A trolley moves in horizontal direction with an acceleration a. A simple pendulum of length l is suspended from the roof of the trolley. The time period of the pendulum will be

Options

A. infinity
B. zero
C. `2pisqrt((l)/((g+a)))`
D. `2pisqrt((l)/(sqrt(a^(2)+g^(2))))`

Explanation

To find the time period of a simple pendulum suspended from a trolley that is accelerating horizontally, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a trolley moving with horizontal acceleration \( a \).
- A simple pendulum of length \( L \) is suspended from the roof of the trolley.

2. **Identifying Forces**:
- The pendulum experiences two accelerations:
- The gravitational acceleration \( g \) acting downwards.
- The pseudo force due to the trolley's acceleration \( a \) acting horizontally in the opposite direction of the trolley's motion.

3. **Effective Gravitational Acceleration**:
- The effective gravitational acceleration \( g_{\text{effective}} \) can be found by considering the resultant of the two accelerations (since they are perpendicular to each other).
- Using the Pythagorean theorem, we have:
\[ g_{\text{effective}} = \sqrt{g^2 + a^2} \]

4. **Time Period of the Pendulum**:
- The formula for the time period \( T \) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} \]
- Substituting \( g_{\text{effective}} \) into the formula, we get:
\[ T = 2\pi \sqrt{\frac{L}{\sqrt{g^2 + a^2}}} \]

5. **Final Expression**:
- Simplifying further, we can express the time period as:
\[ T = 2\pi \frac{\sqrt{L}}{\sqrt{g^2 + a^2}} \]

### Final Answer:
Thus, the time period of the pendulum is:
\[ T = 2\pi \sqrt{\frac{L}{\sqrt{a^2 + g^2}}} \]