The time period of a `2m` long simple pendulum is 2 seconds. Find the value of `'g'` at that place?

Options

A. `2pi^2`
B. `pi^2`
C. `4pi^2`
D. `pi^/2`

Answer

To find the value of 'g' using the given information about the simple pendulum, we can follow these steps:

### Step 1: Write the formula for the time period of a simple pendulum
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 2: Substitute the known values into the formula
We are given:
- Length \( L = 2 \) m,
- Time period \( T = 2 \) s.

Substituting these values into the formula:

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

### Step 3: Simplify the equation
Divide both sides by \( 2 \):

\[ 1 = \pi \sqrt{\frac{2}{g}} \]

### Step 4: Isolate the square root
To isolate the square root, divide both sides by \( \pi \):

\[ \frac{1}{\pi} = \sqrt{\frac{2}{g}} \]

### Step 5: Square both sides to eliminate the square root
Now, square both sides:

\[ \left(\frac{1}{\pi}\right)^2 = \frac{2}{g} \]

### Step 6: Rearrange to solve for \( g \)
Rearranging gives:

\[ g = \frac{2}{\left(\frac{1}{\pi}\right)^2} \]

### Step 7: Simplify the expression
This can be simplified as:

\[ g = 2\pi^2 \]

### Step 8: Calculate the numerical value
Using \( \pi \approx 3.14 \):

\[ g \approx 2 \times (3.14)^2 \approx 2 \times 9.8596 \approx 19.7192 \, \text{m/s}^2 \]

Thus, the value of \( g \) at that place is approximately \( 19.72 \, \text{m/s}^2 \).

### Final Answer:
\[ g \approx 19.72 \, \text{m/s}^2 \]

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