A railway engine passes by the platform at a speed of 36 km/hr blowing its whistle having a frequency of 660 Hz. The different in the frequencies of the whistle heard by a person standing on the platform as the engine goes past the person is equal to

Options

A. zero
B. 20 Hz
C. 40 Hz
D. 60 Hz

Answer

To solve the problem, we will use the Doppler Effect formula to find the difference in frequencies of the whistle heard by a person standing on the platform as the engine approaches and then moves away.

### Step-by-step Solution:

1. **Identify Given Data:**
- Speed of the railway engine (Vs) = 36 km/hr
- Frequency of the whistle (f0) = 660 Hz
- Velocity of sound (V) = 330 m/s (standard value)

2. **Convert Speed of the Engine:**
- Convert the speed from km/hr to m/s:
\[ Vs = 36 \text{ km/hr} \times \frac{5}{18} = 10 \text{ m/s} \]

3. **Calculate Apparent Frequency when the Engine Approaches:**
- When the source (engine) approaches the observer (person on the platform), the formula for the apparent frequency (f1) is:
\[ f1 = f0 \times \frac{V}{V - Vs} \]
- Substituting the values:
\[ f1 = 660 \times \frac{330}{330 - 10} = 660 \times \frac{330}{320} \]

4. **Calculate Apparent Frequency when the Engine Moves Away:**
- When the source moves away from the observer, the formula for the apparent frequency (f2) is:
\[ f2 = f0 \times \frac{V}{V + Vs} \]
- Substituting the values:
\[ f2 = 660 \times \frac{330}{330 + 10} = 660 \times \frac{330}{340} \]

5. **Calculate the Difference in Frequencies:**
- Now, we need to find the difference (Δf) between the two frequencies:
\[ \Delta f = f1 - f2 \]
- Substituting the expressions for f1 and f2:
\[ \Delta f = \left(660 \times \frac{330}{320}\right) - \left(660 \times \frac{330}{340}\right) \]
- Factor out the common terms:
\[ \Delta f = 660 \times 330 \left(\frac{1}{320} - \frac{1}{340}\right) \]

6. **Calculate the Value of the Difference:**
- To simplify the expression:
\[ \frac{1}{320} - \frac{1}{340} = \frac{340 - 320}{320 \times 340} = \frac{20}{108800} \]
- Therefore:
\[ \Delta f = 660 \times 330 \times \frac{20}{108800} \]
- Calculate the final value:
\[ \Delta f = 660 \times 330 \times \frac{20}{108800} = 20 \text{ Hz} \]

### Final Answer:
The difference in the frequencies of the whistle heard by the person standing on the platform as the engine goes past is **20 Hz**.