Question

A particle executes simple harmonic motion according to the equation \( x(t) = A \sin^2(\alpha t) \). If the time period of the S.H.M is 0.2 s, then the value of \( \alpha \) (in units of rad/s) is:

• A. \( 2\pi \)
• B. \( 10\pi \)
• C. \( 5\pi \)
• D. \( 2.5\pi \)

Hint:

\( \sin^2\theta = \frac{1 - \cos 2\theta}{2} \). The period of \( \cos(2\alpha t) \) is \( \pi/\alpha \), not \( 2\pi/\alpha \).

Answer 1

The Correct Option is C

Concept: Use trigonometric identity to rewrite \( \sin^2(\alpha t) \) in terms of cosine of double angle, then find the period.

Step 1: Rewrite \( \sin^2(\alpha t) \). \[ \sin^2(\alpha t) = \frac{1 - \cos(2\alpha t)}{2} \] Thus: \[ x(t) = A \cdot \frac{1 - \cos(2\alpha t)}{2} = \frac{A}{2} - \frac{A}{2} \cos(2\alpha t). \]

Step 2: Identify the angular frequency of the oscillatory part. The term \( \cos(2\alpha t) \) has angular frequency \( \omega = 2\alpha \).

Step 3: Relate time period to angular frequency. \[ T = \frac{2\pi}{\omega} = \frac{2\pi}{2\alpha} = \frac{\pi}{\alpha}. \]

Step 4: Substitute given \( T = 0.2 \) s. \[ 0.2 = \frac{\pi}{\alpha} \quad \Rightarrow \quad \alpha = \frac{\pi}{0.2} = 5\pi \, \text{rad/s}. \]