A particle is executing simple harmonic motion. Its amplitude is A...

The equation of simple harmonic motion for a particle starting from the extreme position $x = A$ is given by:

$x = A \cos(\omega t)$

We need to find the time $t$ when the particle reaches $x = \dfrac{A}{\sqrt{2}}$. Substituting this value into the equation:

$\dfrac{A}{\sqrt{2}} = A \cos(\omega t)$

$\cos(\omega t) = \dfrac{1}{\sqrt{2}}$

$\omega t = \dfrac{\pi}{4}$

Since the angular frequency $\omega = \dfrac{2\pi}{T}$, we have:

$\dfrac{2\pi}{T} t = \dfrac{\pi}{4}$

$t = \dfrac{T}{8}$

Given the time period $T = 5$ sec, the required time is:

$t = \dfrac{5}{8}$ sec

Answer: $5/8$