Question

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

Options

• with amplitude x₀

• with amplitude 2x₀

• with time period $$\frac{2\pi }{\omega }$$

• with time period $$\frac{\pi }{\omega }.$$

Answer 1

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 = 

$$\frac{2\pi }{\text{ angular frequency }}=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }$$