Question

A particle moves on the X-axis according to the equation x = x₀ sin² ωt. The motion is simple harmonic
(a) with amplitude x₀

(b) with amplitude 2x₀

(c) with time period $\frac{2\pi }{\omega }$

(d) with time period $\frac{\pi }{\omega }.$

Answer 1

(d) with time period $\frac{\pi }{\omega }$

Given equation:
x = x_o sin² ωt

⇒​$x=\frac{x_0}{2}(\cos 2\omega t-1)$

Now, the amplitude of the particle is x_o/2 and the angular frequency of the SHM is 2ω.

Thus, time period of the SHM = $\frac{2\pi }{\text{angular frequency}}=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }$