JEE Main 2014 · 6 Apr · Shift 1, Physics question 14

Starting from rest at an extreme, the distance covered is $x=A(1-\cos\omega t)$.
$$A(1-\cos\omega\tau)=a,\qquad A(1-\cos2\omega\tau)=3a.$$
With $c=\cos\omega\tau$: $2(1-c)(1+c)=3(1-c)$, so $c=\frac12$ and $\omega\tau=\frac\pi3$. Then $T=\frac{2\pi}\omega=6\tau$ (and $A=2a$).