The time period of oscillation of a simple pendulum is `sqrt(2)s`. If its length is decreased to half of initial length, then its new period is

Options

A. `1s`
B. `0.707s`
C. `0.414s`
D. `0.5s`

Explanation

To solve the problem, we need to find the new time period of a simple pendulum when its length is decreased to half of its initial length. The formula for the time period \( T \) of a simple pendulum is given by:

\[ T = 2\pi \sqrt{\frac{L}{g}} \]

where:
- \( T \) is the time period,
- \( L \) is the length of the pendulum,
- \( g \) is the acceleration due to gravity (which remains constant).

### Step 1: Identify the initial time period and length
We are given that the initial time period \( T_1 \) is \( \sqrt{2} \) seconds. Let the initial length of the pendulum be \( L_1 \).

### Step 2: Write the expression for the initial time period
Using the formula for the time period:

\[ T_1 = 2\pi \sqrt{\frac{L_1}{g}} \]

### Step 3: Find the new length
The problem states that the length is decreased to half of the initial length. Therefore, the new length \( L_2 \) is:

\[ L_2 = \frac{L_1}{2} \]

### Step 4: Write the expression for the new time period
Now, we can express the new time period \( T_2 \) using the new length \( L_2 \):

\[ T_2 = 2\pi \sqrt{\frac{L_2}{g}} = 2\pi \sqrt{\frac{L_1/2}{g}} = 2\pi \sqrt{\frac{L_1}{2g}} \]

### Step 5: Relate the new time period to the initial time period
We can relate \( T_2 \) to \( T_1 \):

\[ T_2 = 2\pi \sqrt{\frac{L_1}{g}} \cdot \frac{1}{\sqrt{2}} = T_1 \cdot \frac{1}{\sqrt{2}} \]

### Step 6: Substitute the value of \( T_1 \)
Now substitute \( T_1 = \sqrt{2} \):

\[ T_2 = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \text{ second} \]

### Conclusion
Thus, the new time period \( T_2 \) when the length is decreased to half is:

\[ \boxed{1 \text{ second}} \]