Two S.H.Ms. are represented by equations $\mathrm{y}_1=0.1 \sin \left(100 \pi…
Two S.H.Ms. are represented by equations $\mathrm{y}_1=0.1 \sin \left(100 \pi \mathrm{t}+\frac{\pi}{3}\right)$ and $\mathrm{y}_2=0.1 \cos (100 \pi \mathrm{t})$ The phase difference between the speeds of the two particles is
$\frac{\pi}{3}$
$-\frac{\pi}{6}$
$+\frac{\pi}{6}$
$-\frac{\pi}{3}$
Solution
Given: $y_1=0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right)$ and
$\begin{aligned}
\mathrm{y}_2= & 0.1 \cos (100 \pi \mathrm{t}) \\
\therefore \quad \mathrm{v}_1=\frac{\mathrm{dy}_1}{\mathrm{dt}} & =0.1 \times 100 \pi \cos \left(100 \pi \mathrm{t}+\frac{\pi}{3}\right) \\
\mathrm{v}_2=\frac{\mathrm{dy}_2}{\mathrm{dt}} & =-0.1 \times 100 \pi \sin (100 \pi \mathrm{t}) \\
& =(0.1 \times 100 \pi) \cos \left(100 \pi \mathrm{t}+\frac{\pi}{2}\right)
\end{aligned}$
Phase difference of velocity of first particle with respect to the velocity of second particle at $t=0$ is
$\therefore \quad \Delta \phi=\phi_1-\phi_2=\frac{\pi}{3}-\frac{\pi}{2}=\frac{-\pi}{6}$