0
$\begingroup$

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?

$\endgroup$
1
  • 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
2
$\begingroup$

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.

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

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.