A person normally weighing `60 kg` stands on a platform which oscillates up and down harmonically at a frequency `2.0 sec^(-1)` and an amplitude `5.0 cm` . If a machine on the platform gives the person's weight against time deduce the maximum and minimum reading it will shown, `Take g = 10 m//sec^(2)`.

Options

A. maximum reading of the machine is 90 kg
B. minimum reading of the machine is zero
C. minimum reading of the machine is 32 kg
D. maximum reading of the machine is 108 kg

Answer

D

Explanation

The machine records the maximum and minimum weights. The maximum acceleration `=A omega^(2)`

At the highest point, the weight is maximum and the maximum reading
`= mg + m Aomega^(2)=m (g+A omega^(2))=m[g+4pi^(2)n^(2)A)`
`R_("max")=60[10+4xx10xx4xx5xx10^(-2)]`
=60[10+8]=1080 N
`therefore` Reading `=(1080)/(10)=108 kg`
Similarly Minimum Reading
`=m(g-A omega^(2))=60[10-8]=120 N`
`therefore R_("min")=12 kg`