A block of mass `4kg` hangs from a spring of force constant `k = 400 N//m`. The block is pulled down `15 cm` below equilibrium and relesed. How long does it take block to go from `12 cm` below equilibrium (on the way up) to `9cm` above equilibrium ?

Answer

A, B

Explanation

`omega = sqrt((k)/(m)) = sqrt((400)/(4)) = 10 rad//s`
Let `t_(1)` be the time from `x = 0` to `x = 12 cm`and `t_(2)` the time from `x = 0` to `x = 9cm`. Then, `12 = 15 sin (10t_(1))`
or `t_(1) = 0.093s`
`9 = 15 sin (10t_(2))`
or `t_(2) = 0.064 s`
`:.` Total time `= t_(1) + t_(2) = 0.157 s`