The velocities of a particle performing linear SHM are $0.13$ m/s and $0.12$ m/s when it is at $0.12m$ and $0.13$ m from mean position respectively, the period is:

$V=\omega \sqrt{A^{2}-x^{2}}$

$V_{1}=\omega\sqrt{A^{2}-x^{2}_{1}}$

$V_{2}=\omega\sqrt{A^{2}-x^{2}_{2}}$

$\dfrac{0.13}{0.12}$ = $\dfrac{\sqrt{A^{2}-(0.12)^{2}}}{\sqrt{A^{2}-(0.13)^{2}}}$

$\dfrac{0.13^2}{0.12^2}$ = $\dfrac{{A^{2}-(0.12)^{2}}}{{A^{2}-(0.13)^{2}}}$

Cross-multiplying,

$\left ((0.13)^{2}-(0.12)^{2}\right) A^2 =(0.13)^{4}-(0.12)^{4}$

$A^{2}=(0.13)^{2}+(0.12)^{2}$

$A=\sqrt{(0.13)^{2}+(0.12)^{2}}$

Using, $x$ = $\omega \sqrt{A^2 - x^2}$

$0.13=\omega \sqrt{(0.13)^{2}+(0.12)^{2}-(0.12)^{2}}$

$\therefore \omega =1$

$T=2\pi /\omega$

$T=2\pi \sec$