A man measures the period of a simple pendulum inside a stationary lift and finds it to be T sec. if the lift accelerates upwards with an acceleration g/4, then the period of the pendulum will be

Options

A. T
B. `(T)/(4)`
C. `(2T)/(sqrt(5))`
D. `2Tsqrt(5)`

Explanation

To solve the problem, we need to determine the period of a simple pendulum when the lift it is in accelerates upwards with an acceleration of \( g/4 \).

### Step-by-Step Solution:

1. **Understand the Initial Condition**:
When the lift is stationary, the period \( T \) of the simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

2. **Consider the Effect of Lift's Acceleration**:
When the lift accelerates upwards with an acceleration of \( g/4 \), the effective gravitational acceleration \( g_{\text{effective}} \) acting on the pendulum bob changes. The effective acceleration is the sum of the gravitational acceleration and the lift's upward acceleration:
\[ g_{\text{effective}} = g + \frac{g}{4} = g + 0.25g = \frac{5g}{4} \]

3. **Calculate the New Period**:
The new period \( T' \) of the pendulum in the accelerating lift can be calculated using the modified effective gravitational acceleration:
\[ T' = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} = 2\pi \sqrt{\frac{L}{\frac{5g}{4}}} \]
Simplifying this expression:
\[ T' = 2\pi \sqrt{\frac{4L}{5g}} = 2\sqrt{\frac{4}{5}} \cdot \pi \sqrt{\frac{L}{g}} = \frac{2}{\sqrt{5}} \cdot T \]

4. **Final Expression for the Period**:
Therefore, the new period of the pendulum when the lift is accelerating upwards is:
\[ T' = \frac{2T}{\sqrt{5}} \]

### Conclusion:
The period of the pendulum when the lift accelerates upwards with an acceleration of \( g/4 \) is:
\[ \frac{2T}{\sqrt{5}} \]