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 2πω
(d) with time period πω.

Answer 1

Option A is correct.

Explanation:

1. Generalise equation of simple harmonic equation are

   $\mathbf{X =A\sin \omega t}$     ..1)

   Where

   X = displacement of particle at any instant time t

   A = Maximum displacement of particle = Amplitude

   $\omega$ = Angular frequency of oscillation

   t = time in second

2. Equation of question given is

   $\mathbf{X =X_{o}\sin 2\omega t}$     ...2)

3. Comparing equation 1) and equation 2), we get

  Amplitude = A $\mathbf{=X_{o}}$

  Angular frequency of oscillation =$2\omega$

  $\mathbf{\textrm{Time period of oscillation}(T)=\frac{2\pi }{angular \ frequency}=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }}$

4. So option A is correct.

Answer 2

The motion of the particle moving on the X-axis will execute the simple harmonic motion with time period $\frac{\pi}{\omega}$.

Explanation:

• It is given that the particle moves according to the equation $x=x_{0} \sin ^{2} \omega t$.  
• Thus   $x=\left(\frac{x_{0}}{2}\right)(\cos 2 \omega t-1)$ which shows that amplitude of the particle $A=\frac{x_{0}}{2}$  and the angular frequency of the particle is $2 \omega$.  
• From this, the particle can have the time period  $\frac{2 \pi}{\text { angular frequency }}=\frac{2 \pi}{2 \omega}=\frac{\pi}{\omega}$  Therefore the particle will execute the simple harmonic motion with the time period $\frac{\pi}{\omega}$.