# Spring and damper in series, equation of motion

For the motion in fig 4.1 (a), please check the description of this system in the images above, everything should be there. I get it, assuming the positive direction is downward, the spring tension is k(x-y), as (x-y) is the amount of extension by the spring. However, as the entire system is moving downwards (as the positive direction) and the base of the system is fixed to the ground, shouldn't the force of the damper point upwards (against the motion of the system)?

Therefore, I think the motion equation should be: $$-cx(dot)+kx=ky$$ alternatively $$-cx(dot)+k(x-y)=0$$

By the way, I was viewing the motion at point A.Thus the tension of spring exerts a force downwards $$(k(x-y))$$. The damper exerts a force upwards $$(-cx(dot))$$.

• Resisting force of both damper and spring is in the same direction,hence the equation given seems correct. Apr 3 at 7:12
• Hi, tada655. I am aware that both spring and damper are in the same direction. But for compression, the spring force formula should be k(y-x), not k(x-y). So I think it should be k(x-y)+cxdot =0. If you get me. Thanks for replying!
– S R
Apr 7 at 17:07

Let $$\delta$$ be the no-load length of the spring. The problem indicates that the distance between $$x$$ and $$y$$ is equal to $$\delta$$ when both $$x$$ and $$y$$ are zero. I.e. the spring force is zero when both $$x$$ and $$y$$ are zero.
If $$x$$ remains zero and $$y>0$$, then the spring compresses and $$F_s$$ acts in the positive $$x$$ direction.
If $$y$$ remains zero and $$x>0$$, then the spring extracts and $$F_s$$ acts in the negative $$x$$ direction.
Equivalently, when $$y-x>0$$, the spring compresses and $$F_s$$ acts downward on $$B$$.
When using the right figure, writing Newton's 2nd law, we get $$m\ddot{x}=F_s-F_d=0$$ That is, $$m\ddot{x}=0$$ because the mass is zero.
So, if we define $$F_s=k(y-x)$$ then $$F_s>0$$ when $$y>x$$ fits the direction of $$F_s$$ in the figure.
Furthermore, $$F_d=c\dot{x}$$ such that Newton's 2nd law gives $$F_s=F_d\hspace{0.5cm}\Rightarrow\hspace{0.5cm}k(y-x)=c\dot{x}$$