I have a spring-damper system like so:

spring-damper in series

When $x_1$ is the length of the spring and $x_2$ is the length of the damper. The forces are given by: $$ \\ x=x_1+x_2\\F_d=-\sigma\ \frac{dx}{dt}\ , F_k=-k(x_1-x_0)$$ Which means that the damping force depends on the velocity of the mass.
From Newton we get: $$m\frac{d^2x}{dt^2}= -k(x_1-x_0)-\sigma\ \frac{dx}{dt}\ $$ Which can't be solved unless there is another relationship between $x_1$ and $x$.
Am I missing something? Is there a way to solve it?

  • 2
    $\begingroup$ Your equation of motion is incorrect. The forces are identical, not additive. You can show this by drawing a free-body diagram of each component. $\endgroup$ Nov 13 '21 at 15:01

This is the FBD

enter image description here

From here you can obtain the equation of motion, you should get second order differential equation plus first order differential equation.


$$m_1\,\ddot x_2=F_\sigma$$

and put dummy mass between the spring and the damper

$$m_d\,\ddot x_1=F_k-F_\sigma$$

with $~m_d=0~$ and

$$F_k-F_\sigma=0\quad,F_\sigma=\sigma\,(\dot x_1-\dot x_2)\quad, F_k=-k\,x_1\quad, \quad \Rightarrow\\ -k\,x_1-\sigma\,(\dot x_1-\dot x_2)=0 $$ hence $$\sigma\,\dot x_1=-k\,x_1+\sigma\,\dot x_2\tag 1$$


$$m\,\ddot x_2=\sigma\,(\dot x_1-\dot x_2)=-k\,x_1\tag 2$$

equation (1) and (2) are the EOM's

  • $\begingroup$ That is exactly what I got. Thank you! $\endgroup$ Nov 15 '21 at 18:06
  • $\begingroup$ @OfirShukrun you got spring damper parallel. I will write you the equations $\endgroup$
    – Eli
    Nov 15 '21 at 19:42

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