# Spring-damper in series

I have a spring-damper system like so:

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?

• 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. Nov 13 '21 at 15:01

This is the FBD

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

Edit

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

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 $$\sigma\,\dot x_1=-k\,x_1+\sigma\,\dot x_2\tag 1$$

and

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

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

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