The displacement of a particle is represented by the equation  \(y =\sin^{3}~\omega t\). The motion is:
1. non-periodic.
2. periodic but not simple harmonic.
3. simple harmonic with period \(2\pi / \omega\).
4. simple harmonic with period \(\pi / \omega\).

(b) Hint: In SHM, the acceleration is directly proportional to the displacement.
Step 1: Find the acceleration of the particle.
Given the equation of motion is,
y=sin3ωt
  =14(3sinωt-sin3ωt)      [sin3θ=3sinθ-4sin3θ]                    y
dydt=14ddt(3sinωt)-ddt(sin3ωt)
4dydt=3ωcosωt-3ωcos3ωt
4d2ydt2=-3ω2sinωt+9ω2sin3ωt
4d2ydt2=-3ω2sinωt-3sin3ωt   
Step 2: Find if the motion is SHM.
d2ydt2 is not proportional to y.
Hence the motion is not SHM.
As the expression is involving a sine function, the motion is periodic.