A simple harmonic oscillator has an amplitude A and time period T. The time required by it to travel from $x=A\text{ to }x=A/2$ is

For S.H.M., $x=A\sin \left(\frac{2\pi }{T}\right)$

When $x=A,A=A\sin \left(\frac{2\pi }{T}\cdot t\right)$

$\begin{array}{l}\sin \left(\frac{2\pi }{\mathrm{T}}\mathrm{t}\right)=1\\ \Rightarrow \sin \left(\frac{2\pi }{\mathrm{T}}\cdot \mathrm{t}\right)=\sin \left(\frac{\pi }{2}\right)\Rightarrow \mathrm{t}=(\mathrm{T}/4)\end{array}$

When $x=\frac{A}{2},\frac{A}{2}=A\sin \left(\frac{2\pi }{T},t\right)$

or, $\frac{\sin \pi }{6}=\sin \left(\frac{2\pi }{T}t\right)\text{ or }t=(T/12)$

Now, time taken to travel from $x=A\text{ tox }=A/2\text{ is }(T/4-\mathrm{T}/12)=\mathrm{T}/6$