Ncert Exercises & Solutions

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 $\pi / \omega.$

Hint: For SHM, acceleration is directly proportional to

y .

$y.$

Step 1: Find the acceleration of the particle.
Acceleration of a particle is given by;

a_{y} = \frac{d^{2} y}{d t^{2}}

$a_y =\frac{d^2 y}{d t^2}$
Given, the equation of the motion is;

\Rightarrow y = \sin^{3} ω t [\sin 3 θ = 3 \sin θ - 4 \sin^{3} θ]

$\Rightarrow y =\sin^{3}~\omega t~~~~~~\left[\sin 3 \theta=3 \sin \theta-4 \sin ^3 \theta\right]$

\Rightarrow y = \frac{1}{4} (3 \sin ω t - \sin 3 ω t)

$\Rightarrow y=\frac{1}{4}(3 \sin \omega t-\sin 3 \omega t)$

\Rightarrow \frac{d y}{d t} = \frac{1}{4} [\frac{d}{d t} (3 \sin n ω t) - \frac{d}{d t} (\sin 3 ω t)]

$\Rightarrow \frac{d y}{d t}=\frac{1}{4}\left[\frac{d}{d t}(3 \sin n \omega t)-\frac{d}{d t}(\sin 3 \omega t)\right]$

\Rightarrow 4 \frac{d^{2} y}{d t^{2}} = - 3 ω^{2} \sin ω t + 9 ω^{2} \sin 3 ω t

$\Rightarrow 4 \frac{d^2 y}{d t^2}=-3 \omega^2 \sin \omega t+9 \omega^2 \sin 3 \omega t$

\Rightarrow 4 a_{y} = - 3 ω^{2} (\sin ω t - 3 \sin 3 ω t) [a_{y} = \frac{d^{2} y}{d t^{2}}]

$\Rightarrow 4a_y=-3 \omega^2(\sin \omega t-3 \sin 3 \omega t) ~~~[a_y = \frac{d^2y}{dt^2}]$

Step 2: Find if the motion is SHM.

a_{y}

$a_y$ is not proportional to

y .

$y.$ Thus the motion is periodic but not SHM.
Hence, option (2) is the correct option.