Question

Sol. [Key Concept: In SHM $x = Acos\omega\ t$; moving from $x = A$ to $x = \frac{A}{\sqrt{2}}$ covers phase $\frac{\pi}{4}$, so $t = \frac{\pi/4}{\omega} = \frac{T}{8}$.]

$${x = 0 \rightarrow \theta = 0^{\circ} }{x = \frac{A}{\sqrt{2}} \rightarrow \theta = \frac{\pi}{4} }$$∴ phase covered $= \frac{\pi}{4}$

time $= \frac{\text{~}\text{phase}\text{~}}{\text{~}\text{w}\text{~}} = \frac{\pi}{4\text{~}\text{w}\text{~}} = \frac{\pi}{4} \times \frac{T}{2\pi} = \frac{T}{8}$

$\therefore t = \frac{5}{8}$ secs.