Question

The displacement of a particle is given as a function of time \(t\) by: \[ x = A\sin(2\omega t) + B\sin^2(\omega t) \] Then,

• A. the motion of the particle is SHM with an amplitude of \(\sqrt{A^2 + \frac{B^2}{4}}\)
• B. the motion of the particle is not SHM, but oscillatory with a time period of \(T=\pi/\omega\)
• C. the motion of the particle is oscillatory with a time period of \(T=\pi/2\omega\)
• D. the motion of the particle is aperiodic

Hint:

Presence of constant terms or multiple trigonometric components generally destroys pure SHM, though motion may remain periodic.

Answer 1

The Correct Option is B

Step 1: Use identity: \[ \sin^2(\omega t) = \frac{1-\cos(2\omega t)}{2} \]
Step 2: Substitute: \[ x = A\sin(2\omega t) + \frac{B}{2} - \frac{B}{2}\cos(2\omega t) \]
Step 3: The displacement contains both sine and cosine terms of same angular frequency but with a constant shift.
Step 4: Hence motion is oscillatory but not simple harmonic.
Step 5: The fundamental angular frequency is \(2\omega\), so: \[ T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega} \]