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
\(\left[\sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}, \sin 60^{\circ}=\cos 30^{\circ}=\frac{\sqrt{3}}{2}\right]\)

Lets take simplest equation of SHM,
\(y=A \sin (\omega t)\)


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