Question

The displacement of a particle is represented by the equation $y={\sin }^3\omega t$. the motion is

• A. Non-periodic
• B. Simple harmonic with period $\pi /\omega$
• C. Simple harmonic with period $2\pi /\omega$
• D. Periodic but not simple harmonic

Answer 1

The correct option is D Periodic but not simple harmonic

The displacement of a particle is represented by the equation

$y=si{n}^3\omega t$

$y=\frac{(3sin\omega t-4sin3\omega t)}{4}$ ($sin3\theta =3sin\theta -4si{n}^3\theta$)

$4\frac{dy}{dt}=3\omega cos\omega t-4\times \left[3\omega cos3\omega t\right]$

$4\frac{d^{2}y}{d{t}^2}=-3{\omega }^{2}sin\omega t+12{\omega }^{2}sin3\omega t$

$\frac{d^{2}y}{d{t}^2}=\frac{-3{\omega }^{2}sin\omega t+12{\omega }^{2}sin3\omega t}{4}$

From the above equation we can conclude that

$\frac{d^{2}y}{d{t}^2}\propto y$

Hence, motion is not S.H.M

The above expression involing sin function, hence it will be periodic.

$si{n}^3\omega t=(sin\omega t{)}^3$

$=\left[sin\right(\omega t+2\pi ){]}^3=[sin\omega (t+2\pi /\omega ){]}^3$

The time periodic $T=\frac{2\pi }{\omega }$

The correct option is D.