Time period of a simple pendulum of length L is `T_(1)` and time period of a uniform rod of the same length L pivoted about an end and oscillating in a vertical plane is `T_(2)`. Amplitude of osciallations in both the cases is small. Then `T_(1)/T_(2)` is

Options

A. `sqrt(4/3)`
B. 1
C. `sqrt(3/2)`
D. `sqrt(1/3)`

Answer

To find the ratio of the time periods \( \frac{T_1}{T_2} \) for a simple pendulum and a uniform rod pivoted at one end, we can follow these steps:

### Step 1: Determine the time period of the simple pendulum

The time period \( T_1 \) of a simple pendulum of length \( L \) is given by the formula:
\[ T_1 = 2\pi \sqrt{\frac{L}{g}} \]
where \( g \) is the acceleration due to gravity.

### Step 2: Determine the time period of the uniform rod

For a uniform rod of length \( L \) pivoted about one end, the time period \( T_2 \) can be calculated using the formula:
\[ T_2 = 2\pi \sqrt{\frac{I}{MgL}} \]
where:
- \( I \) is the moment of inertia of the rod about the pivot point,
- \( M \) is the mass of the rod,
- \( g \) is the acceleration due to gravity,
- \( L \) is the length of the rod.

The moment of inertia \( I \) of a uniform rod about an end is given by:
\[ I = \frac{1}{3}ML^2 \]

### Step 3: Substitute the moment of inertia into the time period formula

Substituting \( I \) into the equation for \( T_2 \):
\[ T_2 = 2\pi \sqrt{\frac{\frac{1}{3}ML^2}{MgL}} \]
This simplifies to:
\[ T_2 = 2\pi \sqrt{\frac{L}{3g}} \]

### Step 4: Calculate the ratio \( \frac{T_1}{T_2} \)

Now we can find the ratio \( \frac{T_1}{T_2} \):
\[ \frac{T_1}{T_2} = \frac{2\pi \sqrt{\frac{L}{g}}}{2\pi \sqrt{\frac{L}{3g}}} \]
The \( 2\pi \) cancels out:
\[ \frac{T_1}{T_2} = \frac{\sqrt{\frac{L}{g}}}{\sqrt{\frac{L}{3g}}} \]
This simplifies to:
\[ \frac{T_1}{T_2} = \sqrt{\frac{3g}{g}} = \sqrt{3} \]

### Step 5: Final result

Thus, the ratio \( \frac{T_1}{T_2} \) is:
\[ \frac{T_1}{T_2} = \sqrt{3} \]