The length of a simple pendulum is increased four times of its initial valuel, its time period with respect to its previous value will

Options

A. become twice
B. not be different
C. be halved
D. be `sqrt(2)` times

Explanation

To solve the problem, we need to analyze the relationship between the length of a simple pendulum and its time period. The time period \( T \) of a simple pendulum 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: Identify the initial conditions
Let the initial length of the pendulum be \( L \). Therefore, the initial time period \( T \) can be expressed as:

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

### Step 2: Determine the new length
According to the problem, the length of the pendulum is increased to four times its initial value. Thus, the new length \( L' \) is:

\[ L' = 4L \]

### Step 3: Calculate the new time period
Now, we need to find the new time period \( T' \) using the new length \( L' \):

\[ T' = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{4L}{g}} \]

### Step 4: Simplify the expression for the new time period
We can simplify \( T' \):

\[ T' = 2\pi \sqrt{\frac{4L}{g}} = 2\pi \cdot 2 \sqrt{\frac{L}{g}} = 2 \cdot (2\pi \sqrt{\frac{L}{g}}) \]

### Step 5: Relate the new time period to the initial time period
From the previous steps, we see that:

\[ T' = 2 \cdot T \]

This means the new time period \( T' \) is twice the initial time period \( T \).

### Conclusion
Thus, when the length of the simple pendulum is increased four times, its time period becomes twice its previous value.

### Final Answer
The time period with respect to its previous value will **become twice**.

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