Question

A particle is executing simple harmonic motion with instantaneous displacement: $$ x = A \sin^2\left( \omega t - \frac{\pi}{4} \right) $$ The time period of oscillation of the particle is:

• A. \( \dfrac{2\pi}{\omega} \)
• B. \( \dfrac{\pi}{\omega} \)
• C. \( \dfrac{\pi}{2\omega} \)
• D. \( \dfrac{\omega}{2\pi} \)

Hint:

Use identity \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \) to convert to standard SHM form and determine period.

Answer 1

The Correct Option is B

Given: \[ x = A \sin^2\left( \omega t - \frac{\pi}{4} \right) \] Using identity: \[ \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \Rightarrow x = \frac{A}{2} \left[1 - \cos\left(2\omega t - \frac{\pi}{2} \right) \right] \] This is a cosine function with angular frequency \( 2\omega \), hence: \[ T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega} \]