Two particles execute simple harmonic motion | vedclass

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(B) Let the displacements of the two particles be $x_1 = a \sin \omega t$ and $x_2 = a \sin (\omega t + \phi)$, where $a$ is the amplitude, $\omega$ i

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$2 \pi / 3$

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(B) Let the displacements of the two particles be $x_1 = a \sin \omega t$ and $x_2 = a \sin (\omega t + \phi)$, where $a$ is the amplitude, $\omega$ is the angular frequency, and $\phi$ is the phase difference.When they cross each other, their displacements are equal: $x_1 = x_2 = a/2$.For the first particle: $a \sin \omega t = a/2 \Rightarrow \sin \omega t = 1/2$. Thus, $\omega t = \pi/6$ (assuming the particle is moving in the positive direction).For the second particle: $a \sin (\omega t + \phi) = a/2 \Rightarrow \sin (\omega t + \phi) = 1/2$.This implies $\omega t + \phi = \pi/6$ or $\omega t + \phi = 5\pi/6$.If $\omega t + \phi = \pi/6$, then $\phi = 0$, which means they are moving in the same direction, not opposite.If $\omega t + \phi = 5\pi/6$, then $\phi = 5\pi/6 - \pi/6 = 4\pi/6 = 2\pi/3$.At this phase, the velocity $v_2 = a \omega \cos (\omega t + \phi) = a \omega \cos (5\pi/6) = -a \omega \sqrt{3}/2$, which is negative, confirming they are moving in opposite directions.Therefore, the phase difference is $2\pi/3$.

Two particles execute simple harmonic motion $(SHM)$ along close parallel lines. Both particles have the same frequency and same amplitude. When they pass each other moving in opposite directions, their displacement is half their amplitude. Their phase difference is:

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