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 \(y.\)

Step 1: Find the acceleration of the particle.
Acceleration of a particle is given by; \(a_y =\frac{d^2 y}{d t^2}\)
Given, the equation of the motion is;
\(\Rightarrow y =\sin^{3}~\omega t~~~~~~\left[\sin 3 \theta=3 \sin \theta-4 \sin ^3 \theta\right]\)
\(\Rightarrow y=\frac{1}{4}(3 \sin \omega t-\sin 3 \omega t)\)
\(\Rightarrow \frac{d y}{d t}=\frac{1}{4}\left[\frac{d}{d t}(3 \sin n \omega t)-\frac{d}{d t}(\sin 3 \omega t)\right]\)
\(\Rightarrow 4 \frac{d^2 y}{d t^2}=-3 \omega^2 \sin \omega t+9 \omega^2 \sin 3 \omega t\)
\(\Rightarrow 4a_y=-3 \omega^2(\sin \omega t-3 \sin 3 \omega t) ~~~[a_y = \frac{d^2y}{dt^2}]\)

Step 2: Find if the motion is SHM.
\(a_y\) is not proportional to \(y.\) Thus the motion is periodic but not SHM.
Hence, option (2) is the correct option.

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