At time t = 0, one particle is at maximum positive amplitude and the other particle is at half of the positive amplitude. Their amplitudes and time periods T are same. If they are approaching, find the time by which they cross each other:

${\mathrm{x}}_1= \mathrm{A} \cos \mathrm{\omega t} (\mathrm{at} \mathrm{t} = 0, {\mathrm{x}}_1 = \mathrm{A})$

${\mathrm{x}}_2 = \mathrm{A} \sin (\mathrm{\omega t}+\frac{\mathrm{\pi }}{6})$              $(\mathrm{at} \mathrm{t} =0, {\mathrm{x}}_2 = +\frac{\mathrm{A}}{2})$

For ${\mathrm{x}}_1 = {\mathrm{x}}_2,$

$\cos \mathrm{\omega t} = \sin (\mathrm{\omega t}+\frac{\mathrm{\pi }}{6})$

     = $\sin \mathrm{\omega t} \cos \frac{\mathrm{\pi }}{6}+\cos \mathrm{\omega t} \sin \frac{\mathrm{\pi }}{6}$

$\cos \mathrm{\omega t}(1-\sin \frac{\mathrm{\pi }}{6}) = \sin \mathrm{\omega t} \cos \frac{\mathrm{\pi }}{6}$

$\tan \mathrm{\omega t} = \frac{(1-\sin {\displaystyle \frac{\mathrm{\pi }}{6}})}{\cos {\displaystyle \frac{\mathrm{\pi }}{6}}} = \frac{1}{\sqrt{3}}$

$\mathrm{\omega t} = \left(\frac{2\mathrm{\pi }}{\mathrm{T}}\right)\mathrm{t} = \frac{\mathrm{\pi }}{6}$

Will give $\mathrm{t} = \frac{\mathrm{T}}{12}$