Question

A particle, doing simple harmonic motion, at a distance \(3\,cm\) from mean position has acceleration \(12\,cm/s^2\). What is its time period?

• A. \(0.5\,s\)
• B. \(1\,s\)
• C. \(2\,s\)
• D. \(3.14\,s\)

Hint:

Use \( a = \omega^2 x \) to directly find angular frequency in SHM.

Answer 1

The Correct Option is D

To solve the problem, we need to find the time period of a particle executing Simple Harmonic Motion (SHM). The particle has an acceleration of \(12 \, \text{cm/s}^2\) at a distance of \(3 \, \text{cm}\) from the mean position.

The acceleration in SHM is given by the formula:

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

where:

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

Given, \(a = 12 \, \text{cm/s}^2\) and \(x = 3 \, \text{cm}\).

Substituting these values into the formula:

\(12 = -\omega^2 \times 3\)

Simplifying, we get:

\(12 = -3 \omega^2\)

\(\omega^2 = -\frac{12}{3} = -4\)

\(\omega^2 = 4\)

\(\omega = 2 \text{ (as negative value indicates direction)}\)

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

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

Substitute \(\omega = 2\) into the equation:

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

\(T = \pi \approx 3.14 \, \text{s}\)

Therefore, the time period is \(3.14 \, \text{s}\), which matches with option \(3.14 \, \text{s}\).

Conclusion: The time period of the particle in SHM is \(3.14 \, \text{s}\).