N connected spring in vertical direction (with gravity)

What will happened if the spring is released. Click link for Answer

This center spring in this applet simulate the above situation.
The spring force $F(x)=-k (x-x_0)$ where $x_0$ is the equilibrium position.
The damping force is assumed to be $-b *\vec{v}$ , where $b$ is the damping constant.
In the following there are n springs in the simulation, the mass at two ends only experience one force from the spring.
Howerer, the other particles in between experience two forces from two springs at different side.
For the nth particle (n!=0 and n!=N-1), where N is the total number of the spring
Assume y for the n-th particle is $y_n$ ,
The n-th sprint force F_n =-k (y_{n}-y_{n+1}-L_0) -k (y_{n-1}-y_{n}-L_0)= k(2y_n- y_{n+1}-y_{n-1})

If you unchecked the fixed checkbox, the center spring will be released and fall down.
You can adjust b value, adjust mass or spring constant to find new equilibrium positions.
You can drag any particles up/down, too!

uncheck fixed checkbox and click play to find the answer to the above question.
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Embed a running copy link(show simulation in a popuped window)

Full screen applet or Problem viewing java?Add http://www.phy.ntnu.edu.tw/ to exception site list
Press the Alt key and the left mouse button to drag the applet off the browser and onto the desktop. This work is licensed under a Creative Commons Attribution 2.5 Taiwan License

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