# Homework Help: Spring-Damper System Question

1. Oct 17, 2009

### aerandir4

1. The problem statement, all variables and given/known data

Spring-Damper system has a force applied such that the formula f(t)= kx(t)+c(dx/dt) holds
Determine the resulting displacement x(t) and sketch the function.
What is the displacement at t=1 sec?

2. Relevant equations
k=10^4 Nm^-1
c=1.25x10^4 Nsm^-1
x(0)=0

the is also a graph that has time on the horizontal axis and f(t) on the vertical axis f(t). It goes up linearly from 0 to f(0.5)=100 and then linearly drops with negative gradient to 0 at f(1)=0 and then carries on at infinitely at f(t)=0 for t>1

3. The attempt at a solution

I think we have to define the graph first in three steps

0<t<0.5 f(t)=200t
0.5<t<1 f(t)=200(1-t)
t>1 f(t)=0

after this I'm confused on how to proceed. Initially I took each set of conditions and converted them to X(s) via laplace transforms then solved by partial fractions and then converted back to X(t). I'm not sure if this is the right method.

Another way would be to solve each condition by integration,
i.e. when 0<t<0.5 then f(t)=integral between 0 and 0.5 of exp^-st*200t dt
then substitute this value back into f(t) in the equation of motion.

some guidance would be nice.

thanks

2. Oct 18, 2009

### fantispug

You broke up f(t) into three steps; can you guess an x(t) for each step that satisfies the constraint equation?
Once you've got this what is the most general solution of the constraint equation? You should be able to solve the problem now using the initial condition and continuity of x(t).

(Though I'm not sure if x(t) will satisfy the equation at .5 and 1).

3. Oct 18, 2009

### aerandir4

between 0<t<0.5
i plugged in 200t=kx+c(dx/dt)
now---> 200/s^2=kX(s)+c[sX(s)-X(0)]
I re arranged in terms of X(s)=200/s^2*(k+cs)
after using partial fractions i get X(s)=-1/40s+1/50*s^2+1/40*(k/c+s)
which would give X(t)=-1/40+t/50+exp^(-k/c*t)/40

doing the same for 0.5<t<1

I get X(t)=1/200-t/250-exp^(-k/c*t)/200
i assume the third condition to be 0 as f(t)is 0 for t>1

am I going in the right direction up to now?

I didn't understand your last tip fully what do you mean by "general solution of the constraint equation"
what do I do for each x(t) equation?

many thanks

4. Oct 22, 2009

### fantispug

Yeah, it certainly looks like you're moving in the right direction. When I was talking about "general solution" I meant that a generic first order differential equation has a parameter family of solutions (i.e. you typically get a constant of integration that can be any number).

Now suppose you've got an inhomogeneous linear differential equation; that is if D is some linear differential operator (here D=k+ c d/dt) and you want to solve
D X(t) = f(t) for some function f(t).
Now suppose we've got one solution X and we want to find another solution Y,
D X = f
D Y = f and so by linearity
D (Y-X) = 0. Thus any other solution is of the form X + U where D(U) = 0.

(To give you an example: Solve f'(t) = t. Well we know X(t)=t^2/2 is a solution, and any other solution is X + U, where U'(t) = 0, i.e. U = constant. So the general solution is t^2/2 + constant).

Now it looks like you've got one solution: given the above hints can you find the most general solution (i.e. the one with the arbitrary constant of integration). Continuity of X(t) at 0 and 0.5 should force a particular value of this constant of integration.