Time period of a pendulum in a stationary lift is 2 seconds. Now the lift is moving up at a speed of 30 m/s but slowing down at a rate of 5 m/`s^2`. What is the time period of the pendulum now?

Options

A. 1.6s
B. `2.8s`
C. `1.1s`
D. `2.3s`

Answer

To solve the problem, we need to determine the time period of a pendulum when the lift is moving upwards and decelerating. Here are the steps to arrive at the solution:

### Step 1: Understand the initial conditions
The time period of the pendulum in a stationary lift is given as \( T = 2 \) seconds. The formula for the time period of a simple pendulum is:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

### Step 2: Calculate the effective gravity in the moving lift
When the lift is moving upwards with a speed of \( 30 \, \text{m/s} \) and decelerating at \( 5 \, \text{m/s}^2 \), we need to find the effective acceleration due to gravity \( g_{\text{effective}} \). The effective gravity is given by:
\[ g_{\text{effective}} = g - a \]
where \( a \) is the deceleration of the lift. Here, \( g \approx 10 \, \text{m/s}^2 \) and \( a = 5 \, \text{m/s}^2 \).

Substituting the values:
\[ g_{\text{effective}} = 10 - 5 = 5 \, \text{m/s}^2 \]

### Step 3: Calculate the new time period
Now, we can calculate the new time period \( T' \) using the effective gravity:
\[ T' = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} = 2\pi \sqrt{\frac{L}{5}} \]

### Step 4: Relate the new time period to the original time period
From the original time period equation, we have:
\[ T = 2\pi \sqrt{\frac{L}{10}} = 2 \, \text{seconds} \]
Squaring both sides:
\[ T^2 = 4 = 4\pi^2 \frac{L}{10} \implies L = \frac{10}{\pi^2} \cdot 4 \]

Now substituting \( L \) back into the equation for \( T' \):
\[ T' = 2\pi \sqrt{\frac{\frac{10 \cdot 4}{\pi^2}}{5}} = 2\pi \sqrt{\frac{8}{\pi^2}} = 2\sqrt{8} = 4\sqrt{2} \]

### Step 5: Calculate the numerical value
Now, calculating \( 4\sqrt{2} \):
\[ \sqrt{2} \approx 1.41 \implies T' \approx 4 \times 1.41 = 5.64 \, \text{seconds} \]

### Conclusion
The new time period of the pendulum when the lift is moving upwards and slowing down is approximately \( 5.64 \) seconds.