$m{\omega ^2}a = m \times {(2\pi f)^2}a = 60 \times {\left( {2\pi \times \frac{2}{\pi }} \right)^2} \times 0.1\,N$

$ = 60 \times 16 \times 0.1 = 96N = \frac{{96}}{{9.8}} \approx 10\,kgf$ and this force is towards mean position.

The reaction of the force on the platform away from the mean position. It reduces the weight of man on upper extreme i.e. net weight $= (60 -10)\, kgf.$

This force adds to the weight at lower extreme position i.e. net weight becomes $= (60 + 10)\, kgf.$

Therefore, the reading the weight recorded by spring balance fluctuates between $50\, kgf$ and $70\, kgf.$