Consider a simple pendulam of length 1 m. Its bob performs a circular motion is horizontal plane with its string making an angle `60^@` with the vertical. The centripetal accleration experienced by the bob is

Options

A. `17.3m//s^2`
B. `5.8m//s^2`
C. `10m//s^2`
D. `5m//s^2`

Answer

To find the centripetal acceleration experienced by the bob of a simple pendulum, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a simple pendulum of length \( L = 1 \, \text{m} \).
- The string makes an angle \( \theta = 60^\circ \) with the vertical.

2. **Identify Forces Acting on the Bob**:
- The weight of the bob \( Mg \) acts vertically downward.
- The tension \( T \) in the string acts along the string, making an angle \( \theta \) with the vertical.

3. **Resolve the Tension into Components**:
- The vertical component of tension:
\[ T \cos \theta = Mg \]
- The horizontal component of tension (which provides the centripetal force):
\[ T \sin \theta = F_c \]

4. **Substituting the Angle**:
- For \( \theta = 60^\circ \):
- \( \cos 60^\circ = \frac{1}{2} \)
- \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)

5. **Calculate Tension Using the Vertical Component**:
- From \( T \cos 60^\circ = Mg \):
\[ T \left(\frac{1}{2}\right) = Mg \implies T = 2Mg \]

6. **Calculate Centripetal Force Using the Horizontal Component**:
- From \( T \sin 60^\circ = F_c \):
\[ T \left(\frac{\sqrt{3}}{2}\right) = M a_c \]
- Substitute \( T = 2Mg \):
\[ 2Mg \left(\frac{\sqrt{3}}{2}\right) = M a_c \]
- Simplifying gives:
\[ \sqrt{3} g = a_c \]

7. **Final Calculation**:
- Using \( g \approx 10 \, \text{m/s}^2 \):
\[ a_c = \sqrt{3} \cdot 10 \approx 17.3 \, \text{m/s}^2 \]

### Conclusion:
The centripetal acceleration experienced by the bob is approximately \( 17.3 \, \text{m/s}^2 \).