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September 30, 2020, 04:43:59 am

Give me a standpoint and I will move the earth. ...Archimedes (287-212BC)

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 Author Topic: N connected spring in vertical direction (with gravity)  (Read 13602 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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 « Embed this message on: April 27, 2006, 10:31:06 am » posted from:Taipei,T'ai-pei,Taiwan

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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