A sound source emitting sound of frequency 450 Hz is approaching a stationary observer with velocity `30 ms^(-1)` and another identical source is going away from the observer with the same velocity. If the velocity of sound is `330 ms^(-1)`, then the difference of frequencies heard by observer is

Options

A. 103. 5 Hz
B. 82.5 Hz
C. 33. 5 Hz
D. 92.5 Hz

Answer

To solve the problem, we will use the Doppler effect formula to find the apparent frequencies of the sound emitted by the two sources, one approaching and the other receding from the observer. We will then calculate the difference between these two frequencies.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Frequency of the sound source, \( f = 450 \, \text{Hz} \)
- Velocity of sound, \( v = 330 \, \text{m/s} \)
- Velocity of the source, \( v_s = 30 \, \text{m/s} \)
- Velocity of the observer, \( v_o = 0 \, \text{m/s} \) (since the observer is stationary)

2. **Calculate the Apparent Frequency for the Source Approaching the Observer:**
The formula for the apparent frequency when the source is moving towards a stationary observer is given by:
\[ f_A = f \left( \frac{v + v_o}{v - v_s} \right) \]
Substituting the values:
\[ f_{A1} = 450 \left( \frac{330 + 0}{330 - 30} \right) = 450 \left( \frac{330}{300} \right) = 450 \times 1.1 = 495 \, \text{Hz} \]

3. **Calculate the Apparent Frequency for the Source Moving Away from the Observer:**
The formula for the apparent frequency when the source is moving away from a stationary observer is given by:
\[ f_A = f \left( \frac{v + v_o}{v + v_s} \right) \]
Substituting the values:
\[ f_{A2} = 450 \left( \frac{330 + 0}{330 + 30} \right) = 450 \left( \frac{330}{360} \right) = 450 \times 0.9167 \approx 412.5 \, \text{Hz} \]

4. **Calculate the Difference in Frequencies:**
The difference in frequencies heard by the observer is:
\[ \Delta f = f_{A1} - f_{A2} = 495 \, \text{Hz} - 412.5 \, \text{Hz} = 82.5 \, \text{Hz} \]

### Final Answer:
The difference of frequencies heard by the observer is **82.5 Hz**.

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