# 2 DOF oscillator max force response

1. Apr 25, 2012

### saxymon

Hello,

I am mounting a component onto structure and I need to determine the maximum force input into the component.

My system can be represented by a base driven two degree of freedom oscillator:

I need to determine the force applied to m2:

F = k2*(x2-x1) + c2*(x2-x1)

This force needs to be a function of (m1,m2,c1,c2,k1,k2,y) and not of (x1,x2).
Basically, for a given input y, what will the be force response on m2.

Every time I solve the system of equations, my result is a function of x1 and/or x2.

p.s. If it is easier, feel free to remove the dampers from the system. An un-damped system will work for my purposes.

2. Apr 25, 2012

### MisterX

This is somewhat unfamiliar territory for me. A real spring has some length, right? Let that be l2 or l1 when the springs are unloaded.

F2 = -k2d
d = x2 - x1 - l2
F2 = m2dx2/dt = k2(x1 + l2 - x2)

Then the force on m1 would be the negation of that plus the force from the spring connecting it to the base

F1 = m1dx1/dt = k2(x2- x1 - l2) + k1(y + l1 - x1)

This should be a system of differential equations which has solutions dependent on initial values of x1 and x2 as well as the entire input function y(t). You must further specify the problem by placing constraints on the initial positions, and the input function y(t). Then you may still be left with a seemingly large space of y(t) functions over which to maximize the force.

Last edited: Apr 25, 2012
3. Apr 26, 2012

### saxymon

The nominal length of the spring can be ignored as this is an oscillatory system (and as long as the stiffness is the same, the response will be the same, regardless of length).

Also, I am seeing the force on m2 (not m1), which is defined as
F = k2*(x2-x1)

I am looking for a steady state response due to a stationary random input which means initial conditions are not needed (they effect only the transient response). Think of this as a "Force" Frequency Response Function

Here is what I've done...

my matrix system of equations is...

[m1 0 ; 0 m2] [x''1 ; x''2] + [k1+k2 -k2 ; -k2 k2] [x1 ; x2] = [0 ; k1] [0 y]

Interestingly enough, the force I need to recover is naturally in the lower portion of the equation giving:

F = k1*y - m2*x''2

I make the reduction to

F = k1*y + Wn^2*m2*x2

but still was not able to remove all of the x2's and x1's (after making a number of substitutions).