When a train is approaching the observer, the frequency of the whistle is 100 cps. When it has passed observer, it is 50 cps. Calculate the frequncy when the observer moves with the train.

Options

A. 66.67cps
B. 33.33cps
C. 30cps
D. 25cps

Explanation

To solve the problem, we need to apply the Doppler effect principles for sound waves. Here's a step-by-step solution:

### Step 1: Understand the Given Information
- Frequency of the whistle when the train is approaching the observer (F1) = 100 cps (cycles per second)
- Frequency of the whistle when the train is moving away from the observer (F2) = 50 cps

### Step 2: Set Up the Doppler Effect Equations
For a source approaching the observer:
\[ F_1 = \frac{v}{v - v_s} \cdot f \]
For a source receding from the observer:
\[ F_2 = \frac{v}{v + v_s} \cdot f \]
Where:
- \( F_1 \) = frequency heard when approaching
- \( F_2 \) = frequency heard when receding
- \( v \) = speed of sound
- \( v_s \) = speed of the source (train)
- \( f \) = actual frequency of the whistle

### Step 3: Rearranging the Equations
From the two equations, we can express the actual frequency \( f \) in terms of \( F_1 \) and \( F_2 \):
1. Rearranging for \( f \) from the first equation:
\[ f = F_1 \cdot \frac{v - v_s}{v} \]
2. Rearranging for \( f \) from the second equation:
\[ f = F_2 \cdot \frac{v + v_s}{v} \]

### Step 4: Set the Two Expressions for \( f \) Equal
Since both expressions represent the same frequency \( f \):
\[ F_1 \cdot \frac{v - v_s}{v} = F_2 \cdot \frac{v + v_s}{v} \]

### Step 5: Substitute Known Values
Substituting \( F_1 = 100 \) cps and \( F_2 = 50 \) cps into the equation:
\[ 100 \cdot (v - v_s) = 50 \cdot (v + v_s) \]

### Step 6: Simplify the Equation
Expanding both sides:
\[ 100v - 100v_s = 50v + 50v_s \]
Rearranging gives:
\[ 100v - 50v = 100v_s + 50v_s \]
\[ 50v = 150v_s \]
Thus, we find:
\[ v = 3v_s \]

### Step 7: Calculate the Speed of the Source
Assuming the speed of sound \( v \) is approximately 343 m/s (at room temperature):
\[ 343 = 3v_s \]
So:
\[ v_s = \frac{343}{3} \approx 114.33 \text{ m/s} \]

### Step 8: Calculate the Actual Frequency \( f \)
Now, substituting \( v_s \) back into one of the equations to find \( f \):
Using the first equation:
\[ f = 100 \cdot \frac{343 - 114.33}{343} \]
Calculating:
\[ f = 100 \cdot \frac{228.67}{343} \approx 66.67 \text{ cps} \]

### Final Answer
The frequency when the observer moves with the train is approximately **66.67 cps**.

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