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 time periods of a simple pendulum under different conditions: when the lift is stationary, when it descends at a steady speed, and when it descends with constant downward acceleration.

### Step-by-Step Solution:

1. **Time Period in Stationary Lift (T₀)**:
The time period of a simple pendulum in a stationary lift is given by the formula:
\[ T_0 = 2\pi \sqrt{\frac{L}{g}} \]
where \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity.

2. **Time Period in Lift Descending at Steady Speed (T₁)**:
When the lift descends at a steady speed, the effective gravitational acceleration acting on the pendulum does not change. Thus, the time period remains the same as in the stationary case:
\[ T_1 = T_0 = 2\pi \sqrt{\frac{L}{g}} \]

3. **Time Period in Lift Descending with Constant Downward Acceleration (T₂)**:
When the lift descends with a constant downward acceleration \(a\), the effective gravitational acceleration becomes:
\[ g_{\text{effective}} = g - a \]
Therefore, the time period of the pendulum in this case is:
\[ T_2 = 2\pi \sqrt{\frac{L}{g - a}} \]

4. **Comparison of Time Periods**:
- We have established that \(T_0 = T_1\).
- For \(T_2\), since \(g - a < g\) (because \(a\) is positive), it follows that:
\[ T_2 = 2\pi \sqrt{\frac{L}{g - a}} > 2\pi \sqrt{\frac{L}{g}} = T_0 \]
Therefore, we can conclude:
\[ T_2 > T_0 = T_1 \]

### Conclusion:
From the analysis, we find that:
- \(T_0 = T_1\)
- \(T_2 > T_0\) and \(T_2 > T_1\)

Thus, the correct relationship is:
\[ T_0 = T_1 < T_2 \]