(N/A) Given: Mass $m = 50\, kg$,frequency $\nu = 2.0\, s^{-1}$,amplitude $A = 5.0\, cm = 0.05\, m$.
Angular frequency $\omega = 2\pi\nu = 2 \times \pi \times 2 = 4\pi\, rad/s$.
Maximum acceleration $a_{max} = \omega^2 A = (4\pi)^2 \times 0.05 = 16 \times 9.8696 \times 0.05 \approx 7.896\, m/s^2$.
$(a)$ Yes,the weight changes because the platform is accelerating.
$(b)$ The apparent weight $N$ is given by $N = m(g + a)$,where $a$ is the acceleration of the platform (positive upwards).
At the lowest point,acceleration is upward: $a = +\omega^2 A$.
$N_{max} = m(g + \omega^2 A) = 50(9.8 + 7.896) = 50(17.696) = 884.8\, N$.
At the highest point,acceleration is downward: $a = -\omega^2 A$.
$N_{min} = m(g - \omega^2 A) = 50(9.8 - 7.896) = 50(1.904) = 95.2\, N$.