A particle executing linear S.H.M has period 3 second and amplitude $6 \mathrm{~cm}$. The time required by it to travel a distance of $3 \mathrm{~cm}$ from positive extreme position is $\left[\sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}, \sin 60^{\circ}=\cos 30^{\circ}=\frac{\sqrt{3}}{2}\right]$

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Aicharya · Accepted Answer

Lets take simplest equation of SHM, $y=A \sin (\omega t)$ Given, $A=6 \mathrm{~cm}, \mathrm{~T}=3 \mathrm{sec}$ $	herefore \omega=\frac{2 \pi}{T}=\frac{2 \pi}{3} \sec ^{-1} .$ Time when particle reaches extreme position $\mathrm{P}$ is $1.5 \mathrm{sec}$. Let time to reach $\mathrm{Q}$ is $\delta$. $\begin{aligned} & \Rightarrow \mathrm{y}=3 \mathrm{~cm}=6 \mathrm{~cm} \sin \left(\frac{\pi}{2}+\omega \deltaight) \ & \Rightarrow \sin \left(\frac{\pi}{2}+\omega \deltaight)=\frac{1}{2} \ & \Rightarrow \cos (\omega \delta)=\frac{1}{2} \ & \Rightarrow \delta=\frac{\pi}{3 \omega}=\frac{\pi(3)}{3(2 \pi)}=0.5 \mathrm{~s} \end{aligned}$

A particle executing linear S.H.M has period 3 second and amplitude $6 \mathrm{~cm}$. The time required by…

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