Appearance of the Jerk Term in Dynamics of Mass-Spring-Damper System

Question

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?!

Answer 1

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

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

Thus

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

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

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

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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

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

where

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

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.

Answer 2

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.