Critically Damped Oscillator Spring Constant and Damping Parameter

  • Thread starter JordanGo
  • Start date
  • #1
73
0

Homework Statement



A mass of 1000 kg drops from a height of 10.0 m onto a platform of negligible mass.
It is desired to design a spring and damper on which to mount the platform so that it
will settle to a new equilibrium position 2.00 m below its original position as quickly
as possible without overshooting.
Find the spring constant k and the damping parameter
if the system is critically
damped.

Homework Equations



ω^2(frequency squared)=γ^2(damping parameter squared)
E=U=mgh at equilibrium
E=1/2kA^2
x(t)=(A1+A2t)e^(-γt)

The Attempt at a Solution



First, I solved for energy:
E=U=mgh=19400

Then for the spring constant:
k=2E/A^2

But now I need amplitude, so this is where I taking a shot in the dark:

x(t)=(A1+A2t)e^(-γt)
Now I was thinking to say that if t goes to infinity, x is 2, but it gave me no information... I need help! please and thank you
 

Answers and Replies

  • #2
rude man
Homework Helper
Insights Author
Gold Member
7,856
792
1. Realize that there is no actual "equlibrium" position. That position is realized only after infinite time has passed.

2. The moment the mass hits the platform, you have a classical mass-spring-damper situation with an initial velocity x'(0+) easily determined from energy conservation. Just solve the 2nd order diff. eq. mx'' + kx + cx' = 0. When you do, you'll look for the solution for which c/m, where c is the damping coefficient, yields barely two real roots for the attendant algebraic equation. k, the spring constant, is easily derived form the 2m criterion.

By 'barely' I mean just avoiding a complex-conjugate solution of your algebraic equation, which implies oscillatory behavior. So your time constants will be as short as possible without incurring oscillations.
 
  • #3
73
0
Ok then, well I am having problems with the DE since we have two unknowns, the spring constant k and the damping parameter...
 
  • #4
rude man
Homework Helper
Insights Author
Gold Member
7,856
792
Think about it for a second. When the system is at rest, the spring supports 1000 kg at a depth of 2m. So what's k?
 
  • #5
73
0
I want to use F=-mg=-kx, yes?
 
Last edited:
  • #7
73
0
ok so k=4900, thus damping parameter is 2.21, and the final equation is:
x(t)=(10+22.1t)exp(-2.21t)
Right?
 
  • #8
rude man
Homework Helper
Insights Author
Gold Member
7,856
792
Right! But you didn't even need to solve the whole equation.

If you write the equation in the standard form mx'' + nx' + p2x = 0 then
critical damping → n = p

But p = √(k/m) = 2.21 = n so that's all you were asked to do. (Assuming the instructor meant "n" as the "damping parameter", which he apparently did.)

(I plead guilty to not solving the whole equation like you did. I never do what's not necessary!).
 
  • #9
73
0
Haha Thanks so much, I really appreciate it!
 

Related Threads on Critically Damped Oscillator Spring Constant and Damping Parameter

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
5K
Replies
1
Views
8K
  • Last Post
Replies
1
Views
2K
Replies
2
Views
664
  • Last Post
Replies
3
Views
21K
Replies
2
Views
6K
  • Last Post
Replies
1
Views
1K
Replies
15
Views
2K
Replies
4
Views
1K
Top