A particle is executing SHM of amplitude A ab | vedclass

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(C) The displacement equation for $SHM$ is $x = A \cos(	heta)$,where $	heta = \omega t + \phi$.For $x = +A/2$,$\cos(	heta_1) = 1/2$,so $	heta_1 =

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$135$

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(C) The displacement equation for $SHM$ is $x = A \cos(	heta)$,where $	heta = \omega t + \phi$.For $x = +A/2$,$\cos(	heta_1) = 1/2$,so $	heta_1 = 60^o$ or $300^o$.For $x = -A/\sqrt{2}$,$\cos(	heta_2) = -1/\sqrt{2}$,so $	heta_2 = 135^o$ or $225^o$.The possible phase differences $\Delta	heta = |	heta_2 - 	heta_1|$ are:$|135^o - 60^o| = 75^o$$|225^o - 60^o| = 165^o$$|135^o - 300^o| = 165^o$$|225^o - 300^o| = 75^o$Alternatively,using $x = A \sin(	heta)$:For $x = +A/2$,$\sin(	heta_1) = 1/2$,so $	heta_1 = 30^o$ or $150^o$.For $x = -A/\sqrt{2}$,$\sin(	heta_2) = -1/\sqrt{2}$,so $	heta_2 = 225^o$ or $315^o$.The possible phase differences are $|225^o - 30^o| = 195^o$ or $|315^o - 150^o| = 165^o$.Comparing these results,$135^o$ is not a possible phase difference.

$A$ particle is executing $SHM$ of amplitude $A$ about the mean position $x = 0.$ Which of the following cannot be a possible phase difference between the positions of the particle at $x = +A/2$ and $x = -A/\sqrt{2}$ (in $^o$)?

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