A simple pendulum suspended from the celling of a stationary lift has period `T_(0)`. When the lift descends at steady speed, the period is `T_(1)`, and when it descends with constant downward acceleration, the period is `T_(2)` which one of the following is true?

Options

A. `T_(0) = T_(1) = T_(2)`
B. `T_(0) = T_(1) lt T_(2)`
C. `T_(0) = T_(1) gt T_(2)`
D. `T_(0) lt T_(1) lt T_(2)`

Answer

To solve the problem, we need to analyze the behavior of a simple pendulum in different scenarios involving a lift. We will use the formula for the period of a simple pendulum and consider the effects of the lift's motion on the effective gravitational acceleration.

### 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. **Case 1: Stationary Lift**:
When the lift is stationary, the period is:
\[ T_0 = 2\pi \sqrt{\frac{L}{g}} \]

3. **Case 2: Lift Descending at Steady Speed**:
When the lift descends at a steady speed, there is no acceleration acting on the pendulum. Thus, the effective gravitational acceleration remains the same:
\[ T_1 = 2\pi \sqrt{\frac{L}{g}} \]
Therefore, we have:
\[ T_1 = T_0 \]

4. **Case 3: Lift Descending with Constant Downward Acceleration**:
When the lift descends with a constant downward acceleration \( a \), we need to consider the effective gravitational acceleration. The effective gravitational acceleration \( g_{\text{effective}} \) becomes:
\[ g_{\text{effective}} = g - a \]
Thus, the period in this case is:
\[ T_2 = 2\pi \sqrt{\frac{L}{g - a}} \]

5. **Comparing the Periods**:
- Since \( a \) is positive (the lift is accelerating downwards), \( g - a < g \). This means that the denominator in the expression for \( T_2 \) is less than that for \( T_0 \) and \( T_1 \), leading to:
\[ T_2 > T_0 \quad \text{and} \quad T_2 > T_1 \]
- Since \( T_0 = T_1 \), we conclude:
\[ T_0 = T_1 < T_2 \]

### Conclusion:
The final relationships between the periods are:
\[ T_0 = T_1 < T_2 \]

### Final Answer:
The correct option is \( T_0 = T_1 \) and both are less than \( T_2 \).

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