Question

A spring has a natural length of 50 cm and a force constant of `2.0 xx 10^(3)` N`m^(-1)`. A body of mass 10 kg is suspended from it and the spring is stretched. If the body is pulled down to a length of 58 cm and released, it executes simple harmonic motion. The net force on the body when it is at its lowermost position of its oscillation is (10x) newton. Find value of x. (Take g=10m/`s^(2)`)
A. 160 N
B. 40 N
C. 60 N
D. 80 N

Answer 1

Correct Answer - C
`l=A+x+L`
`58=A+50+x`
`A+x=8`
`F=Kx " "therefore x=(F)/(K)`
`x=0.05m`
`A=0.03m`