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

Options

A. `5%`
B. `20%`
C. `0%`
D. `0.5%`

Explanation

To solve the problem of finding the percentage increase in the apparent frequency when an observer moves towards a stationary source of sound, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Let the velocity of sound be \( v \).
- The observer's velocity \( v_0 \) is given as \( \frac{v}{5} \).
- The source's velocity \( v_s \) is \( 0 \) (since it is stationary).

2. **Use the Doppler Effect formula:**
The formula for the apparent frequency \( n' \) when the observer is moving towards a stationary source is given by:
\[ n' = n \cdot \frac{v + v_0}{v + v_s} \]
Here, \( n \) is the original frequency.

3. **Substitute the values into the formula:**
Since \( v_s = 0 \), the formula simplifies to:
\[ n' = n \cdot \frac{v + \frac{v}{5}}{v + 0} \]
This can be rewritten as:
\[ n' = n \cdot \frac{v + \frac{v}{5}}{v} = n \cdot \frac{1 + \frac{1}{5}}{1} \]

4. **Simplify the expression:**
\[ n' = n \cdot \frac{6}{5} \]

5. **Calculate the increase in frequency:**
The increase in frequency is given by:
\[ \text{Increase} = n' - n = \left( \frac{6}{5}n - n \right) = \left( \frac{6}{5}n - \frac{5}{5}n \right) = \frac{1}{5}n \]

6. **Calculate the percentage increase:**
The percentage increase in frequency is given by:
\[ \text{Percentage Increase} = \left( \frac{\text{Increase}}{n} \right) \times 100 = \left( \frac{\frac{1}{5}n}{n} \right) \times 100 = 20\% \]

7. **Conclusion:**
The percentage increase in the apparent frequency is \( 20\% \).

### Final Answer:
The percentage increase in the apparent frequency is **20%**.