A simple pendulum has time period `2s`. The point of suspension is now moved upward accoding to relation `y = (6t - 3.75t^(2))m` where `t` is in second and y is the vertical displacement in upward direction. The new time period of simple pendulum will be

Options

A. `2s`
B. `1s`
C. `4s`
D. None of these

Explanation

To solve the problem, we need to find the new time period of a simple pendulum when the point of suspension is moved upward according to the given relation \( y = 6t - 3.75t^2 \).

### Step-by-Step Solution:

1. **Identify the given parameters:**
- The initial time period \( T = 2 \) seconds.
- The vertical displacement \( y(t) = 6t - 3.75t^2 \).

2. **Calculate the first derivative of \( y(t) \):**
\[ \frac{dy}{dt} = \frac{d}{dt}(6t - 3.75t^2) = 6 - 7.5t \]

3. **Calculate the second derivative of \( y(t) \):**
\[ \frac{d^2y}{dt^2} = \frac{d}{dt}(6 - 7.5t) = -7.5 \]
This indicates that the acceleration \( a = -7.5 \, \text{m/s}^2 \).

4. **Determine the effective acceleration due to gravity:**
The effective acceleration \( g_{\text{effective}} \) when the pendulum is moving upward is given by:
\[ g_{\text{effective}} = g + a \]
Assuming \( g = 10 \, \text{m/s}^2 \):
\[ g_{\text{effective}} = 10 - 7.5 = 2.5 \, \text{m/s}^2 \]

5. **Relate the time period to the effective gravity:**
The time period of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
For the new time period \( T' \):
\[ T' = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} = 2\pi \sqrt{\frac{L}{2.5}} \]

6. **Express \( T' \) in terms of the original time period \( T \):**
Since the original time period \( T = 2\pi \sqrt{\frac{L}{10}} \):
\[ T' = 2\pi \sqrt{\frac{L}{2.5}} = 2\pi \sqrt{\frac{L}{10}} \cdot \sqrt{4} = 2T \]
Therefore:
\[ T' = 2 \times 2 = 4 \, \text{seconds} \]

### Final Answer:
The new time period of the simple pendulum will be \( 4 \) seconds.