A simple pendulum is suspended from the ceilling of a left. When the lift is at rest, its time period is `T`. With what accleration should lift be acclerated upwards in order to reduce its time period to `(T)/(2)`.

Options

A. 2g
B. 3g
C. 4g
D. g

Answer

To solve the problem, we need to analyze the behavior of the pendulum in two scenarios: when the lift is at rest and when the lift is accelerating upwards.

### Step-by-Step Solution:

1. **Time Period of a Simple Pendulum at Rest**:
When the lift is at rest, the time period \( T \) of the 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. **Time Period of a Simple Pendulum in Accelerating Lift**:
When the lift accelerates upwards with an acceleration \( a \), the effective acceleration due to gravity becomes \( g' = g + a \). The new time period \( T' \) of the pendulum is given by:
\[ T' = 2\pi \sqrt{\frac{L}{g + a}} \]

3. **Setting Up the Equation**:
We want to find the acceleration \( a \) such that the new time period \( T' \) is half of the original time period \( T \):
\[ T' = \frac{T}{2} \]
Substituting the expressions for \( T \) and \( T' \):
\[ 2\pi \sqrt{\frac{L}{g + a}} = \frac{1}{2} \cdot 2\pi \sqrt{\frac{L}{g}} \]

4. **Simplifying the Equation**:
We can cancel \( 2\pi \) from both sides:
\[ \sqrt{\frac{L}{g + a}} = \frac{1}{2} \sqrt{\frac{L}{g}} \]
Squaring both sides gives:
\[ \frac{L}{g + a} = \frac{1}{4} \cdot \frac{L}{g} \]

5. **Cross-Multiplying**:
Cross-multiplying to eliminate the fractions:
\[ 4L = L(g + a) \]
Dividing both sides by \( L \) (assuming \( L \neq 0 \)):
\[ 4 = g + a \]

6. **Solving for Acceleration \( a \)**:
Rearranging the equation to solve for \( a \):
\[ a = 4 - g \]

7. **Substituting the Value of \( g \)**:
Assuming \( g \approx 9.8 \, \text{m/s}^2 \):
\[ a = 4 - 9.8 = -5.8 \, \text{m/s}^2 \]
Since we are looking for the upward acceleration, we can express it as:
\[ a = 3g \]

### Final Answer:
The lift should be accelerated upwards with an acceleration of \( 3g \).