A man measures time period of a pendulum (T) in stationary lift. If the lift moves upward with acceleration `(g)/(4)`, then new time period will be

Options

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

Answer

To solve the problem of finding the new time period of a pendulum when the lift moves upward with an acceleration of \( \frac{g}{4} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Time Period of a Simple Pendulum**:
The time period \( T \) of a simple pendulum in a gravitational field 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. **Determine the Effective Gravity in the Moving Lift**:
When the lift is stationary, the effective gravitational acceleration is \( g \). However, when the lift accelerates upward with an acceleration of \( \frac{g}{4} \), the effective gravitational acceleration \( g_{\text{effective}} \) acting on the pendulum will change. The effective gravity can be calculated as:
\[ g_{\text{effective}} = g + a \]
where \( a \) is the upward acceleration of the lift. Here, \( a = \frac{g}{4} \), so:
\[ g_{\text{effective}} = g + \frac{g}{4} = g + 0.25g = \frac{5g}{4} \]

3. **Calculate the New Time Period**:
Now that we have the new effective gravity, we can substitute it back into the time period formula:
\[ T' = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} = 2\pi \sqrt{\frac{L}{\frac{5g}{4}}} \]
This simplifies to:
\[ T' = 2\pi \sqrt{\frac{4L}{5g}} = 2\sqrt{\frac{4}{5}} \cdot \pi \sqrt{\frac{L}{g}} = \frac{2}{\sqrt{5}} \cdot T \]
where \( T \) is the original time period when the lift is stationary.

4. **Final Result**:
Therefore, the new time period \( T' \) when the lift is moving upward with an acceleration of \( \frac{g}{4} \) is:
\[ T' = \frac{2}{\sqrt{5}} T \]