A train standing at the outer signal of a railway station, blows a whistle of frequency 400 Hz in stil air. (i) What is the frequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of `10 ms^(-1)`. (b) recedes from the platform with a speed of 10 ms ? What is the speed of sound in each case? `(V=340ms^(-1))`.

Answer

To solve the problem, we will use the Doppler effect formula for sound waves. The frequency observed by a stationary observer when the source of sound is moving can be calculated using the following formulas:

1. When the source is approaching the observer:
\[ f' = f \frac{v + v_0}{v - v_s} \]
where:
- \( f' \) = observed frequency
- \( f \) = source frequency (400 Hz)
- \( v \) = speed of sound in air (340 m/s)
- \( v_0 \) = speed of the observer (0 m/s for a stationary observer)
- \( v_s \) = speed of the source (10 m/s for the approaching train)

2. When the source is receding from the observer:
\[ f' = f \frac{v - v_0}{v + v_s} \]

### Step-by-Step Solution:

**(a) When the train approaches the platform:**

1. **Identify the parameters:**
- Source frequency \( f = 400 \, \text{Hz} \)
- Speed of sound \( v = 340 \, \text{m/s} \)
- Speed of the observer \( v_0 = 0 \, \text{m/s} \)
- Speed of the source \( v_s = 10 \, \text{m/s} \)

2. **Apply the Doppler effect formula for approaching source:**
\[ f' = f \frac{v + v_0}{v - v_s} \]
Substituting the values:
\[ f' = 400 \frac{340 + 0}{340 - 10} \]

3. **Calculate the frequency:**
\[ f' = 400 \frac{340}{330} = 400 \times 1.0303 \approx 412.12 \, \text{Hz} \]

**(b) When the train recedes from the platform:**

1. **Identify the parameters:**
- Source frequency \( f = 400 \, \text{Hz} \)
- Speed of sound \( v = 340 \, \text{m/s} \)
- Speed of the observer \( v_0 = 0 \, \text{m/s} \)
- Speed of the source \( v_s = 10 \, \text{m/s} \)

2. **Apply the Doppler effect formula for receding source:**
\[ f' = f \frac{v - v_0}{v + v_s} \]
Substituting the values:
\[ f' = 400 \frac{340 - 0}{340 + 10} \]

3. **Calculate the frequency:**
\[ f' = 400 \frac{340}{350} = 400 \times 0.9714 \approx 388.57 \, \text{Hz} \]

### Summary of Results:
- Frequency when the train approaches: \( f' \approx 412.12 \, \text{Hz} \)
- Frequency when the train recedes: \( f' \approx 388.57 \, \text{Hz} \)
- Speed of sound in both cases: \( v = 340 \, \text{m/s} \)