Question
A mass of 1.66 kg stretches a vertical spring 0.312 m. If the spring is stretched an additional 0.136 m and released, how long does it take to reach the (new) equilibrium position again?
Answers
bobpursley
Hmmm
Ke initial= mg v^2
k .136^2=mg v^2 solve for v.
Now half of v, or v/2 is the average velocity. time= distance/avgvelocity=.136*2/v
Ke initial= mg v^2
k .136^2=mg v^2 solve for v.
Now half of v, or v/2 is the average velocity. time= distance/avgvelocity=.136*2/v