(D) The displacement of a particle in simple harmonic motion starting from the equilibrium position is given by $x = a \sin(\omega t)$.Given that the

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(D) The displacement of a particle in simple harmonic motion starting from the equilibrium position is given by $x = a \sin(\omega t)$.Given that the particle travels half of the amplitude,we set $x = a/2$.Substituting this into the equation: $a/2 = a \sin(\omega t) \Rightarrow \sin(\omega t) = 1/2$.Since $\sin(\pi/6) = 1/2$,we have $\omega t = \pi/6$.Substituting $\omega = 2\pi/T$,we get $(2\pi/T) \cdot t = \pi/6$.Solving for $t$,we find $t = T/12$.

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$A$ particle executes simple harmonic oscillation with an amplitude $a$. The period of oscillation is $T$. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

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A
$T/2$
B
$T/4$
C
$T/8$
D
$T/12$

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