A pendulum is hung the roof of a sufficiently high huilding and is moving freely to and fro like a simple harmonic oscillator .The acceleration of the bob of the pendulum is `20m//s^(2)` at a distance of `5m` from the meanposition .The time period of oscillation is

Options

A. 2S
B. `piS`
C. `2piS`
D. 1S

Answer

To solve the problem, we need to find the time period of a pendulum that is oscillating in simple harmonic motion (SHM). We are given the acceleration of the bob at a certain displacement from the mean position.

### Step-by-Step Solution:

1. **Understand the relationship between acceleration and displacement in SHM**:
In simple harmonic motion, the acceleration \( a \) of the bob is given by the formula:
\[ a = -\omega^2 x \]
where:
- \( \omega \) is the angular frequency,
- \( x \) is the displacement from the mean position.

2. **Identify the given values**:
From the problem, we know:
- The acceleration \( a = 20 \, \text{m/s}^2 \)
- The displacement \( x = 5 \, \text{m} \)

3. **Use the formula to find \( \omega \)**:
Since we are given the magnitude of acceleration, we can write:
\[ a = \omega^2 x \]
Substituting the known values:
\[ 20 = \omega^2 \cdot 5 \]

4. **Solve for \( \omega^2 \)**:
Rearranging the equation gives:
\[ \omega^2 = \frac{20}{5} = 4 \]

5. **Calculate \( \omega \)**:
Taking the square root of both sides:
\[ \omega = \sqrt{4} = 2 \, \text{rad/s} \]

6. **Find the time period \( T \)**:
The time period \( T \) of oscillation is related to the angular frequency by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{2} = \pi \, \text{seconds} \]

### Final Answer:
The time period of the oscillation is \( \pi \) seconds.

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