Search any question & find its solution
Question:
Answered & Verified by Expert
Two particle are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is
Options:
Solution:
1646 Upvotes
Verified Answer
The correct answer is:
$\frac{2 \pi}{3}$
$$
\begin{aligned}
y_1 & =\frac{A}{2}=A \sin \omega t \\
\Rightarrow \quad \omega t & =30^{\circ} \\
y_2 & =\frac{A}{2}=A \sin \left(\omega t+\frac{\pi}{2}\right) \\
\Rightarrow \quad \omega t+\frac{\pi}{2} & =150^{\circ}
\end{aligned}
$$
$\therefore$ phase difference
$$
\begin{aligned}
\phi & =150^{\circ}-30^{\circ} \\
& =120=\frac{2 \pi}{3} \mathrm{rad}
\end{aligned}
$$
\begin{aligned}
y_1 & =\frac{A}{2}=A \sin \omega t \\
\Rightarrow \quad \omega t & =30^{\circ} \\
y_2 & =\frac{A}{2}=A \sin \left(\omega t+\frac{\pi}{2}\right) \\
\Rightarrow \quad \omega t+\frac{\pi}{2} & =150^{\circ}
\end{aligned}
$$
$\therefore$ phase difference
$$
\begin{aligned}
\phi & =150^{\circ}-30^{\circ} \\
& =120=\frac{2 \pi}{3} \mathrm{rad}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.