Shouldn't the values in the absolute value be just the norm of x1-x2, ie. ||x1-x2||? Since we are talking about vectors, absolute values don't make much sense either. Alternatively this can be written as \sqrt{(x1-x2)^T(x1-x2)}.
So in total, it should be 1/2k(\sqrt{(x1-x2)^T(x1-x2)} - L_0)^2
This is a good reminder to always check that the dimensions of terms match.
mdsavage
@kc1: I think the expression on the slides is correct as written, since the potential energy in a spring is proportional to the square of the displacement of the length of the spring from its length under no forces. (So replacing the value in the parenthesis with ||x1 - x2|| would only be correct if the spring's rest length is 0.)
kc1
@mdsavage as written |x1-x2|^2 has dimensions of meters^2 ie. it can be written as (x1-x2)^T(x1-x2). Then you subtract length which is in meters which doesn't make sense. It can't possibly be correct.
keenan
@kc1 Yes, that square shouldn't be there. Just a bug in the slide. Also note that single bars $|\cdot|$ are quite often used for the norm of a vector, rather than double bars $||\cdot||$.
Shouldn't the values in the absolute value be just the norm of x1-x2, ie. ||x1-x2||? Since we are talking about vectors, absolute values don't make much sense either. Alternatively this can be written as \sqrt{(x1-x2)^T(x1-x2)}.
So in total, it should be 1/2k(\sqrt{(x1-x2)^T(x1-x2)} - L_0)^2
This is a good reminder to always check that the dimensions of terms match.
@kc1: I think the expression on the slides is correct as written, since the potential energy in a spring is proportional to the square of the displacement of the length of the spring from its length under no forces. (So replacing the value in the parenthesis with ||x1 - x2|| would only be correct if the spring's rest length is 0.)
@mdsavage as written |x1-x2|^2 has dimensions of meters^2 ie. it can be written as (x1-x2)^T(x1-x2). Then you subtract length which is in meters which doesn't make sense. It can't possibly be correct.
@kc1 Yes, that square shouldn't be there. Just a bug in the slide. Also note that single bars $|\cdot|$ are quite often used for the norm of a vector, rather than double bars $||\cdot||$.