Question

A spring of spring constant `200N//m` has a block of mass `1kg` hanging at its one end and other end of spring is attached to ceiling of an elevator. The elevator is rising upwards with an acceleration `g//3`. What should be the angular frequency and elongation during the time when the elevator is accelerating?

A. `14.14`rad/s, `0.07`m
B. 14 rad/s, `0.1`m
C. `14.14` rad/s, `0.05` m
D. 10 rad/s, `0.07`m

Answer 1

Correct Answer - A
The angular frequency under all circumstance is
`omega=sqrt(((k)/(m)))=sqrt(((200)/(1)))=14.14rad//s`
When elevator is moving up, the equation of motion is
`T-mg=(mg)/(3)rArrT=(4mg)/(3)`
This tension elongates the spring by x
T=kx
`rArr" "x=(4mg)/(3k)=0.07m`