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Time calculations in SHM are a frequent JEE question type. The key insight is that SHM velocity is not constant, so equal distances do NOT take equal times. The particle moves fastest at the mean position and slowest near the extremes.

Standard time intervals (starting from mean position): mean to $A/2$ takes $T/12$; $A/2$ to $A$ takes $T/6$; mean to $A/\sqrt{2}$ takes $T/8$; mean to $A$ (extreme) takes $T/4$. From one extreme to the other takes $T/2$. Method: use $x = A\sin(\omega t)$ (starting from mean) or $x = A\cos(\omega t)$ (starting from extreme), solve for $t$. For starting from an arbitrary position, use the phase form: $x_0 = A\sin\phi_0$, find $\phi_0$, then solve for the phase at the target position.

A useful symmetry: the time from $x_1$ to $x_2$ through the mean position is NOT the same as the time from $x_1$ to $x_2$ through the extreme (unless symmetric about the mean). Always check which path the particle takes. Drawing the reference circle (phasor diagram) is the most reliable method for complex time calculations — convert the problem to finding the angle swept on the circle, then time $= \text{angle}/\omega$.