The time period of a simple pendulum is T remaining at rest inside a lift. Find the time period of pendulum when lift starts to move up with an acceleration of g/4

Options

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

Explanation

To solve the problem step by step, we will follow the concepts of the time period of a simple pendulum and how it changes under different conditions of acceleration.

### Step 1: Understand the time period of a simple pendulum
The time period \( T \) of a simple pendulum at rest is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where:
- \( L \) is the length of the pendulum,
- \( g \) is the acceleration due to gravity.

### Step 2: Determine the effective acceleration when the lift accelerates upwards
When the lift starts moving upwards with an acceleration of \( \frac{g}{4} \), the effective acceleration due to gravity \( g' \) inside the lift becomes:
\[ g' = g + \frac{g}{4} = g + 0.25g = \frac{5g}{4} \]

### Step 3: Write the new time period formula for the pendulum in the moving lift
The new time period \( T' \) of the pendulum when the lift is accelerating upwards is given by:
\[ T' = 2\pi \sqrt{\frac{L}{g'}} \]
Substituting \( g' \) into the equation:
\[ T' = 2\pi \sqrt{\frac{L}{\frac{5g}{4}}} \]

### Step 4: Simplify the expression for the new time period
We can simplify \( T' \):
\[ T' = 2\pi \sqrt{\frac{4L}{5g}} = 2\pi \cdot \sqrt{\frac{4}{5}} \cdot \sqrt{\frac{L}{g}} \]
Since \( T = 2\pi \sqrt{\frac{L}{g}} \), we can express \( T' \) in terms of \( T \):
\[ T' = T \cdot \sqrt{\frac{4}{5}} \]

### Step 5: Calculate the ratio of the new time period to the original time period
Now, we can express \( T' \) as:
\[ T' = T \cdot \frac{2}{\sqrt{5}} \]

### Final Result
Thus, the time period of the pendulum when the lift starts to move up with an acceleration of \( \frac{g}{4} \) is:
\[ T' = \frac{2T}{\sqrt{5}} \]