Question

A simple harmonic motion is represented by, $x(t) = \sin^2 \omega t - 2\cos^2 \omega t$. The angular frequency of oscillation is given by

• A. $\omega$
• B. $2\omega$
• C. $4\omega$
• D. $\omega/2$
• E. $\omega/4$

Hint:

Convert trigonometric expressions into $\cos(\omega t)$ form to identify frequency.

Answer 1

The Correct Option is B

Concept:
Use trigonometric identities to convert into standard SHM form.

Step 1: Apply identities.
\[ \sin^2 \omega t = \frac{1 - \cos 2\omega t}{2}, \quad \cos^2 \omega t = \frac{1 + \cos 2\omega t}{2} \]

Step 2: Substitute.
\[ x(t) = \frac{1 - \cos 2\omega t}{2} - 2\cdot \frac{1 + \cos 2\omega t}{2} \]

Step 3: Simplify.
\[ x(t) = \frac{1 - \cos 2\omega t - 2 - 2\cos 2\omega t}{2} = \frac{-1 - 3\cos 2\omega t}{2} \]

Step 4: Identify angular frequency.
\[ x(t) \propto \cos(2\omega t) \] \[ \Rightarrow \text{Angular frequency} = 2\omega \]