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I'm working on a mass-spring-dashpot system and I have a system of coupled differential equations I'm not sure how to solve. Below I've given a picture of the problem setup.

enter image description here

There is a mass $m$ attached to two elements in parallel: a spring $k$ and a dashpot $d$. These are connected to another dashpot $c$. A force $F(t)$ is applied to the junction between $c$ and $k, d$. I've defined some distances such as $x(t)$ and $y(t)$.

I'll write my first equation at the yellow point. There are four forces applied here (from $F(t), c, k, d$). Since there is no mass at the yellow dot, we can write:

$$0 = F + k(x - y) + d(\dot{x} - \dot{y}) - c\dot{y}$$

I'll write my second equation at mass $m$. There are two forces applied here (from $k, d$).

$$m\ddot{x} = -k(x - y) - d(\dot{x} - \dot{y})$$

I now have two equations for two unknowns, $x$ and $y$. What I would like is an equation independent of $y$: it may contain $x, F(t)$, any derivatives of $x$, and any constants.

I'm not sure how to solve this system. Both of my equations contain derivatives of both variables, so I don't see a substitution I can make to simplify one to just having $x$.

What I attempted to do (from a physics, not a math perspective) was to redraw the diagram using capacitors, inductors, and current sources - this way I could sum up the components using complex impedance. I didn't get very far with this method, though.

Any ideas on what I can do? Both equations are linear, so I conjecture a nice solution should exist, but I can't seem to find it.

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It's a simple one which mostly arises in control theory I suggest use Laplace Transformation for solving this type. you can find the method in the following Link:

(Modern Control Systems by Richard C. Dorf) http://www.amazon.com/Modern-Control-Systems-12th-Edition/dp/0136024580

if u just need the solution it's like this:

taking Laplace Transform of the first equation gives: $$F(s) +k(X(s)-Y(s))+d(sX(s)-x(0)-sY(s)+y(0))-c(sY(s)-y(0))=0$$ also taking Laplace Transform from the second equation gives: $$m(s^2X(s)-sx(0)-x'(0))=-k(X(s)-Y(s))-d(sX(s)-x(0)+sY(s)-y(0))$$

the lower case x means x as a function of time x(t) and upper case X is the Laplace Transform of x which is a function of complex frequency variable s which is X(s). $x'$ is derivative of x with respect to t

now differential equation system is transformed to linear equation system in Laplace Domain (S_Domain).

usually the rest condition response of system to the input is needed . under this condition we consider $$x(0)=x'(0) = 0 \; \; and \; \; y(0)=y'(0)=0 $$ so the linear equation system burns down to: $$X(s)(k+ds)-Y(s)((d-c)s+k)=-F(s)$$ $$X(s)(ms^2+ds+k)-Y(s)(ds+k)=0$$

with X(s) and Y(s) as unknown. now if you have the equation of input f(t) by taking Laplace Transform of f(t) and obtaining F(s) as a function of s; you can solve for X(s) and Y(s). then taking the inverse Laplace Transform of X(s) and Y(s) gives you x(t) and y(t).

in control Theory the response to unit step function is usually needed and is enough to fully characterize a system(u can see the proof of this claim in so called reference above). so if we consider $f(t) = u(t)$ the unit step function (AKA Heaviside Step) then $F(s) = 1/s$; so the solution to equations becomes:

$$X = \frac{-(k+ds)}{(s^2(ms^2(c-d) + (cd-km)s+ck))}$$ $$Y = \frac{-( ms^2 + ds +k)}{(s^2(ms^2(c-d) + (cd-km)s+ck))}$$

for finding the time domain functions we must take inverse Laplace transform using Partial Fraction Expansion of X and Y. the solutions will be different according to the values of m ,d ,c and k and it can be sinusoidal or exponential or a combination of both. this is because the behavior of roots second order polynomial in the denominator and the determinant of the equation $\Delta$.

another way to solve this is by State Space representation of the equations.

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  • $\begingroup$ ...I'm in absolute shock that I completely ignored converting it into Laplace-space to begin with. My logic was that it would be easier to just find the single ODE and then convert it to Laplace space, but this method is so much simpler. Thank you. ^^ $\endgroup$ – anonymouse Feb 29 '16 at 19:38
  • $\begingroup$ anytime dude :) $\endgroup$ – K.K.McDonald Feb 29 '16 at 19:55

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