Physics Asked by Lachie James on November 7, 2020
So every spring-damper system I’ve found online has the equation of motion:
$$mx”+cx’+kx=0$$
I can understand how this is derived when downwards is positive, but what about when upwards is positive? Wouldn’t it be
$$mx”-cx’-kx=0$$
Let $hat d$ and $hat u$ be unit vectors in the down and up direction.
Looking at your right (correct) derivation you have made down $hat d$ as positive and the displacement of the mass from the equilibrium position is $x hat d$ leading to the velocity being $dot x hat d$ and the acceleration being $ddot x hat d$.
So your equation of motion is $mg hat d - ks hat d - kx hat d - c dot x hat d = m ddot x hat d Rightarrow mddot x hat d + c dot x hat d + k hat d = 0$
To change from having down as positive to having up $hat u$ as positive all you have to do is note that $hat d = - hat u$ and substitute for $hat d$ into the equation which was derived was having down as positive.
You will find that you get the same equation of motion.
So what was wrong with your left hand derivation?
You made the displacement of the mass $-x hat u$ and the velocity of the mass $- dot x hat u$ but what you failed to do was to make the acceleration $-ddot x hat u$
Answered by Farcher on November 7, 2020
Yes.
This is because your 2nd equation should read as
$$ -mx''-cx'-kx=0 $$
since the sense where positive $x$ is has switched. This means a sign change for $x$, $x'$ and $x''$.
Answered by John Alexiou on November 7, 2020
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