A particle executing $SHM$ of amplitude $4\,cm$ and $T = 4s.$ The time taken by it to move from $+2\,cm$ to $+2\sqrt 3\,cm$ is
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$\mathrm{x}=\mathrm{a} \sin \omega \mathrm{t}$
$2=4 \sin \omega t_{1} \Rightarrow \frac{1}{2}=\sin \omega t_{1} \Rightarrow \omega t_{1}=\frac{\pi}{6}$
and $2 \sqrt{3}=4 \sin \omega \mathrm{t}_{2} \quad \Rightarrow \frac{\sqrt{3}}{2}=\omega \mathrm{t}_{2}=\frac{2 \pi}{6}$
$\therefore \omega\left(t_{2}-t_{1}\right)=\frac{2 \pi}{6}-\frac{\pi}{6}=\frac{\pi}{6}$
$\Rightarrow \mathrm{t}_{2}-\mathrm{t}_{1}=\frac{\pi}{6} \times \frac{1}{\omega}=\frac{\pi}{6} \times \frac{\mathrm{T}}{2 \pi}=\frac{\mathrm{T}}{12}=\frac{4}{12}=\frac{1}{3} \mathrm{s}$
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