A simple pendulum has time period `T_(1)` The point of suspension is now moved upward according to the relation `y = kt^(2)(k = 1 m//s^(2))` where `y` is vertical displacement, the time period now becomes `T_(2)`. The ratio of `((T_(1))/(T_(2)))^(2)` is : `(g = 10 m//s^(2))`

Options

A. `(4)/(5)`
B. `(6)/(5)`
C. `(5)/(60)`
D. 1

Answer

To solve the problem, we need to find the ratio of the squares of the time periods \( T_1 \) and \( T_2 \) of a simple pendulum when the point of suspension is moved upward with a specific acceleration. Let's break down the solution step by step.

### Step 1: Understand the Time Period of a Simple Pendulum
The time period of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

### Step 2: Determine the Acceleration
The point of suspension is moved upward according to the relation:
\[ y = kt^2 \]
where \( k = 1 \, \text{m/s}^2 \). The acceleration \( a \) is the second derivative of displacement with respect to time:
\[ a = \frac{d^2y}{dt^2} = \frac{d^2(kt^2)}{dt^2} = 2k = 2 \times 1 = 2 \, \text{m/s}^2 \]

### Step 3: Modify the Time Period Formula for \( T_2 \)
When the point of suspension is moving upward with an acceleration \( a \), the effective acceleration acting on the pendulum becomes \( g - a \). Therefore, the new time period \( T_2 \) is given by:
\[ T_2 = 2\pi \sqrt{\frac{L}{g - a}} \]

### Step 4: Write the Time Periods in Terms of \( L \) and \( g \)
Now we have:
\[ T_1 = 2\pi \sqrt{\frac{L}{g}} \]
\[ T_2 = 2\pi \sqrt{\frac{L}{g - 2}} \]

### Step 5: Find the Ratio of the Time Periods
Now we need to find the ratio \( \left(\frac{T_1}{T_2}\right)^2 \):
\[ \frac{T_1}{T_2} = \frac{2\pi \sqrt{\frac{L}{g}}}{2\pi \sqrt{\frac{L}{g - 2}}} = \frac{\sqrt{\frac{L}{g}}}{\sqrt{\frac{L}{g - 2}}} \]
This simplifies to:
\[ \frac{T_1}{T_2} = \sqrt{\frac{g - 2}{g}} \]
Squaring both sides gives:
\[ \left(\frac{T_1}{T_2}\right)^2 = \frac{g - 2}{g} \]

### Step 6: Substitute the Value of \( g \)
Given \( g = 10 \, \text{m/s}^2 \):
\[ \left(\frac{T_1}{T_2}\right)^2 = \frac{10 - 2}{10} = \frac{8}{10} = \frac{4}{5} \]

### Final Answer
Thus, the ratio \( \left(\frac{T_1}{T_2}\right)^2 \) is:
\[ \frac{4}{5} \]