Mass-spring-dampened system

  • Thread starter phenalor
  • Start date
  • #1
6
0

Homework Statement


A mass-spring-dampener system is applied a force [itex]mg[/itex] and is immediatly removed, setting the system in motion. The system is constantly applied force [itex]Mg[/itex] and is static at [itex]y=y_0[/itex].
Find a formula for both [itex]A[/itex] and [itex]\phi[/itex]


Homework Equations



[itex]\ddot{y}+2\delta\dot{y}+w_0^2y=0[/itex]
[itex]\frac{2\pi}{w_0}=T_0[/itex]
[itex]\delta = \frac{3}{5}w_0[/itex]
[itex]F_f=-b\dot{y}[/itex]
[itex]Mg=ky_0[/itex]

The Attempt at a Solution



from this i find [itex]k[/itex] and [itex]b[/itex]. No problem, not part of my question.

when the force is applied, the system 'moves' in y direction and is set in motion, given function:

[itex]y(t)=Ae^{-\delta t}cos\left(w_d t+\phi\right)[/itex]
[itex]w_d=\sqrt{w_0^2+\delta^2}[/itex]

I'm to find [itex]A[/itex] and [itex]\phi[/itex]

my try:

I understand [itex]\dot{y}(0)=0[/itex] gives:
[itex]\dot{y}=-\delta Ae^{-\delta t}cos(w_d t+\phi)-w_d A e^{-\delta t}sin(w_d t + \phi)[/itex]
gives:
[itex]\phi=\arctan{\frac{-\delta}{w_d}}[/itex]

however i do not find a substitute for [itex]y(0)[/itex]. The solution says [itex]y(0)=\frac{m}{M}y_0[/itex], but i don't see the logic in that at all

Sorry if its a bit caotic, this is only part of the assignment. ask and i will provide!
 

Answers and Replies

  • #2
6
0
nvm, found the solution
 

Related Threads on Mass-spring-dampened system

Replies
6
Views
2K
  • Last Post
Replies
2
Views
529
  • Last Post
Replies
2
Views
975
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
16
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
9
Views
7K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
6
Views
10K
Top