Question

If the function $\sin^2 \omega t$ (where $t$ is time in seconds) represents a periodic motion, then the period of the motion is

• A. \(\sqrt{\frac{\pi}{\omega}}\) s
• B. \(\frac{\pi}{\omega}\) s
• C. \(\frac{2\pi}{\omega}\) s
• D. \(\sqrt{\frac{2\pi}{\omega}}\) s

Hint:

When dealing with squared sine or cosine functions, use trigonometric identities to find the fundamental frequency and period. The period usually halves compared to the original sine or cosine function.

Answer 1

The Correct Option is B

The function is \(\sin^2 \omega t\). Using the identity: \[ \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \] So, \[ \sin^2 \omega t = \frac{1 - \cos 2\omega t}{2} \] The angular frequency of \(\cos 2 \omega t\) is \(2\omega\). Hence, its period is: \[ T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega} \] Therefore, the period of \(\sin^2 \omega t\) is \(\frac{\pi}{\omega}\).