An engine giving off whistle is moving towards a stationary observer with 50m/s speed. What will be the ratio of the frequencies of the whistle heard when engine is approaching and receding from the observer? (speed of sound = 350 m/s)

Options

A. `2:1`
B. `4:5`
C. `4:3`
D. `3:4`

Answer

To solve the problem, we need to calculate the frequencies of the whistle when the engine is approaching and receding from a stationary observer, and then find the ratio of these frequencies.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Speed of sound (V) = 350 m/s
- Speed of the source (Vs) = 50 m/s
- Let the real frequency of the sound be \( f_0 \).

2. **Calculate Frequency When the Engine is Approaching (f1):**
- The formula for the frequency heard by the observer when the source is approaching is:
\[ f_1 = f_0 \times \frac{V}{V - V_s} \]
- Substituting the known values:
\[ f_1 = f_0 \times \frac{350}{350 - 50} = f_0 \times \frac{350}{300} \]
- Simplifying:
\[ f_1 = f_0 \times \frac{7}{6} \]

3. **Calculate Frequency When the Engine is Receding (f2):**
- The formula for the frequency heard by the observer when the source is receding is:
\[ f_2 = f_0 \times \frac{V}{V + V_s} \]
- Substituting the known values:
\[ f_2 = f_0 \times \frac{350}{350 + 50} = f_0 \times \frac{350}{400} \]
- Simplifying:
\[ f_2 = f_0 \times \frac{7}{8} \]

4. **Calculate the Ratio of Frequencies (f1/f2):**
- Now we find the ratio of the frequencies:
\[ \frac{f_1}{f_2} = \frac{f_0 \times \frac{7}{6}}{f_0 \times \frac{7}{8}} \]
- The \( f_0 \) cancels out:
\[ \frac{f_1}{f_2} = \frac{\frac{7}{6}}{\frac{7}{8}} = \frac{7}{6} \times \frac{8}{7} \]
- Simplifying further:
\[ \frac{f_1}{f_2} = \frac{8}{6} = \frac{4}{3} \]

5. **Final Result:**
- The ratio of the frequencies when the engine is approaching and receding from the observer is:
\[ \text{Ratio} = 4:3 \]