NEET Physics Question

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: (2007)

$T/8$

$T/12$

$T/2$

$T/4$

Solution:

Equation from equilibrium is $x = a\sin(\omega t)$. For $x = a/2$, $a/2 = a\sin(\omega t) \Rightarrow \sin(\omega t) = 1/2$. This gives $\omega t = \pi/6$. Substituting $\omega = 2\pi/T$, we get $\frac{2\pi}{T}t = \pi/6 \Rightarrow t = T/12$.

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