A simple pendulu has a time period `T_(1)`. The point of suspension of the pendulum is moved upword according to the relation `y= (3)/(2)t^(2)`, Where y is the vertical displacement . If the new time period is `T_(2)` , The ratio of `(T_(1)^(2))/(T_(2)^(2))` is (`g=10m//s^(2))`

Options

A. `(7)/(10)`
B. `(10)/(7)`
C. `(13)/(10)`
D. `(10)/(13)`

Explanation

To solve the problem, we need to find the ratio of the squares of the time periods of a simple pendulum before and after the point of suspension is moved upward. Let's break this down step by step.

### Step 1: Write the expression for the time period of the pendulum before the upward movement.
The time period \( T_1 \) of a simple pendulum is given by the formula:
\[ T_1 = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

### Step 2: Determine the new effective acceleration due to gravity after the upward movement.
The point of suspension is moved upward according to the relation:
\[ y = \frac{3}{2} t^2 \]
To find the acceleration of the point of suspension, we differentiate the displacement with respect to time:
\[ v = \frac{dy}{dt} = 3t \]
Differentiating again gives us the acceleration:
\[ a = \frac{dv}{dt} = 3 \, \text{m/s}^2 \]
This upward movement introduces a pseudo force acting downward on the pendulum bob, which modifies the effective acceleration due to gravity. The effective acceleration \( g_{\text{effective}} \) is given by:
\[ g_{\text{effective}} = g + a \]
Substituting \( g = 10 \, \text{m/s}^2 \) and \( a = 3 \, \text{m/s}^2 \):
\[ g_{\text{effective}} = 10 + 3 = 13 \, \text{m/s}^2 \]

### Step 3: Write the expression for the time period of the pendulum after the upward movement.
The new time period \( T_2 \) is given by:
\[ T_2 = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} = 2\pi \sqrt{\frac{L}{13}} \]

### Step 4: Find the ratio of the squares of the time periods.
Now, we can find the ratio of the squares of the time periods:
\[ \frac{T_1^2}{T_2^2} = \frac{\left(2\pi \sqrt{\frac{L}{g}}\right)^2}{\left(2\pi \sqrt{\frac{L}{g_{\text{effective}}}}\right)^2} \]
This simplifies to:
\[ \frac{T_1^2}{T_2^2} = \frac{\frac{L}{g}}{\frac{L}{g_{\text{effective}}}} = \frac{g_{\text{effective}}}{g} \]
Substituting the values we found:
\[ \frac{T_1^2}{T_2^2} = \frac{13}{10} \]

### Final Answer:
Thus, the ratio of \( \frac{T_1^2}{T_2^2} \) is:
\[ \frac{T_1^2}{T_2^2} = \frac{13}{10} \]

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