Recent content by questionas

I figured out what I was doing wrong. I forgot to take the square root of the 0.5 . Silly mistake. Thank you for the help.

Intreasting. I'll try the math again.

I thought it would be 2mx''(t) = k(x1-x2)-kx2 2mx''(t) = kx1 - 2kx2 mx''(t) = 0.5kx1 - kx2 ?

I proceeded with 0.5k. I do not understand why it should be negative?

if your referring to the one where I applied Newton's second law, I assumed the system ws horizontal and that the mass 1 was pulled a distance x1 to the right and mass 2 was pulled a distance x2 to the right. Then the spring between the two masses would be stretched and that spring will pull...

This is a the representation of the two masses. . Using Newton's second law I got the following equations assuming x1>>x2: m*x1''(t) = -k(x1-x2) m*x2''(t) = 0.5kx1-kx2 I put it in matrix form m| x1''| = | -k k| *|x1| |x2''| =| 0.5k -k| |x2| After some simplification assuming...