A source of sound attached to the bob of a simple pendulum execute `SHM` . The difference between the apparent frequency of sound as received by an observer during its approach and recession at the mean frequency of the source . The velocity of the source at the mean position is ( velocity of sound in the air is `340 m//s`)
[Assume velocity of sound `lt lt` velocity of sound in air ]

Options

A. (a)`1.4 m//s`
B. (b)`3.4 m//s`
C. (c)`1.7 m//s`
D. (d)`2.1 m//s`

Explanation

To solve the problem, we need to analyze the situation where a source of sound is attached to a simple pendulum executing simple harmonic motion (SHM). The observer perceives different frequencies of sound during the approach and recession of the pendulum.

### Step-by-Step Solution:

1. **Understanding the Doppler Effect**:
When the source of sound moves towards the observer, the frequency increases, and when it moves away, the frequency decreases. We denote:
- \( f_0 \): the natural frequency of the source.
- \( v \): the speed of sound in air (given as \( 340 \, \text{m/s} \)).
- \( v_0 \): the velocity of the source at the mean position.

2. **Apparent Frequencies**:
The apparent frequency when the source approaches the observer is given by:
\[ f_1 = f_0 \frac{v + v_0}{v} \]
The apparent frequency when the source recedes from the observer is given by:
\[ f_2 = f_0 \frac{v - v_0}{v} \]

3. **Difference in Frequencies**:
The difference between the apparent frequencies during approach and recession is:
\[ f_1 - f_2 = f_0 \left( \frac{v + v_0}{v} - \frac{v - v_0}{v} \right) \]
Simplifying this, we get:
\[ f_1 - f_2 = f_0 \left( \frac{(v + v_0) - (v - v_0)}{v} \right) = f_0 \left( \frac{2v_0}{v} \right) \]

4. **Given Condition**:
According to the problem, the difference in apparent frequency is \( 2\% \) of the natural frequency:
\[ f_1 - f_2 = 0.02 f_0 \]
Setting the two expressions for the frequency difference equal gives:
\[ f_0 \left( \frac{2v_0}{v} \right) = 0.02 f_0 \]

5. **Canceling \( f_0 \)**:
Since \( f_0 \) is not zero, we can cancel it out:
\[ \frac{2v_0}{v} = 0.02 \]

6. **Solving for \( v_0 \)**:
Rearranging gives:
\[ 2v_0 = 0.02v \]
\[ v_0 = \frac{0.02v}{2} = 0.01v \]
Substituting \( v = 340 \, \text{m/s} \):
\[ v_0 = 0.01 \times 340 = 3.4 \, \text{m/s} \]

### Final Answer:
The velocity of the source at the mean position is \( 3.4 \, \text{m/s} \).

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