Hint: There will be a pseudo force on the person due to oscillations of the platform.

Step 1: Find if the weight of the person will be variable.

In this case, the acceleration is variable. In accelerated motion, the weight of the body depends on the magnitude and direction of acceleration for upward or downward motion.

Step 2: Find the minimum weight of the person.

$\mathrm{Angular} \mathrm{frequency}, \mathrm{\omega }=2\mathrm{\pi n}=2\times 3.14\times 2=12.56 \mathrm{rad}/\sec$
$\mathrm{Maximum} \mathrm{acceleration}, {\mathrm{a}}_{\max }={\mathrm{A\omega }}^2=\frac{5}{100}\times {\left(12.56\right)}^2=7.89 \mathrm{m}/{\mathrm{s}}^2$

The weight of the person will be minimum when the force on the person is minimum at the upper extreme position.

$\mathrm{Minimum} \mathrm{weight}=\mathrm{W}-{\mathrm{ma}}_{\max }=50\times \left(9.8-7.89\right)=95.5 \mathrm{N}$

Step 3: Find the minimum weight of the person.

The weight of the person will be maximum when the force on the person is maximum at the lower extreme position.

$\mathrm{Minimum} \mathrm{weight}=\mathrm{W}+{\mathrm{ma}}_{\max }=50\times \left(9.8+7.89\right)=884.5 \mathrm{N}$