Mass attached to a horizontal spring

  • Thread starter pvpkillerx
  • Start date
  • #1
pvpkillerx
22
0
A particle with a mass of 0.210kg is attached to a horizontal spring with a force constant of 0.84 N/m. At the moment t=0, the particle has its maximum speed of 5m/s and is moving to the left. (Assume that the positive direction is to the right.)

Determine the particle's equation of motion, specifying its position as a function of time (use the following as necessary: t)

F = -kx
F = ma




What I did was:
-0.84x = 0.210a
-0.84x = 0.210[(v - 5)/t)]
-0.84x = 0.210[(x/t - 5)/t]


I still can't isolate for x, and i dont know if what i am doing is right. Please help.
 

Answers and Replies

  • #2
pvpkillerx
22
0
nvm, i figured it out, its 5/2Sin(2t+pi)
 

Suggested for: Mass attached to a horizontal spring

Replies
20
Views
426
Replies
10
Views
270
  • Last Post
Replies
20
Views
848
Replies
4
Views
513
  • Last Post
Replies
10
Views
617
Replies
1
Views
1K
Replies
3
Views
699
Top