JEE Main · 2014 · 06 Apr 2014 · Offline · Elasticity & SHM

Starting from rest means starting at an extreme of the motion, so measure displacement from there as $x=A(1-\cos \omega t)$. After the first interval $\tau$, $a=A(1-\cos \omega \tau )$; after $2\tau$ the total distance is $3a=A(1-\cos 2\omega \tau )$. Writing $\cos 2\omega \tau =2{\cos }^2\omega \tau -1$ and substituting $\cos \omega \tau =1-\frac{a}{A}$ gives $1-\frac{3a}{A}=2{\left(1-\frac{a}{A}\right)}^2-1$, which solves to $\frac{a}{A}=\frac{1}{2}$, so $A=2a$ and $\cos \omega \tau =\frac{1}{2}$. Then $\omega \tau =\frac{\pi }{3}$, and the period is $T=\frac{2\pi }{\omega }=6\tau$.