A particle is executing simple harmonic motion. Its amplitude is \(A\) and time period is \(5\) sec. The time required by it to move from \(x = A\) to \(x = \dfrac{A}{\sqrt{2}}\) is _______ sec.

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: \(5/8\)