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

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, \(\quad \frac{2 \pi}{T} \tau=\frac{\pi}{3} \Rightarrow \quad \mathrm{T}=6 \tau\)