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/s² at a distance of 5 m from the mean position. The time period of oscillation is

• A. 2πs
• B. 2s
• C. πs
• D. 1s

Answer 1

The Correct Option is C

 To find the time period of oscillation of a pendulum moving like a simple harmonic oscillator, we must understand the relationship between acceleration, displacement, and time period in simple harmonic motion (SHM).

For a simple harmonic oscillator, the acceleration \(a\) is given by:

\(a = -\omega^2 x\)

where:

• \(\omega\) is the angular frequency,
• \(x\) is the displacement from the mean position.

Given \(a = 20 \, \text{m/s}^2\) and \(x = 5 \, \text{m}\), we can substitute these values into the equation:

\(20 = -\omega^2 \times 5\)

Solving for \(\omega^2\) gives:

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

So, \(\omega = \sqrt{4} = 2 \, \text{rad/s}\).

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

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

Substituting the value of \(\omega\):

\(T = \frac{2\pi}{2} = \pi \, \text{s}\)

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

Conclusion: The correct answer is

πs

.

 

Other options such as

2πs

,

2s

, and

1s

are incorrect based on the calculation for the time period using angular frequency.