Question

A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is $20 \, m/s^{-2}$ at a distance of $5 \,m$ from the mean position. The time period of oscillation is

• A. 1 s
• B. 2 $\pi$ s
• C. 2s
• D. $\pi s $

Answer 1

The Correct Option is D

To solve this problem, we will use the concept of simple harmonic motion (SHM) for a pendulum. In SHM, the acceleration \(a\) of the bob is given by the formula:

a = -\omega^2 \cdot x

where:

• \(a\) is the acceleration of the bob in m/s^2,
• \(\omega\) is the angular frequency in rad/s,
• \(x\) is the displacement from the mean position in meters.

Given that \(a = 20 \, m/s^2\) and \(x = 5 \, m\), substitute these values into the formula:

20 = -\omega^2 \times 5

Solve for \(\omega^2\):

\omega^2 = \frac{20}{5} = 4

So, the angular frequency \(\omega\) is:

\omega = \sqrt{4} = 2 \, rad/s

The time period \(T\) of the oscillator is related to the angular frequency by the formula:

T = \frac{2\pi}{\omega}

Substitute the value of \(\omega\) into the formula:

T = \frac{2\pi}{2} = \pi \, s

Thus, the time period of the oscillation is \(\pi \, s\).

Therefore, the correct answer is:

\pi \, s