Question

The displacement of a particle is represented by the equation $y=si{n}^3\left(\omega t\right)$. The motion is

• A. non-periodic
• B. periodic but not simple harmonic
• C. simple harmonic with period $\frac{2\pi }{\omega }$
• D. simple harmonic with period $\frac{\pi }{\omega }$

Answer 1

Answer: (B)

Given the equation of displacement of the particle, $y={sin}^3\omega t$

We know $sin3\theta =3sin\theta -4{sin}^3\theta$

Hence, $y=\frac{(3sin\omega t-4sin3\omega t)}{4}\Rightarrow 4\frac{dy}{dt}=3\omega cos\omega t-4\times \left[3\omega cos3\omega t\right]\Rightarrow 4\times \frac{d^{2}y}{{dt}^2}=-3{\omega }^{2}sin\omega t+12\omega sin3\omega t\Rightarrow \frac{d^{2}y}{{dt}^2}=\frac{-3{\omega }^{2}sin\omega t+12\omega sin3\omega t}{4}\Rightarrow \frac{d^{2}y}{{dt}^2}$ is not proportional to y.

Hence, the motion is not SHM.

As the expression is involving sine function, hence it will be periodic.

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

Hence, $y={sin}^3\omega t$ represents a periodic motion with period $2\pi /\omega$.