miscellaneous: Practice Problem & Solution
Two particles 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 paths of the two particles. The phase difference is: (2011 Mains)
Solution Explained:
To solve this problem, we apply the core principles of miscellaneous. Understanding the underlying formula is key to arriving at the correct answer below:
Let displacement be $x = A\sin(\omega t + \phi)$. When they cross, $x = A/2$. $A/2 = A\sin(\phi) \Rightarrow \sin(\phi) = 1/2$. The two phases are $\pi/6$ and $5\pi/6$ (since moving in opposite directions). Phase difference $$= 5\pi/6 - \pi/6 = 4\pi/6 = 2\pi/3$$.
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