I am coming from the computer science territory and have not a long trace in mechanics. My background in derivation of the system dynamics could be summarized with utilization of the Lagrange Mechanics but here is a vague point, with which I have confronted, recently. How does the existence of the the jerk term in dynamics of a Mass-Spring-Damper could be justified, where the Lagrange Dynamics just computes up to the 2nd derivatives of the generalized parameters?!...

If one considers a dynamic system, which (from left to right) consists of a spring with constant $k_1$, a mass $m$, a damper with constant $c$ and the other spring with constant $k_2$, all connected together, respectively, the derived dynamics of the system would be declared as: (The junction of spring $k_2$ and damper $c$ is massless)

$$ \frac{cm}{k_2} · \frac{d^{3} x}{d t^{3}} + m \frac{d^{2} x}{d t^{2}} + c \left( 1 +\frac{k_1}{k_2}\right) \frac{dx}{dt}+ k_1 x= 0 $$

Obviously, the Jerk term has been appeared, up there, noticeably.

Would you please guiding me that how such dynamics could be interpreted by either Lagrange Dynamics or Newtonian Method?!

  • $\begingroup$ Can you provide more of the derivation that gave you a third derivative? It doesn't sound like it should be there. $\endgroup$ Aug 14, 2015 at 11:23
  • $\begingroup$ Concerning difficulties in an action principle for jerk, see this Phys.SE post. $\endgroup$
    – Qmechanic
    Aug 14, 2015 at 12:14

2 Answers 2


If one considers a dynamic system, which (from left to right) consists of a spring with constant k1, a mass m, a damper with constant c and the other spring with constant k2, all connected together, respectively

If I understand your setup correctly, the damper is connected between the mass and the 2nd spring.

Denote the extension of spring 1 with $x_1$ and the compression of spring 2 with $x_2$ (a positive $x_1$ results in a leftward force and likewise for $x_2$).

Since the connection of the damper and spring 2 is massless, the net force there must vanish:

$$0 = -k_2x_2 - c(\dot x_2 - \dot x_1) \Rightarrow c\dot x_2 + k_2x_2 = c\dot x_1$$

At the junction of the mass and spring 1 we have

$$m\ddot x_1 = -k_1x_1 - c(\dot x_1 - \dot x_2) \Rightarrow m\ddot x_1 + c\dot x_1 + k_1x_1 = c\dot x_2$$

Substitute the 1st equation into the 2nd to find

$$m\ddot x_1 + k_1x_1 = -k_2x_2$$


$$c\dot x_2 = -\frac{cm}{k_2}\dddot x_1 - c\frac{k_1}{k_2}\dot x_1$$

Finally, substitute this into the 2nd equation to yield a third order equation of motion in $x_1$:

$$\frac{cm}{k_2}\dddot x_1 + m\ddot x_1 +c\left(1 + \frac{k_1}{k_2}\right)\dot x_1 + k_1 x_1 = 0$$

A Lagrangian for the system is

$$L = \frac{1}{2}m\dot x^2_1 - \frac{1}{2}k_1x^2_1 - \frac{1}{2}k_2x^2_2$$

The Lagrange equations (of the first kind) are

$$\frac{d}{dt}\frac{\partial L}{\partial \dot x_i} - \frac{\partial L}{\partial x_i} + \frac{\partial D}{\partial \dot x_i} = 0$$


$$D = \frac{c}{2}(\dot x_1 - \dot x_2)^2$$

yielding two equations of motion

$$m\ddot x_1 + k_1x_1 + c(\dot x_1 - \dot x_2) = 0$$

$$k_2x_2 - c(\dot x_1 - \dot x_2) = 0$$

which match the first and second equations above.

  • $\begingroup$ The complete trace of the dynamics... It sounds complicated especially in view of the modified Lagrangian... Thank you very much for your contribution... $\endgroup$
    – user89243
    Aug 15, 2015 at 13:24
  • 1
    $\begingroup$ Minor semantic comment to a good answer (v2): The word Euler-Lagrange equations usually implies a variational principle, which is absent because of the presence of a Rayleigh dissipation term. Without a variational principle, they are just called Lagrange equations (of the first kind), i.e. without the name Euler. $\endgroup$
    – Qmechanic
    Aug 15, 2015 at 17:16

I'm not sure where the term with the third derivative came from; taking the equilibrium point of our system to be $x=0$ we find the force due to the leftmost spring is given by: \begin{equation} F_{\textrm{spr,1}}\left(x\right)=-k_1x \end{equation} the force due to the rightmost spring is given by: \begin{equation} F_{\textrm{spr,2}}\left(x\right)=k_2x \end{equation} and the force due to the damper is given by: \begin{equation} F_{\textrm{damp}}\left(x\right)=-c\frac{\textrm{d}x}{\textrm{d}t} \end{equation} where $x$ is the position of the mass. We therefore find from Newton's laws that: \begin{equation} \begin{aligned} \sum_i\textbf{F}_i&=m\frac{\textrm{d}^2x}{\textrm{d}t^2}\\ \implies k_2x-k_1x-c\frac{\textrm{d}x}{\textrm{d}t}&=m\frac{\textrm{d}^2x}{\textrm{d}t^2}\\ \implies m\frac{\textrm{d}^2x}{\textrm{d}t^2}+c\frac{\textrm{d}x}{\textrm{d}t}+\left(k_1-k_2\right)x&=0 \end{aligned} \end{equation} which has no terms with a third derivative. In fact, this can be solved exactly using the method described here to obtain that: \begin{equation} x\left(t\right)=A\textrm{e}^{-\frac{ct}{2m}}\cos\left(\sqrt{\frac{k}{m}-\frac{c^2}{4m^2}}-\phi\right) \end{equation} where $A$ and $\phi$ are constants that depend on the initial conditions of our system.

  • $\begingroup$ Its not the right dynamics... You've missed the mysterious jerk term, is which the subject of this thread... $\endgroup$
    – user89243
    Aug 15, 2015 at 13:13
  • $\begingroup$ If you take the third derivative of my solution with respect to time you'll find the jerk term of the solution. $\endgroup$ Aug 15, 2015 at 13:28
  • $\begingroup$ Within the path of the presented scheme, the third derivative is not reasonable, as your final dynamics is from 2nd order. Also, even derivation will not lead to the right dynamics, because it is just supposed to consisting of 3 terms, instead of 4 terms... $\endgroup$
    – user89243
    Aug 15, 2015 at 13:38

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