The magnitude of displacement of a particle m | vedclass

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(B) Let the particle start from the point $(a, 0)$ at $t = 0$. The position vector at time $t$ is given by $\vec{R}(t) = a \cos(\omega t) \hat{i} + a

Vedclass · Accepted Answer

$2a \sin \frac{\omega t}{2}$

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(B) Let the particle start from the point $(a, 0)$ at $t = 0$. The position vector at time $t$ is given by $\vec{R}(t) = a \cos(\omega t) \hat{i} + a \sin(\omega t) \hat{j}$.The initial position vector is $\vec{R}(0) = a \hat{i}$.The displacement vector $\vec{d}$ is $\vec{R}(t) - \vec{R}(0) = a(\cos \omega t - 1) \hat{i} + a \sin(\omega t) \hat{j}$.The magnitude of displacement $d$ is given by $d = \sqrt{[a(\cos \omega t - 1)]^2 + [a \sin \omega t]^2}$.$d = a \sqrt{\cos^2 \omega t - 2 \cos \omega t + 1 + \sin^2 \omega t}$.Using the identity $\sin^2 	heta + \cos^2 	heta = 1$,we get $d = a \sqrt{2 - 2 \cos \omega t}$.Using the trigonometric identity $1 - \cos 	heta = 2 \sin^2(	heta/2)$,we get $d = a \sqrt{2 	imes 2 \sin^2(\omega t / 2)}$.Therefore,$d = 2a \sin(\omega t / 2)$.

The magnitude of displacement of a particle moving in a circle of radius $a$ with constant angular speed $\omega$ varies with time $t$ as

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