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(C) For the particle to move from the extreme position $(x=A)$ to half of the amplitude $(x=A/2)$: $x = A \cos(\omega t_1) \implies A/2 = A \cos(\omeg

Vedclass · Accepted Answer

$t_1=2 t_2$

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(C) For the particle to move from the extreme position $(x=A)$ to half of the amplitude $(x=A/2)$: $x = A \cos(\omega t_1) \implies A/2 = A \cos(\omega t_1) \implies \cos(\omega t_1) = 1/2$. Thus,$\omega t_1 = \pi/3$,which gives $t_1 = \pi / (3\omega)$. For the particle to move from the mean position $(x=0)$ to half of the amplitude $(x=A/2)$: $x = A \sin(\omega t_2) \implies A/2 = A \sin(\omega t_2) \implies \sin(\omega t_2) = 1/2$. Thus,$\omega t_2 = \pi/6$,which gives $t_2 = \pi / (6\omega)$. Comparing the two times: $t_1 / t_2 = (\pi / 3\omega) / (\pi / 6\omega) = 6/3 = 2$. Therefore,$t_1 = 2 t_2$.

$A$ particle is executing simple harmonic motion. If the minimum time taken by the particle to move from extreme position to half of the amplitude is $t_1$,and the minimum time taken by the particle to move from mean position to half of the amplitude is $t_2$,then

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