Question

The displacement of a particle executing SHM is given by \( X = 0.01 \sin 100\pi (t + 0.05) \). The time period of the particle is:

• A. \(0.1 \, {s}\)
• B. \(0.01 \, {s}\)
• C. \(0.2 \, {s}\)
• D. \(0.02 \, {s}\)

Hint:

For SHM, always remember \(T = \frac{2\pi}{\omega}\), where \(\omega\) is the angular frequency.

Answer 1

The Correct Option is D

Step 1: Identify the angular frequency \(\omega\). The angular frequency \(\omega\) from the equation \(X = 0.01 \sin 100\pi (t + 0.05)\) is \(100\pi \, {rad/s}\). 
Step 2: Calculate the time period \(T\). The time period \(T\) of simple harmonic motion is given by \(T = \frac{2\pi}{\omega}\). \[ T = \frac{2\pi}{100\pi} = 0.02 \, {s} \]