Equilibrium position

  • Thread starter timtng
  • Start date
  • #1
timtng
25
0
A load of mass .2 g is hanging from a light spring whose elastic constant is 20 N/m. The load is pulled down .1 m from its equilibrium position and released.

How long is required for the load to reach its equilibrium position?

T=2pi(square root(m/k))=.628s

Please verify to see if I did it correctly.

Thanks
 

Answers and Replies

  • #2
FZ+
1,599
3
I'm not looking carefully, but you seem to be wrong. It asks the time taken to reach the initial equilibrium position, while you seem to have given the periodic time for an oscillation in SHM, which would be 4*t.
 
  • #3
timtng
25
0
so the answer should be T/4?
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
43,021
970
Yes. The "T" you give is the time to go up to the maximum height, then back down to the initial position: 1 cycle. The weight will take exactly 1/4 of that time to go back to the equilibrium point (1/2 T to reach the highest point, 3/4 T to pass the equilibrim point again and then at T back to the initial point).
 
  • #5
timtng
25
0
so t should equal to .628s/4 = .157 s
 

Suggested for: Equilibrium position

Replies
10
Views
592
Replies
29
Views
402
  • Last Post
Replies
9
Views
304
  • Last Post
Replies
3
Views
225
Replies
5
Views
181
Replies
2
Views
289
Replies
24
Views
511
Replies
2
Views
499
Top