If `T_(1)` and `T_(2)` are the time-periods of oscillation of a simple pendulum on the surface of earth (of radius R) and at a depth d, the d is equal to

Options

A. `(1-(T_(1)^(2))/(T_(2)^(2)))R`
B. `(1-(T_(2)^(2))/(T_(2)^(2)))R`
C. `(1-(T_(2))/(T_(2)))R`
D. `(1-(T_(2))/(T_(1)))R`

Explanation

To solve the problem of finding the depth \( d \) at which the time period of a simple pendulum changes, we can follow these steps:

### Step 1: Understand the formula for the time period of a pendulum
The time period \( T \) 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: Write the time periods for the two scenarios
1. For the pendulum on the surface of the Earth, the time period is:
\[ T_1 = 2\pi \sqrt{\frac{L}{g}} \]
2. For the pendulum at a depth \( d \), the effective acceleration due to gravity \( g' \) is given by:
\[ g' = g \left(1 - \frac{d}{R}\right) \]
Therefore, the time period at depth \( d \) is:
\[ T_2 = 2\pi \sqrt{\frac{L}{g'}} \]

### Step 3: Substitute \( g' \) into the equation for \( T_2 \)
Substituting \( g' \) into the equation for \( T_2 \):
\[ T_2 = 2\pi \sqrt{\frac{L}{g \left(1 - \frac{d}{R}\right)}} \]

### Step 4: Set up the ratio of the time periods
Taking the ratio \( \frac{T_1}{T_2} \):
\[ \frac{T_1}{T_2} = \frac{2\pi \sqrt{\frac{L}{g}}}{2\pi \sqrt{\frac{L}{g \left(1 - \frac{d}{R}\right)}}} \]
This simplifies to:
\[ \frac{T_1}{T_2} = \sqrt{\frac{g \left(1 - \frac{d}{R}\right)}{g}} = \sqrt{1 - \frac{d}{R}} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \left(\frac{T_1}{T_2}\right)^2 = 1 - \frac{d}{R} \]

### Step 6: Solve for \( d \)
Rearranging the equation to solve for \( d \):
\[ \frac{d}{R} = 1 - \left(\frac{T_1}{T_2}\right)^2 \]
Thus,
\[ d = R \left(1 - \left(\frac{T_1}{T_2}\right)^2\right) \]

### Final Answer
The depth \( d \) is given by:
\[ d = R \left(1 - \frac{T_1^2}{T_2^2}\right) \]