A spring is placed in vertical position by suspending it from a hook at its top. A similar hook on the bottom of the spring is at 11 cm above a table top. A mass of 75 g and of negligible size is then suspended from the bottom hook, which is measured to be 4.5 cm above the table top. The mass is then pulled down a distance of 4 cm and released. Find the approximate. Position of the bottom hook after 5 s? Take `g=10(m)/(s^2)` and hook's mass to be negligible.

Options

A. 5 cm above the table top
B. 4.5 cm above the table top
C. 9 cm above the table top
D. 0.5 cm above the table top

Answer

D

Explanation

From equilibrium position of mass, `mg=ky_0` where `y_0=(11-4.5)cm=6.5cm`
So, time period of simple harmonic motion is
`T=2pisqrt((m)/(k))=2pisqrt((y_0)/(g))`
`=2pisqrt((5.6xx10^-2)/(10))=0.5065s=0.5s`
5 s is equivalent to 10 complete time period, so the mass is at its initial position at `t=5s`, i.e, it is at `0.5cm` above the table top at `t=5s`.