When the point of suspendion of pendulum is moved, its period of oscillation

Options

A. decreaes when it moves vertically upwards with an acceleration `a`
B. decreases when it moves vertcally downwards with acceleration greater than 2 g.
C. Increases when it moves horizontally with acceleration `a`
D. all of the above.

Explanation

To solve the question regarding the effect of moving the point of suspension of a pendulum on its period of oscillation, we will analyze each scenario step by step.

### Step-by-Step Solution:

1. **Understanding the Pendulum's Period**:
The period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} \]
where \( L \) is the length of the pendulum and \( g_{\text{effective}} \) is the effective acceleration due to gravity acting on the pendulum.

2. **Case 1: Point of Suspension Moves Vertically Upward with Acceleration \( a \)**:
- When the point of suspension moves upward with acceleration \( a \), the effective acceleration acting on the pendulum becomes:
\[ g_{\text{effective}} = g + a \]
- Since \( g_{\text{effective}} \) increases, the period \( T \) will be:
\[ T = 2\pi \sqrt{\frac{L}{g + a}} \]
- As the denominator increases, the time period \( T \) decreases.

3. **Case 2: Point of Suspension Moves Vertically Downward with Acceleration \( 2g \)**:
- When the point of suspension moves downward with acceleration \( 2g \), the effective acceleration becomes:
\[ g_{\text{effective}} = g - 2g = -g \]
- This implies that the pendulum cannot oscillate in a normal manner, leading to a decrease in the time period (as it effectively becomes undefined).

4. **Case 3: Point of Suspension Moves Horizontally with Acceleration \( a \)**:
- When the point of suspension moves horizontally with acceleration \( a \), the effective acceleration due to gravity can be calculated using vector addition:
\[ g_{\text{effective}} = \sqrt{g^2 + a^2} \]
- Since \( \sqrt{g^2 + a^2} > g \), the effective acceleration increases, leading to a decrease in the period:
\[ T = 2\pi \sqrt{\frac{L}{\sqrt{g^2 + a^2}}} \]

5. **Conclusion**:
- The period of oscillation decreases when the point of suspension moves vertically upward with acceleration \( a \) (Case 1).
- The period of oscillation decreases when the point of suspension moves vertically downward with acceleration \( 2g \) (Case 2).
- The period of oscillation decreases when the point of suspension moves horizontally with acceleration \( a \) (Case 3).

### Final Answer:
The period of oscillation decreases in both Case 1 and Case 2, and also in Case 3.

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