A particle executes a simple harmonic motion of time period $T$. Find the time taken by the particle to go directly from its mean position to half the amplitude
A$T / 2$
B$T / 4$
C$T / 8$
D$T / 12$
Medium
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D$T / 12$
d (d) $y = A\sin \omega t = \frac{{A\sin 2\pi }}{T}t$
==> $\frac{A}{2} = A\sin \frac{{2\pi t}}{T}$
==> $t = \frac{T}{{12}}$.
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