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)` ?
`(ii)` What must be the accleration of the lift for the period of oscillation of the pendulum to be `(T_(0))/(2)` ?

Answer

To solve the problem, we will break it down into two parts as given in the question.

### Part (i): Finding the period of oscillation `T` when the lift accelerates downwards with an acceleration of `(3g)/(4)`.

1. **Understanding the Effective Gravity**:
When the lift accelerates downwards, the effective acceleration due to gravity `g'` acting on the pendulum changes. The effective gravity can be calculated as:
\[ g' = g - a \]
where `a` is the downward acceleration of the lift. In this case, \( a = \frac{3g}{4} \).

2. **Calculating Effective Gravity**:
Substitute the value of `a` into the equation:
\[ g' = g - \frac{3g}{4} = \frac{g}{4} \]

3. **Finding the New Period of Oscillation**:
The period of a pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g'}} \]
Substitute the effective gravity \( g' = \frac{g}{4} \):
\[ T = 2\pi \sqrt{\frac{L}{\frac{g}{4}}} = 2\pi \sqrt{\frac{4L}{g}} = 2 \cdot 2\pi \sqrt{\frac{L}{g}} = 2T_0 \]
where \( T_0 = 2\pi \sqrt{\frac{L}{g}} \) is the period when the lift is stationary.

### Conclusion for Part (i):
The period of oscillation of the pendulum when the lift accelerates downwards with an acceleration of \( \frac{3g}{4} \) is:
\[ T = 2T_0 \]

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### Part (ii): Finding the acceleration of the lift for the period of oscillation of the pendulum to be \( \frac{T_0}{2} \).

1. **Setting Up the Equation**:
We want to find the acceleration `a` such that the new period \( T \) is \( \frac{T_0}{2} \). Using the formula for the period:
\[ T = 2\pi \sqrt{\frac{L}{g'}} \]
We set:
\[ \frac{T_0}{2} = 2\pi \sqrt{\frac{L}{g'}} \]

2. **Rearranging the Equation**:
Squaring both sides gives:
\[ \left(\frac{T_0}{2}\right)^2 = 4\pi^2 \frac{L}{g'} \]
Substitute \( T_0 = 2\pi \sqrt{\frac{L}{g}} \):
\[ \left(\frac{2\pi \sqrt{\frac{L}{g}}}{2}\right)^2 = 4\pi^2 \frac{L}{g'} \]
Simplifying gives:
\[ \frac{L}{g} = \frac{4L}{g'} \]

3. **Finding Effective Gravity**:
Rearranging gives:
\[ g' = 4g \]

4. **Relating Effective Gravity to Lift Acceleration**:
Since \( g' = g - a \), we have:
\[ 4g = g - a \]
Rearranging gives:
\[ a = g - 4g = -3g \]

### Conclusion for Part (ii):
The acceleration of the lift must be \( -3g \) (upwards) for the period of oscillation of the pendulum to be \( \frac{T_0}{2} \).

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