Question 319: Oscillations

Question

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

Accepted Answer

Given: $$\mathrm{y}_1=0.1 \sin \left(100 \pi \mathrm{t}+\frac{\pi}{3}ight)$$ and
$$\begin{aligned}
\mathrm{y}_2=0.1 \cos (100 \pi \mathrm{t}) & \
	herefore \quad \mathrm{v}_1=\frac{\mathrm{dy_{1 }}}{\mathrm{dt}} & =0.1 	imes 100 \pi \cos \left(100 \pi \mathrm{t}+\frac{\pi}{3}ight) \
\mathrm{v}_2=\frac{\mathrm{dy_{2 }}}{\mathrm{dt}} & =-0.1 	imes 100 \pi \sin (100 \pi \mathrm{t}) \
& =(0.1 	imes 100 \pi) \cos \left(100 \pi \mathrm{t}+\frac{\pi}{2}ight)
\end{aligned}$$
Phase difference of velocity of first particle with respect to the velocity of second particle at $$t=0$$ is
$$	herefore \quad \Delta \phi=\phi_1-\phi_2=\frac{\pi}{3}-\frac{\pi}{2}=\frac{-\pi}{6}$$

Answer

$$\frac{\pi}{3}$$

Answer

$$-\frac{\pi}{6}$$

Answer

$$+\frac{\pi}{6}$$

Answer

$$-\frac{\pi}{3}$$

Oscillations

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

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