A simple pendulum has time period `25 .` The point of suspension is now moved upward according to relation `y=(6 t-3.75 t^2) m` where `t` is in second and `y` is the vertical displacement in upward direction. The new time period (in s) of simple pendulum will be `(g=10 ms^(-2))`

Answer

To solve the problem of finding the new time period of a simple pendulum when the point of suspension is moved upward, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Initial time period \( T_1 = 25 \) seconds.
- The vertical displacement of the point of suspension is given by the equation:
\[ y(t) = 6t - 3.75t^2 \]
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \).

2. **Differentiate the Displacement to Find Velocity**:
- Differentiate \( y(t) \) with respect to time \( t \) to find the velocity \( v(t) \):
\[ v(t) = \frac{dy}{dt} = 6 - 7.5t \]

3. **Differentiate the Velocity to Find Acceleration**:
- Differentiate \( v(t) \) with respect to time \( t \) to find the acceleration \( a(t) \):
\[ a(t) = \frac{dv}{dt} = -7.5 \, \text{m/s}^2 \]

4. **Determine the New Time Period**:
- The formula for the time period of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
- When the point of suspension moves upward, the effective acceleration due to gravity changes. The new effective acceleration \( g' \) is given by:
\[ g' = g + a \]
- Substitute \( a = -7.5 \, \text{m/s}^2 \) into the equation:
\[ g' = 10 - 7.5 = 2.5 \, \text{m/s}^2 \]

5. **Calculate the New Time Period \( T_2 \)**:
- The new time period \( T_2 \) can be expressed as:
\[ T_2 = 2\pi \sqrt{\frac{L}{g'}} \]
- To find the ratio of the time periods \( \frac{T_1}{T_2} \):
\[ \frac{T_1}{T_2} = \sqrt{\frac{g'}{g}} = \sqrt{\frac{2.5}{10}} = \sqrt{0.25} = 0.5 \]
- Thus, we can express \( T_2 \) in terms of \( T_1 \):
\[ T_2 = 2T_1 = 2 \times 25 = 50 \, \text{seconds} \]

6. **Final Calculation**:
- The new time period of the simple pendulum is:
\[ T_2 = 50 \, \text{seconds} \]

### Final Answer:
The new time period of the simple pendulum is \( 50 \, \text{seconds} \).