A source is moving with constant speed `upsilon_(s) = 20 m//s` towards a stationary observer due east of the source. Wind is blowing at the speed of `20 m//s` at `60^(@)` north of east. The source has frequency `500 H_(Z)`. Speed of sound `= 300 m//s`. The frequency resgistered by the observer is approximately

Options

A. `541 H_(Z)`
B. `552 H_(Z)`
C. `534 H_(Z)`
D. `512 H_(Z)`

Explanation

To solve the problem, we need to calculate the frequency registered by the observer when a sound source is moving towards them while wind is blowing in a specific direction. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Speed of the source, \( v_s = 20 \, \text{m/s} \)
- Frequency of the source, \( f_s = 500 \, \text{Hz} \)
- Speed of sound, \( v = 300 \, \text{m/s} \)
- Speed of wind, \( v_w = 20 \, \text{m/s} \)
- Wind direction: \( 60^\circ \) north of east

### Step 2: Resolve the wind speed into components
The wind speed needs to be resolved into its eastward and northward components. Since the wind is blowing at \( 60^\circ \) north of east:
- Eastward component of wind, \( v_{w, \text{east}} = v_w \cos(60^\circ) = 20 \cos(60^\circ) = 20 \times \frac{1}{2} = 10 \, \text{m/s} \)
- Northward component of wind, \( v_{w, \text{north}} = v_w \sin(60^\circ) = 20 \sin(60^\circ) = 20 \times \frac{\sqrt{3}}{2} \approx 17.32 \, \text{m/s} \)

### Step 3: Calculate the effective speed of sound towards the observer
The effective speed of sound towards the observer combines the speed of sound and the eastward component of the wind:
\[ v' = v + v_{w, \text{east}} = 300 \, \text{m/s} + 10 \, \text{m/s} = 310 \, \text{m/s} \]

### Step 4: Apply the Doppler effect formula
The frequency observed by the observer can be calculated using the Doppler effect formula:
\[ f' = f_s \cdot \frac{v' - 0}{v' - v_s} \]
Where:
- \( f' \) is the observed frequency,
- \( f_s \) is the source frequency,
- \( v' \) is the effective speed of sound towards the observer,
- \( v_s \) is the speed of the source.

Substituting the values:
\[ f' = 500 \cdot \frac{310 - 0}{310 - 20} = 500 \cdot \frac{310}{290} \]

### Step 5: Calculate the observed frequency
Now, calculate the observed frequency:
\[ f' = 500 \cdot \frac{310}{290} \approx 500 \cdot 1.068965517 = 534.48 \, \text{Hz} \]
Rounding off, we can say the frequency registered by the observer is approximately:
\[ f' \approx 534 \, \text{Hz} \]

### Conclusion
The frequency registered by the observer is approximately **534 Hz**.