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 find the difference in frequencies heard by the observer from two sound sources, one approaching and one receding, we can use the Doppler effect formula for sound.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Frequency of the source, \( f = 450 \, \text{Hz} \)
- Velocity of the sound, \( v = 330 \, \text{m/s} \)
- Velocity of the source, \( u = 30 \, \text{m/s} \)

2. **Calculate the Apparent Frequency when the Source is Approaching**:
The formula for the apparent frequency when the source is approaching is given by:
\[ f' = \frac{v}{v - u} \cdot f \]
Substituting the values:
\[ f' = \frac{330}{330 - 30} \cdot 450 \]
\[ f' = \frac{330}{300} \cdot 450 \]
\[ f' = 1.1 \cdot 450 = 495 \, \text{Hz} \]

3. **Calculate the Apparent Frequency when the Source is Receding**:
The formula for the apparent frequency when the source is moving away is given by:
\[ f'' = \frac{v}{v + u} \cdot f \]
Substituting the values:
\[ f'' = \frac{330}{330 + 30} \cdot 450 \]
\[ f'' = \frac{330}{360} \cdot 450 \]
\[ f'' = 0.917 \cdot 450 \approx 412.65 \, \text{Hz} \]

4. **Calculate the Difference in Frequencies**:
Now, we find the difference between the two frequencies:
\[ \Delta f = f' - f'' = 495 - 412.65 \approx 82.35 \, \text{Hz} \]

5. **Final Result**:
The difference of frequencies heard by the observer is approximately \( 82.35 \, \text{Hz} \).