times 100$

$= \frac{\left( \frac{6\upsilon_{0}}{5} - \upsilon_{0} \right)}{\upsilon_{0}} \times 100\frac{1}{5} \times 100 = 20\%$