The correct option is
D The spring balance reading fluctuates between
50 kg and
70 kg.
The maximum force acting on the body executing simple harmonic motion is
Fmax=|kA|=mω2A=m×(2πf)2A
From the data given in the question,
F=60×(2π×2π)2×0.1=96 N or
969.8≈10 kgf
This force always acts towards the mean position.
Case -I :
In Equilibrium position , As shown in the figure, Normal reaction force balances the weight of the man
So, the spring balance reading measures
60 kg.
Case -II :
At Upper extreme position , The force
Fmax is acting downwards (towards mean position). So a pseudo force acts on it in the opposite direction.
From FBD,
N+Fmax=mg⇒N=mg−Fmax=60−10=50 kgf
The reaction of the force on the platform is away from the mean position. It reduces the weight of man at upper extreme i.e. Net weight
=50 kgf.
Case- III :
At lower extreme position, the force
Fmax is acting upwards (towards mean position). So, a pseudo force acts on it in the opposite direction.
From FBD,
N=mg+Fmax=60+10=70 kgf
The reaction of the force on the platform is away from the mean position. This force adds to the weight at lower extreme position i.e. net weight
=70 kgf.
Therefore, the reading of the weight recorded by the spring balance flutuates between
50 kgf and
70 kgf.
Thus, option (d) is the correct answer.