A violin player riding on a slow train plays a 440 Hz note. Another violin player standing near the track plays the same note. When the two are close by and the train approaches the person on the ground, he hears 4'0 beats per second. The speed of sound in air` = 340 ms^-1` (a) Calculate the speed of the train. (b) What beat frequency is heard by the player in the train ?

Explanation

To solve the problem, we will break it down into two parts as specified in the question.

### Part (a): Calculate the speed of the train.

1. **Identify the given values:**
- Frequency of the note played by both violin players, \( f_0 = 440 \, \text{Hz} \)
- Beat frequency heard by the observer on the ground, \( n = 4 \, \text{Hz} \)
- Speed of sound in air, \( v = 340 \, \text{m/s} \)

2. **Determine the apparent frequency heard by the observer on the ground:**
- The observer hears a beat frequency of 4 Hz, which means the apparent frequency \( f \) is:
\[ f = f_0 + n = 440 \, \text{Hz} + 4 \, \text{Hz} = 444 \, \text{Hz} \]

3. **Apply the Doppler effect formula:**
- Since the train is approaching the observer, we use the formula for the apparent frequency when the source is moving towards a stationary observer:
\[ f = f_0 \frac{v + v_o}{v - v_s} \]
Where:
- \( f \) is the apparent frequency (444 Hz)
- \( f_0 \) is the original frequency (440 Hz)
- \( v_o = 0 \) (the observer is stationary)
- \( v_s \) is the speed of the source (the train, which we need to find)

4. **Substituting the known values into the formula:**
\[ 444 = 440 \frac{340 + 0}{340 - v_s} \]

5. **Rearranging the equation to solve for \( v_s \):**
\[ 444(340 - v_s) = 440 \times 340 \]
\[ 444 \times 340 - 444 v_s = 440 \times 340 \]
\[ 444 v_s = 444 \times 340 - 440 \times 340 \]
\[ 444 v_s = (444 - 440) \times 340 \]
\[ 444 v_s = 4 \times 340 \]
\[ v_s = \frac{4 \times 340}{444} \approx 3.06 \, \text{m/s} \]

### Part (b): Calculate the beat frequency heard by the player in the train.

1. **Determine the frequency heard by the player in the train:**
- For the player in the train, he is moving towards the stationary source (the player on the ground). We will use the Doppler effect formula again:
\[ f' = f_0 \frac{v + v_o}{v - v_s} \]
Where:
- \( f_0 = 440 \, \text{Hz} \)
- \( v_o = v_s = 3.06 \, \text{m/s} \) (the player in the train is moving towards the stationary source)
- \( v = 340 \, \text{m/s} \)

2. **Substituting the values into the formula:**
\[ f' = 440 \frac{340 + 3.06}{340 - 0} \]
\[ f' = 440 \frac{343.06}{340} \]
\[ f' \approx 440 \times 1.009 = 443.96 \, \text{Hz} \]

3. **Calculate the beat frequency:**
- The beat frequency \( n' \) is the absolute difference between the frequencies heard by the player in the train and the frequency he is playing:
\[ n' = |f' - f_0| = |443.96 - 440| \approx 3.96 \, \text{Hz} \]

### Summary of Results:
- (a) The speed of the train is approximately \( 3.06 \, \text{m/s} \).
- (b) The beat frequency heard by the player in the train is approximately \( 3.96 \, \text{Hz} \).