The displacement of a particle is given at ti | vedclass

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(A) The displacement of the particle is given by: $x = A \sin (-2 \omega t) + B \sin^2 \omega t$. Using the trigonometric identity $\sin^2 	heta = \f

Vedclass · Accepted Answer

the motion of the particle is $SHM$ with an amplitude of $\sqrt{A^2 + \frac{B^2}{4}}$

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(A) The displacement of the particle is given by: $x = A \sin (-2 \omega t) + B \sin^2 \omega t$. Using the trigonometric identity $\sin^2 	heta = \frac{1 - \cos 2 	heta}{2}$,we get: $x = -A \sin 2 \omega t + \frac{B}{2} (1 - \cos 2 \omega t)$. $x = -A \sin 2 \omega t - \frac{B}{2} \cos 2 \omega t + \frac{B}{2}$. Let $x' = x - \frac{B}{2} = -(A \sin 2 \omega t + \frac{B}{2} \cos 2 \omega t)$. This is of the form $x' = -R \sin (2 \omega t + \phi)$,where $R = \sqrt{A^2 + (B/2)^2} = \sqrt{A^2 + \frac{B^2}{4}}$. Since the displacement can be expressed as a single sine function about a mean position $x = B/2$,the motion is Simple Harmonic Motion $(SHM)$ with amplitude $\sqrt{A^2 + \frac{B^2}{4}}$.

The displacement of a particle is given at time $t$,by: $x = A \sin (-2 \omega t) + B \sin^2 \omega t$. Then,

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