A pendulum bob has a speed of `3ms^(-1)` at its lowest position. The pendulum is `0.5` m long. The speed of the bob, when string makes an angle of `60^(@)` to the vertical is `("take, g"=10ms^(-1))`

Options

A. `2ms^(-1)`
B. `(1)/(2)ms^(-1)`
C. `1 ms^(-1)`
D. `2.5ms^(-1)`

Explanation

To solve the problem, we will use the principles of energy conservation and some basic trigonometry.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Speed at the lowest position, \( u = 3 \, \text{m/s} \)
- Length of the pendulum, \( R = 0.5 \, \text{m} \)
- Angle with the vertical, \( \theta = 60^\circ \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Height (h) when the Pendulum is at 60°:**
- The height \( h \) can be calculated using the formula:
\[ h = R - R \cos(60^\circ) \]
- Since \( \cos(60^\circ) = \frac{1}{2} \):
\[ h = R - R \cdot \frac{1}{2} = R \left(1 - \frac{1}{2}\right) = R \cdot \frac{1}{2} = 0.5 \cdot \frac{1}{2} = 0.25 \, \text{m} \]

3. **Use Energy Conservation to Find the Speed (v) at 60°:**
- According to the conservation of mechanical energy:
\[ \text{K.E. at lowest point} + \text{P.E. at lowest point} = \text{K.E. at 60°} + \text{P.E. at 60°} \]
- At the lowest point, potential energy (P.E.) is zero, and kinetic energy (K.E.) is:
\[ \text{K.E.} = \frac{1}{2} m u^2 \]
- At the point where the pendulum makes an angle of 60°, the kinetic energy is:
\[ \text{K.E.} = \frac{1}{2} m v^2 \]
- The potential energy at this height \( h \) is:
\[ \text{P.E.} = mgh = mg(0.25) \]
- Setting the energies equal gives:
\[ \frac{1}{2} m u^2 = \frac{1}{2} m v^2 + mg(0.25) \]
- We can cancel \( m \) from the equation:
\[ \frac{1}{2} u^2 = \frac{1}{2} v^2 + g(0.25) \]
- Rearranging gives:
\[ \frac{1}{2} v^2 = \frac{1}{2} u^2 - g(0.25) \]
- Substituting \( u = 3 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \):
\[ \frac{1}{2} v^2 = \frac{1}{2} (3^2) - 10 \cdot 0.25 \]
\[ \frac{1}{2} v^2 = \frac{1}{2} (9) - 2.5 \]
\[ \frac{1}{2} v^2 = 4.5 - 2.5 = 2 \]
\[ v^2 = 4 \]
\[ v = \sqrt{4} = 2 \, \text{m/s} \]

4. **Final Answer:**
- The speed of the bob when the string makes an angle of \( 60^\circ \) to the vertical is \( v = 2 \, \text{m/s} \).