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 \(\pi / \omega\).

(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 = \sin^{3} ω t

= \frac{1}{4} (3 \sin ω t - \sin 3 ω t) [\sin 3 θ = 3 \sin θ - 4 \sin^{3} θ]

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

\Rightarrow 4 \frac{d y}{d t} = 3 ω \cos ω t - 3 ω \cos 3 ω t

\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 ω^{2} (\sin ω t - 3 \sin 3 ω t)

Step 2: Find if the motion is SHM.

\Rightarrow \frac{d^{2} y}{d t^{2}}

is not proportional to y.

Hence the motion is not SHM.

As the expression is involving a sine function, the motion is periodic.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions