At each of two stations A and B, a siren is sounding with a constant frequency of 250 cycle `s^-1`. A cyclist from A proceeds straight towards B with a velocity of `12kmh^-1` and hear `5 beats//s`. The velocity of sound is nearly:

Options

A. `328 ms^(-1)`
B. `320 ms^(-1)`
C. `333 ms^(-1)`
D. `336 ms^(-1)`

Explanation

To solve the problem, we will use the principles of the Doppler effect and the concept of beat frequency. Here’s a step-by-step solution:

### Step 1: Convert the cyclist's speed to meters per second
The cyclist's speed is given as \(12 \, \text{km/h}\). To convert this to meters per second, we use the conversion factor \(1 \, \text{km/h} = \frac{5}{18} \, \text{m/s}\):

\[ \text{Speed of cyclist} = 12 \, \text{km/h} \times \frac{5}{18} = \frac{60}{18} = \frac{10}{3} \, \text{m/s} \]

### Step 2: Write the Doppler effect formulas for both sources
Let the velocity of sound be \(V\). The frequency emitted by both sources A and B is \(F = 250 \, \text{Hz}\).

- For the cyclist moving towards source B:
\[ F_B' = F \frac{V + v_o}{V} = 250 \frac{V + \frac{10}{3}}{V} \]

- For the cyclist moving away from source A:
\[ F_A' = F \frac{V - v_o}{V} = 250 \frac{V - \frac{10}{3}}{V} \]

### Step 3: Set up the equation for beat frequency
The beat frequency is given as \(5 \, \text{beats/s}\), which is the difference between the two apparent frequencies:

\[ F_B' - F_A' = 5 \]

Substituting the expressions for \(F_B'\) and \(F_A'\):

\[ 250 \frac{V + \frac{10}{3}}{V} - 250 \frac{V - \frac{10}{3}}{V} = 5 \]

### Step 4: Simplify the equation
Factoring out \(250\) and simplifying:

\[ 250 \left( \frac{(V + \frac{10}{3}) - (V - \frac{10}{3})}{V} \right) = 5 \]

This simplifies to:

\[ 250 \left( \frac{\frac{20}{3}}{V} \right) = 5 \]

### Step 5: Solve for \(V\)
Rearranging the equation gives:

\[ \frac{5000}{3V} = 5 \]

Multiplying both sides by \(3V\):

\[ 5000 = 15V \]

Dividing both sides by \(15\):

\[ V = \frac{5000}{15} = \frac{1000}{3} \approx 333.33 \, \text{m/s} \]

### Conclusion
Thus, the velocity of sound is approximately \(333 \, \text{m/s}\).