A train moving with a velocity of 18 km/hr blows a whistle which is heard by a motor cyclist driving with speed of 6 km/h with the frequency received as 200 cps. The actual frequency when train and motor cyclist are moving in opposite directions and approaching towards each other will be (Speed of sound is 340 m/s, cps is cycle per second)

Options

A. 150 cps
B. 180.5 cps
C. 196 cps
D. 200 cps

Explanation

To solve the problem of finding the actual frequency of the whistle blown by the train, we will use the Doppler effect formula. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Information
- The velocity of the train (source) \( v_s = 18 \) km/h
- The velocity of the motorcyclist (observer) \( v_o = 6 \) km/h
- The frequency received by the motorcyclist \( f = 200 \) Hz
- The speed of sound \( v = 340 \) m/s

### Step 2: Convert Speeds from km/h to m/s
To use the Doppler effect formula, we need to convert the speeds from km/h to m/s:
- \( v_s = 18 \) km/h \( = 18 \times \frac{5}{18} = 5 \) m/s
- \( v_o = 6 \) km/h \( = 6 \times \frac{5}{18} = 1.67 \) m/s (approximately)

### Step 3: Apply the Doppler Effect Formula
When the source and observer are moving towards each other, the Doppler effect formula is given by:

\[ f = f_0 \frac{v + v_o}{v - v_s} \]

Where:
- \( f_0 \) is the actual frequency of the source,
- \( f \) is the observed frequency,
- \( v \) is the speed of sound,
- \( v_o \) is the speed of the observer,
- \( v_s \) is the speed of the source.

### Step 4: Rearrange the Formula to Solve for \( f_0 \)
Rearranging the formula to isolate \( f_0 \):

\[ f_0 = f \frac{v - v_s}{v + v_o} \]

### Step 5: Substitute the Known Values
Now, we can substitute the known values into the rearranged formula:

\[ f_0 = 200 \, \text{Hz} \cdot \frac{340 \, \text{m/s} - 5 \, \text{m/s}}{340 \, \text{m/s} + 1.67 \, \text{m/s}} \]

Calculating the numerator and denominator:

\[ f_0 = 200 \cdot \frac{335}{341.67} \]

### Step 6: Calculate \( f_0 \)
Now we compute the value:

\[ f_0 \approx 200 \cdot 0.980 \]
\[ f_0 \approx 196 \, \text{Hz} \]

### Conclusion
The actual frequency of the whistle blown by the train is approximately **196 Hz**.

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