Question

An observer moves towards a stationary source of sound with a velocity equal to one-fifth of the velocity of sound. The percentage change in the frequency will be:

• A. 20%
• B. 10%
• C. 5%
• D. 0%

Answer 1

The Correct Option is A

The problem is about the Doppler Effect, which describes how the frequency of a wave changes for an observer moving relative to the source of the wave. Here, the observer is moving towards a stationary source of sound.

The formula for the perceived frequency (\nu') when an observer moves towards a stationary source is:

\nu' = \nu \left( \frac{v + v_o}{v} \right)

where:

• \nu is the original frequency of the source,
• v is the velocity of sound in the medium,
• v_o is the velocity of the observer towards the source.

Given that the observer moves at a velocity that is one-fifth of the speed of sound, we have:

v_o = \frac{v}{5}

Substituting this into the formula gives:

\nu' = \nu \left( \frac{v + \frac{v}{5}}{v} \right) = \nu \left( \frac{6v}{5v} \right) = \nu \left( \frac{6}{5} \right)

Thus, the perceived frequency is \frac{6}{5} times the original frequency. To find the percentage change in frequency:

\text{Percentage change} = \left( \frac{\nu' - \nu}{\nu} \right) \times 100\% = \left( \frac{\frac{6}{5}\nu - \nu}{\nu} \right) \times 100\%

This simplifies to:

\text{Percentage change} = \left( \frac{\nu}{5\nu} \right) \times 100\% = \frac{1}{5} \times 100\% = 20\%

Therefore, the percentage change in the frequency observed by the observer is 20%.

Hence, the correct answer is: 20%.