Recent content by brotof

Thanks a lot, that makes sense! Im really sorry, I was a bit hasty and have made an error. It's supposed to be \begin{equation} b^2 = ac \end{equation} without the factor 4. I don't think I'm wrong when taking that into account that I meant b^2 = ac, even though I wrote it a bit more...

I see that the Lagrangian then can be written: \begin{equation} L = \frac{m}{2c}(b\dot{x} + c\dot{y})^2 -\frac{k}{2c}(bx + cy)^2 \end{equation} Also, bot my equations of motion reduces to a single one: \begin{equation} \frac{b^2}{c}\dot{x} + b\dot{y} + \frac{k}{m}(\frac{b^2}{c} x + by) = 0...

Homework Statement Consider the following Lagrangian: \begin{equation} L = \frac{m}{2}(a\dot{x}^2 + 2b\dot{x}\dot{y} + c\dot{y}^2)- \frac{k}{2}(ax^2 + 2bxy + cy^2)\end{equation} Assume that \begin{equation} b^2 - 4ac \ne 0 \end{equation}Find the equations of motion and examine the cases...