The magnitude of displacement of a particle m | vedclass

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Question

If a particle is moving with angular velocity $=\omega$

Its angle of rotation is given by $\omega t$

Now displacement $=$ length of line $A B$

Vedclass · Accepted Answer

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

Vedclass · Answer

If a particle is moving with angular velocity $=\omega$

Its angle of rotation is given by $\omega t$

Now displacement $=$ length of line $A B$

Position vector of a particle is given by

$\vec{R}=i a \cos \omega t+j a \sin \omega t$

$\vec{R}_{0}=a i$

displacement$, \vec{d}=\vec{R}-\vec{R}_{\mathrm{o}}$

$=a(\cos \omega t-1) i+a \sin \omega j$

$d=\sqrt{(a(\cos \omega t-1))^{2}+(a \sin \omega)^{2}}$

$=a \sqrt{2(1-\cos \omega t)}=a \sqrt{2 	imes 2(\sin \omega t / 2)^{2}}=2 a \sin \omega t / 2$

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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