A boy sitting on a swing which is moving to an angle of `30^(@)` from the vertical is blowing a whistle which is of frequency `1000Hz`. The whistle is `2m` from the point of support of the swing. If a girl stands in front of the swing, the maximum and minimum frequencies she will hear are
(velocity of sound `=330m//s`, `g=9.8m//s^(2)`)

Options

A. `1000,990Hz`
B. `1007,1000Hz`
C. `1007,993Hz`
D. `1100,900Hz`

Explanation

To solve the problem, we need to find the maximum and minimum frequencies that the girl will hear when the boy on the swing blows the whistle. The swing is moving at an angle of 30 degrees from the vertical, and we need to consider the Doppler effect due to the motion of the source (the boy on the swing) relative to the observer (the girl).

### Step-by-Step Solution:

1. **Determine the velocity of the swing (Vs)**:
The boy on the swing is moving in a circular path. The vertical height (h) from which the swing is at an angle of 30 degrees can be calculated using the formula:
\[ h = L(1 - \cos(\theta)) \]
where \(L\) is the length of the swing (2 m) and \(\theta\) is the angle (30 degrees).

First, calculate \(h\):
\[ h = 2(1 - \cos(30^\circ)) = 2(1 - \frac{\sqrt{3}}{2}) = 2(1 - 0.866) = 2 \times 0.134 = 0.268 \text{ m} \]

Now, using conservation of energy:
\[ \frac{1}{2} m V_s^2 = mgh \]
This simplifies to:
\[ V_s^2 = 2gh \]
Substituting \(g = 9.8 \, \text{m/s}^2\) and \(h = 0.268 \, \text{m}\):
\[ V_s^2 = 2 \times 9.8 \times 0.268 \approx 5.25 \implies V_s \approx \sqrt{5.25} \approx 2.29 \, \text{m/s} \]

2. **Calculate the maximum frequency (f_max)**:
When the source is moving towards the observer, the apparent frequency increases. The formula for the apparent frequency when the source is moving towards the observer is:
\[ f' = f \frac{v + v_o}{v - v_s} \]
where:
- \(f = 1000 \, \text{Hz}\) (actual frequency)
- \(v = 330 \, \text{m/s}\) (speed of sound)
- \(v_o = 0 \, \text{m/s}\) (the girl is stationary)
- \(v_s = V_s \approx 2.29 \, \text{m/s}\)

Substituting the values:
\[ f' = 1000 \frac{330 + 0}{330 - 2.29} = 1000 \frac{330}{327.71} \approx 1006.67 \, \text{Hz} \]

3. **Calculate the minimum frequency (f_min)**:
When the source is moving away from the observer, the apparent frequency decreases. The formula for the apparent frequency when the source is moving away is:
\[ f' = f \frac{v + v_o}{v + v_s} \]
Using the same values as before:
\[ f' = 1000 \frac{330 + 0}{330 + 2.29} = 1000 \frac{330}{332.29} \approx 992.15 \, \text{Hz} \]

4. **Final Results**:
- Maximum frequency \(f_{max} \approx 1006.67 \, \text{Hz}\)
- Minimum frequency \(f_{min} \approx 992.15 \, \text{Hz}\)

### Summary:
The maximum frequency the girl will hear is approximately **1006.67 Hz**, and the minimum frequency is approximately **992.15 Hz**.