A rectangular block of mass 5kg attached to a horizontal spiral...

By Newton's second law,

F = m{a}_\max.

The maximum acceleration for a SHM is given by

{a}_\max = {\omega}^{2}A . . . ( i )

The time period is given by

$T = \dfrac{2\pi}{\omega} = 3 . 14s$

$⇒ \omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{3 . 14} = 2rad{s}^{- 1}$

Thus, the maximum acceleration from equation (i)

{a}_\max = 4 \times 1 = 4m{s}^{- 2}

The maximum force is

$F = 5kg \times 4m{s}^{- 2} = 20N$