An observer is moving with a constant speed of 20m/s on a circular track of radius 50m. A source kept at centre of track emits a sound of frequency 200 Hz. Then the frequency received by the observer is ` x xx10^2` Hz, what is value of x. (V= 340m/s)

Answer

To solve the problem step by step, we will use the Doppler effect formula for sound waves, considering the observer is moving in a circular path around a stationary source.

### Step 1: Identify the given values
- Speed of sound, \( v = 340 \, \text{m/s} \)
- Speed of observer, \( v_0 = 20 \, \text{m/s} \)
- Frequency of the source, \( f_0 = 200 \, \text{Hz} \)
- Speed of the source, \( v_s = 0 \, \text{m/s} \) (since the source is stationary)

### Step 2: Determine the angle between the observer's motion and the line joining the source and observer
Since the observer is moving in a circular path, at any instant, the angle \( \theta \) between the direction of the observer's motion and the line joining the observer to the source is \( 90^\circ \).

### Step 3: Use the Doppler effect formula
The apparent frequency \( f' \) observed by the observer can be calculated using the formula:
\[ f' = \frac{v + v_0 \cos \theta}{v + v_s \sin \theta} f_0 \]
Substituting the values:
- \( \cos 90^\circ = 0 \)
- \( \sin 90^\circ = 1 \)

Thus, the formula simplifies to:
\[ f' = \frac{v + 20 \cdot 0}{v + 0 \cdot 1} f_0 = \frac{v}{v} f_0 = f_0 \]

### Step 4: Substitute the values
Since \( f_0 = 200 \, \text{Hz} \):
\[ f' = 200 \, \text{Hz} \]

### Step 5: Express \( f' \) in terms of \( x \)
According to the problem, the frequency received by the observer can be expressed as:
\[ f' = x \times 10^2 \, \text{Hz} \]
Setting this equal to the calculated frequency:
\[ x \times 10^2 = 200 \]

### Step 6: Solve for \( x \)
To find \( x \):
\[ x = \frac{200}{10^2} = \frac{200}{100} = 2 \]

### Final Answer
The value of \( x \) is \( 2 \).

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