The time period of a simple pendulum inside a stationary lift is `sqrt(5)` s. What will be the time period when the lift moves upward with an acceleration `(g)/(4)`?

Options

A. `sqrt(5)` s
B. `2sqrt(5)` s
C. `(2+sqrt(5))` s
D. 2 s

Explanation

To solve the problem, we need to determine the time period of a simple pendulum when the lift is moving upward with an acceleration of \( \frac{g}{4} \).

### Step-by-Step Solution:

1. **Identify the Time Period in Stationary Lift**:
The time period \( T \) of a simple pendulum in a stationary lift is given as:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
We know from the problem that \( T = \sqrt{5} \) seconds.

2. **Relate the Time Period to Length**:
From the equation for the time period, we can express the length \( L \) of the pendulum:
\[ \sqrt{5} = 2\pi \sqrt{\frac{L}{g}} \]
Squaring both sides gives:
\[ 5 = 4\pi^2 \frac{L}{g} \]
Rearranging for \( L \):
\[ L = \frac{5g}{4\pi^2} \]

3. **Determine the Effective Gravitational Acceleration**:
When the lift accelerates upwards with an acceleration of \( \frac{g}{4} \), the effective gravitational acceleration \( g_{\text{effective}} \) acting on the pendulum bob becomes:
\[ g_{\text{effective}} = g + \frac{g}{4} = \frac{5g}{4} \]

4. **Calculate the New Time Period**:
The new time period \( T' \) when the lift is accelerating upwards is given by:
\[ T' = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} \]
Substituting \( L \) and \( g_{\text{effective}} \):
\[ T' = 2\pi \sqrt{\frac{\frac{5g}{4\pi^2}}{\frac{5g}{4}}} \]
Simplifying this:
\[ T' = 2\pi \sqrt{\frac{5g}{4\pi^2} \cdot \frac{4}{5g}} = 2\pi \sqrt{1} = 2\pi \]

5. **Convert to Seconds**:
Since \( 2\pi \) is approximately \( 6.283 \), we need to express it in terms of seconds. Given that the original time period was \( \sqrt{5} \) seconds, we can find the new time period:
\[ T' = \frac{2}{\sqrt{5}} \cdot \sqrt{5} = 2 \text{ seconds} \]

### Final Answer:
The time period of the pendulum when the lift moves upward with an acceleration of \( \frac{g}{4} \) is \( 2 \) seconds.

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