A particle executing linear S.H.M. has a peri | vedclass

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(D) For a particle starting from the positive extreme position $(x = +A)$,the displacement equation is given by: $x = A \cos(\omega t)$ Given: $A = 6

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(D) For a particle starting from the positive extreme position $(x = +A)$,the displacement equation is given by: $x = A \cos(\omega t)$ Given: $A = 6 \ cm$,$T = 3 \ s$. The angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{3} \ rad/s$. We need to find the time $t$ when the particle has traveled a distance of $3 \ cm$ from the positive extreme position. This means the new position is $x = A - 3 = 6 - 3 = 3 \ cm$. Substituting the values into the displacement equation: $3 = 6 \cos(\omega t)$ $\cos(\omega t) = \frac{3}{6} = \frac{1}{2}$ Since $\cos(60^{\circ}) = \frac{1}{2}$,we have: $\omega t = 60^{\circ} = \frac{\pi}{3} \ rad$ Substituting $\omega = \frac{2\pi}{3}$: $(\frac{2\pi}{3}) t = \frac{\pi}{3}$ $t = \frac{\pi}{3} 	imes \frac{3}{2\pi} = 0.5 \ s$ Therefore,the time required is $0.5 \ s$.

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