Physics - SHM Question with Solution | TestHub

Given: $y_1=0.1\sin \left(100\pi t+\frac{\pi }{3}\right)$ and $\begin{array}{rl}{\mathrm{y}}_2=& 0.1\cos (100\pi \mathrm{t})\\ \therefore {\mathrm{v}}_1=\frac{{\mathrm{dy}}_1}{\mathrm{dt}}& =0.1\times 100\pi \cos \left(100\pi \mathrm{t}+\frac{\pi }{3}\right)\\ {\mathrm{v}}_2=\frac{{\mathrm{dy}}_2}{\mathrm{dt}}& =-0.1\times 100\pi \sin (100\pi \mathrm{t})\\ & =(0.1\times 100\pi )\cos \left(100\pi \mathrm{t}+\frac{\pi }{2}\right)\end{array}$ Phase difference of velocity of first particle with respect to the velocity of second particle at $t=0$ is $\therefore \Delta \varphi ={\varphi }_1-{\varphi }_2=\frac{\pi }{3}-\frac{\pi }{2}=\frac{-\pi }{6}$