A man measures the period of a simple pendulum inside a stationary lift and find it be T. If the lift accelerates downwards with an acceleration of `g//3`, then the period of the pendulum will be

Options

A. `sqrt(2/3) T`
B. `sqrt(3/2) T`
C. `Tsqrt(3)`
D. `T/(sqrt3)`

Answer

To solve the problem, we need to determine the period of a simple pendulum when the lift is accelerating downwards with an acceleration of \( \frac{g}{3} \).

### Step-by-Step Solution:

1. **Understanding the Period of a Simple Pendulum**:
The period \( T \) of a 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. **Identifying the Effective Gravity**:
When the lift is stationary, the effective gravity is simply \( g \). However, when the lift accelerates downwards with an acceleration of \( \frac{g}{3} \), the effective gravity \( g' \) can be calculated as:
\[ g' = g - \frac{g}{3} \]
Simplifying this gives:
\[ g' = \frac{3g}{3} - \frac{g}{3} = \frac{2g}{3} \]

3. **Calculating the New Period**:
Now, we need to find the new period \( T' \) of the pendulum using the effective gravity \( g' \):
\[ T' = 2\pi \sqrt{\frac{L}{g'}} \]
Substituting \( g' \) into the equation:
\[ T' = 2\pi \sqrt{\frac{L}{\frac{2g}{3}}} \]
This can be rewritten as:
\[ T' = 2\pi \sqrt{\frac{3L}{2g}} \]

4. **Relating \( T' \) to \( T \)**:
From the original period \( T \):
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
We can express \( T' \) in terms of \( T \):
\[ T' = \sqrt{\frac{3}{2}} \cdot T \]

5. **Final Result**:
Therefore, the period of the pendulum when the lift accelerates downwards with an acceleration of \( \frac{g}{3} \) is:
\[ T' = T \sqrt{\frac{3}{2}} \]