particle moves with simple harmonic motion in a straight line. In first $\tau\ s$, after starting from rest it travels a distance $a$, and in next $\tau\ s$ it travels $2a$, in same direction, then
JEE MAIN 2014, Diffcult
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In simple harmonic motion, starting from rest,

At $t=0, x=A$

$x= Acos\omega t$                     $...(i)$

$ When\,\,\, t=\tau, x=A-a$

When $t=2 \tau, x=A-3 a$

From equation $( i )$

$A-a=A \cos \omega \tau$                    $...(ii)$

$A-3 a=A \cos 2 \omega \tau$              $...(iii)$

As $\cos 2 \omega \tau=2 \cos ^{2} \omega \tau-1 \ldots(\mathrm{iv})$

From equation $(ii),$ $(iii)$ and $(iv)$

$\frac{A-3 a}{A}=2\left(\frac{A-a}{A}\right)^{2}-1$

$\Rightarrow \quad \frac{A-3 a}{A}=\frac{2 A^{2}+2 a^{2}-4 A a-A^{2}}{A^{2}}$

$\Rightarrow A^{2}-3 a A=A^{2}+2 a^{2}-4 A a$

$\Rightarrow \quad 2 a^{2}=a A \Rightarrow \quad A=2 a$

$\Rightarrow \quad \frac{a}{A}=\frac{1}{2}$

Now, $A-a=A \cos \omega \tau$

$\Rightarrow \quad \cos \omega \tau=\frac{A-a}{A} \Rightarrow \quad \cos \omega \tau=\frac{1}{2}$

or, $\quad \frac{2 \pi}{T} \tau=\frac{\pi}{3} \Rightarrow \quad \mathrm{T}=6 \tau$

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