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. (i),(iii)
B. (ii), (iii)
C. (i), (iv)
D. (ii), (iv)

Answer

A

Explanation

`omega^2A=(2pif)^2A=4pi^2f^2A=4pi^2(2)^2xx5xx10^-2=8`
`mg-N_1=momega^2A`
`N_1=mg-momega^2A=60g-60xxg`
`=60g-0.8xx60g=0.2xx60g=12g`
`N_2-mg=momega^2A`
`N_2=mg+momega^2A=60g+60xxg`
`=60g+0.8xx60g=1.8xx60g=108g`
Maximum reading i.e., `N_2=108kg omegat`
Maximum reading i.e., `N_1=12kgomegat`