Solution:
Using equation for SHM starting from extreme position: $x = A\cos(\omega t)$.\nSubstitute $x = A/2$: $A/2 = A\cos(2\pi t/T) \implies \cos(2\pi t/T) = 1/2$.\n$2\pi t/T = \pi/3 \implies t = T/6$.
A simple harmonic oscillator has an amplitude $A$ and time period $T$. The time required by it to travel from $X = A$ to $X = A/2$ is: (1992)
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