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)/(T)))` where g is equal to

Options

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

Answer

To solve the problem, we need to find the effective acceleration due to gravity when a simple pendulum is suspended from a trolley that is accelerating horizontally with an acceleration \( \alpha \). The time period \( T \) of the pendulum is given by the formula:

\[ T = 2\pi \sqrt{\frac{I}{T}} \]

where \( I \) is the moment of inertia and \( T \) is the tension. However, we need to find the effective gravitational acceleration \( g' \) that will be used in the calculation of the time period.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Pendulum:**
- The weight of the pendulum bob acts downward with a force \( mg \).
- The trolley is accelerating horizontally with acceleration \( \alpha \).

2. **Determine the Effective Acceleration:**
- When the trolley accelerates, the pendulum will not hang vertically downwards. Instead, it will make an angle with the vertical due to the horizontal acceleration.
- The effective acceleration acting on the pendulum bob can be found using vector addition of the gravitational acceleration \( g \) (downward) and the horizontal acceleration \( \alpha \).

3. **Calculate the Resultant Acceleration:**
- The resultant effective acceleration \( g' \) can be calculated using the Pythagorean theorem:
\[ g' = \sqrt{g^2 + \alpha^2} \]

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

5. **Conclusion:**
- The effective gravitational acceleration \( g' \) when the pendulum is in a trolley accelerating horizontally is:
\[ g' = \sqrt{g^2 + \alpha^2} \]

### Final Answer:
The effective gravitational acceleration \( g' \) is given by:
\[ g' = \sqrt{g^2 + \alpha^2} \]