A simple pendulum of time period of oscillation 2.0 s is suspended vertically. The point of suspension of the pendulum now starts moving vertically upwards with a velocity that varies with time t as `v = alphat` where `alpha = 3.5 ms^(-1)`. If the new time period of oscillation of ht pendulum is T, then find the value of 2.2T. (Take `g = 10 ms^(-2)`)

Explanation

To solve the problem, we need to find the new time period \( T \) of a simple pendulum when the point of suspension is moving upwards with a velocity that varies with time. The initial time period of the pendulum is given as 2.0 seconds, and the acceleration of the point of suspension is given as \( \alpha = 3.5 \, \text{ms}^{-2} \).

### Step-by-Step Solution:

1. **Identify the Effective Gravity**:
- When the pendulum is at rest, the effective gravitational acceleration \( g_{\text{effective}} \) is simply \( g \), which is \( 10 \, \text{ms}^{-2} \).

2. **Determine the New Effective Gravity**:
- When the point of suspension is moving upwards with an acceleration \( a = \alpha = 3.5 \, \text{ms}^{-2} \), the effective gravitational acceleration acting on the pendulum bob becomes:
\[ g_{\text{effective}} = g + a = 10 \, \text{ms}^{-2} + 3.5 \, \text{ms}^{-2} = 13.5 \, \text{ms}^{-2} \]

3. **Use the Formula for the Time Period**:
- The time period \( T \) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} \]
- Since we need to find the new time period \( T \) in terms of the original time period \( T_0 = 2.0 \, \text{s} \), we can set up the ratio:
\[ \frac{T}{T_0} = \sqrt{\frac{g}{g_{\text{effective}}}} \]
- Substituting the values:
\[ \frac{T}{2.0} = \sqrt{\frac{10}{13.5}} \]

4. **Calculate the New Time Period**:
- First, calculate \( \frac{10}{13.5} \):
\[ \frac{10}{13.5} \approx 0.74074 \]
- Now take the square root:
\[ \sqrt{0.74074} \approx 0.860 \]
- Therefore, we have:
\[ T \approx 2.0 \times 0.860 \approx 1.72 \, \text{s} \]

5. **Calculate \( 2.2T \)**:
- Now, we need to find \( 2.2T \):
\[ 2.2T \approx 2.2 \times 1.72 \approx 3.784 \, \text{s} \]

6. **Round Off to the Nearest Integer**:
- Rounding \( 3.784 \) to the nearest integer gives:
\[ 2.2T \approx 4 \, \text{s} \]

### Final Answer:
The value of \( 2.2T \) is approximately \( 4 \, \text{s} \).