Two particles are oscillating along two close parallel straight lines side by side, with the same frequency an

Question

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 the paths of the two particles. The phase difference is:

Accepted Answer

1. **Identify the equations of motion:** For two particles executing Simple Harmonic Motion (SHM) with the same amplitude $A$ and frequency $\omega$, their displacements at any time $t$ can be written as $y_1 = A\sin(\omega t + \phi_1)$ and $y_2 = A\sin(\omega t + \phi_2)$.
2. **Determine the phase at the meeting point:** The particles pass each other when their displacement is half of the amplitude, i.e., $y_1 = y_2 = \frac{A}{2}$.
   This gives $\sin(\omega t + \phi_1) = \frac{1}{2}$ and $\sin(\omega t + \phi_2) = \frac{1}{2}$.
3. **Calculate the phase difference:** Since the particles are moving in opposite directions, their velocities must have opposite signs. The possible phases corresponding to a sine value of $\frac{1}{2}$ in one cycle are $\frac{\pi}{6}$ and $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$.
   Let the phase of the first particle be $	heta_1 = \frac{\pi}{6}$ (moving towards the positive extreme) and the phase of the second particle be $	heta_2 = \frac{5\pi}{6}$ (moving towards the mean position).
   The phase difference is $\Delta\phi = 	heta_2 - 	heta_1 = \frac{5\pi}{6} - \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$.

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