Two particles executing S.H.M. of same freque | vedclass

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(D) The displacement equation for $S.H.M.$ is given by $x = A \sin(\omega t + \phi)$.For $x = +A/2$,we have $A/2 = A \sin(\omega t + \phi)$,which impl

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$\frac{2\pi}{3}$

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(D) The displacement equation for $S.H.M.$ is given by $x = A \sin(\omega t + \phi)$.For $x = +A/2$,we have $A/2 = A \sin(\omega t + \phi)$,which implies $\sin(\omega t + \phi) = 1/2$.The possible values for the phase angle $	heta = \omega t + \phi$ are $30^{\circ}$ (or $\pi/6$ radians) and $150^{\circ}$ (or $5\pi/6$ radians).At $30^{\circ}$,the particle is moving in the positive direction (velocity $v = A\omega \cos(30^{\circ}) > 0$).At $150^{\circ}$,the particle is moving in the negative direction (velocity $v = A\omega \cos(150^{\circ}) Since the particles meet at $x = +A/2$ moving in opposite directions,one must have a phase of $30^{\circ}$ and the other $150^{\circ}$.The phase difference is $|150^{\circ} - 30^{\circ}| = 120^{\circ}$.Converting to radians,$120^{\circ} = 120 	imes \frac{\pi}{180} = \frac{2\pi}{3}$ radians.

Column-$I$	Column-$II$
$(a)$ $t = \frac{T}{8}$	$(i)$ $\theta = \frac{5\pi}{4}$
$(b)$ $t = \frac{5T}{8}$	$(ii)$ $\theta = \frac{3\pi}{2}$
	$(iii)$ $\theta = \frac{\pi}{4}$

Two particles executing $S.H.M.$ of same frequency,meet at $x=+A/2$,while moving in opposite directions. Phase difference between the particles is .........

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