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. 4s
B. 1s
C. `sqrt2s`
D. `2 sqrt2 s`

Answer

To solve the problem, we will use the formula for the time period of a simple pendulum. The time period \( T \) is given by the formula:

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

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

### Step 1: Write down the initial time period
We are given that the initial time period \( T \) is \( \sqrt{2} \) seconds.

### Step 2: Express the initial time period in terms of length
Using the formula for the time period, we can express it as:

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

### Step 3: Solve for \( L \)
To find the length \( L \), we can rearrange the equation:

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

Squaring both sides gives:

\[ \frac{L}{g} = \left(\frac{\sqrt{2}}{2\pi}\right)^2 \]

\[ L = g \left(\frac{2}{4\pi^2}\right) = \frac{g}{2\pi^2} \]

### Step 4: Calculate the new length
If the length is decreased to half of the initial length, the new length \( L' \) is:

\[ L' = \frac{L}{2} = \frac{1}{2} \cdot \frac{g}{2\pi^2} = \frac{g}{4\pi^2} \]

### Step 5: Find the new time period
Now, we can find the new time period \( T' \) using the new length \( L' \):

\[ T' = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{\frac{g}{4\pi^2}}{g}} = 2\pi \sqrt{\frac{1}{4\pi^2}} = 2\pi \cdot \frac{1}{2\pi} = 1 \text{ second} \]

### Conclusion
The new time period of the pendulum after its length is decreased to half is:

\[ T' = 1 \text{ second} \]