A pendulum is suspended in a lit and its period of oscillation is `T_(0)` when the lift is stationary.
`(i)` What will be the period `T` of oscillation of pendulum, if the lift begins to accelerates downwards with an acceleration equal to `(3g)/(4)` ?

Answer

To solve the problem, we need to find the new period of oscillation \( T \) of a pendulum when the lift it is suspended in accelerates downwards with an acceleration of \( \frac{3g}{4} \). We start with the known period of oscillation \( T_0 \) when the lift is stationary.

### Step-by-Step Solution:

1. **Identify the initial period of oscillation**:
The period of a simple pendulum when in a stationary lift is given by:
\[ T_0 = 2\pi \sqrt{\frac{l}{g}} \]
where \( l \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

2. **Determine the effective acceleration due to gravity when the lift accelerates downwards**:
When the lift accelerates downwards with an acceleration of \( \frac{3g}{4} \), the effective gravitational acceleration \( g_{\text{effective}} \) acting on the pendulum is given by:
\[ g_{\text{effective}} = g - \text{acceleration of lift} \]
Substituting the value of the lift's acceleration:
\[ g_{\text{effective}} = g - \frac{3g}{4} = g \left(1 - \frac{3}{4}\right) = g \cdot \frac{1}{4} = \frac{g}{4} \]

3. **Calculate the new period of oscillation \( T \)**:
The new period of the pendulum when the lift accelerates downwards can be calculated using the effective gravitational acceleration:
\[ T = 2\pi \sqrt{\frac{l}{g_{\text{effective}}}} = 2\pi \sqrt{\frac{l}{\frac{g}{4}}} \]
Simplifying this expression:
\[ T = 2\pi \sqrt{\frac{l \cdot 4}{g}} = 2\pi \cdot 2 \sqrt{\frac{l}{g}} = 4\pi \sqrt{\frac{l}{g}} \]
Since \( T_0 = 2\pi \sqrt{\frac{l}{g}} \), we can express \( T \) in terms of \( T_0 \):
\[ T = 2 \cdot T_0 \]

4. **Final Result**:
Therefore, the period of oscillation \( T \) when the lift accelerates downwards with an acceleration of \( \frac{3g}{4} \) is:
\[ T = 2T_0 \]