A spring of spring constant `200 N//m` has a block of mass `1 kg` hanging at its one end and form the other and the spring is attached to a celling of an elevator. The elevator rises upwards with an acceleration of `g//3`.
When acceleration is suddenly ceased , then what should be the angular frequency and elongation during the time when the elevator is accelerating?

Explanation

The angular frequency under all circumstances is
`omega = sqrt((k//m)) = sqrt((200//1)) = 14 rad//s`

when elevator is moving up, the equation of mosition
`T = mg = (mg)/(3)`
`implies T = (4 mg)/(3)`
This tension elongates the spring by x
`T = kx`
`implies x = (4 mg)/(3k) = 0.07 m`