A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration `alpha`, then the time period is given by `T = 2pisqrt(((I)/(g)))` where g is equal to

Options

A. g
B. g-a
C. g+a
D. `sqrt(g^(2)+a^(2))`

Explanation

To solve the problem of a simple pendulum suspended from the roof of a trolley that moves horizontally with an acceleration \( \alpha \), we need to determine the effective gravitational acceleration \( g' \) that affects the pendulum's motion and then find the time period \( T \).

### Step-by-Step Solution:

1. **Identify Forces Acting on the Pendulum Bob:**
The forces acting on the pendulum bob of mass \( m \) are:
- The gravitational force acting downward: \( F_g = mg \)
- The pseudo force acting horizontally due to the trolley's acceleration: \( F_p = ma \) (where \( a \) is the acceleration of the trolley)

2. **Determine the Effective Gravitational Force:**
When the trolley accelerates, the pendulum bob experiences an effective force that is a combination of the gravitational force and the pseudo force. This can be visualized as a resultant force acting diagonally.

3. **Calculate the Resultant Force:**
Using the Pythagorean theorem, the effective gravitational force \( g' \) can be expressed as:
\[ g' = \sqrt{g^2 + a^2} \]
Here, \( g \) is the acceleration due to gravity, and \( a \) is the horizontal acceleration of the trolley.

4. **Write the Formula for the Time Period of the Pendulum:**
The time period \( T \) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g'}} \]
Substituting \( g' \) into this formula gives:
\[ T = 2\pi \sqrt{\frac{L}{\sqrt{g^2 + a^2}}} \]

5. **Final Expression for the Time Period:**
Thus, the time period of the pendulum in the accelerating trolley is:
\[ T = 2\pi \sqrt{\frac{L}{\sqrt{g^2 + a^2}}} \]

### Conclusion:
The effective gravitational acceleration \( g' \) is \( \sqrt{g^2 + a^2} \), and the time period \( T \) of the pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{\sqrt{g^2 + a^2}}} \]