The time period of a simple pendulum measured inside a stationary lift is found to be T . If the lift starts accelerating upwards with an acceleration `g //3`, the time period is

Options

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

Answer

To solve the problem, we need to analyze the time period of a simple pendulum in two different scenarios: when the lift is stationary and when it is accelerating upwards.

### Step-by-Step Solution:

1. **Identify the Time Period in the Stationary Lift**:
- The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
- Here, \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

2. **Consider the Lift Accelerating Upwards**:
- When the lift accelerates upwards with an acceleration \( a = \frac{g}{3} \), the effective gravitational acceleration \( g' \) experienced by the pendulum bob changes.
- The effective gravitational acceleration when the lift is accelerating upwards is given by:
\[ g' = g + a = g + \frac{g}{3} = g + \frac{g}{3} = \frac{3g}{3} + \frac{g}{3} = \frac{4g}{3} \]

3. **Calculate the New Time Period**:
- The new time period \( T' \) of the pendulum in the accelerating lift is given by:
\[ T' = 2\pi \sqrt{\frac{L}{g'}} \]
- Substituting \( g' \) into the equation:
\[ T' = 2\pi \sqrt{\frac{L}{\frac{4g}{3}}} = 2\pi \sqrt{\frac{3L}{4g}} = 2\pi \cdot \frac{1}{2} \sqrt{\frac{3L}{g}} = \frac{\sqrt{3}}{2} \cdot 2\pi \sqrt{\frac{L}{g}} \]
- Since \( 2\pi \sqrt{\frac{L}{g}} = T \), we can rewrite the equation as:
\[ T' = \frac{\sqrt{3}}{2} T \]

4. **Conclusion**:
- The new time period of the pendulum when the lift is accelerating upwards is:
\[ T' = \frac{\sqrt{3}}{2} T \]

### Final Answer:
The time period of the pendulum when the lift accelerates upwards with an acceleration of \( \frac{g}{3} \) is \( \frac{\sqrt{3}}{2} T \).