Recent content by ckelly94

Goddamn.

...Yes... That's why the last term is negative.

I got that last result by just plugging the first (positive) eigenvalue into the first row of the matrix, setting it equal to zero. Just curious, how would you go about using, say \lambda_{2} if \lambda_{1} is positive? Do you plug that into the second row, or can you plug it into the first...

Normal modes of oscillation are therefore -\omega_{2}^{2} and (5+\sqrt{10})\omega_{2}^{2}

Got it! For some reason I forgot to properly square the first-order term. Solution should read \lambda_{1,2}=-2\omega_{2}^{2}\pm\sqrt{10}\omega_{2}^{2}

Okay so I tried it over, first by doing the quadratic equation, then writing the result in terms of \omega_{2}^{2} I got \lambda=-2\omega_{2}^{2}\pm\sqrt{\omega_{2}^{2}+3\omega_{2}^{2}} Still though, not sure why the dimensions won't work.

So I solved for \lambda by multiplying that first equation in your reply: \omega_{1}^{2}\omega_{2}^{2}+\omega_{2}^{4}-\omega_{2}^{2}\lambda-\omega_{1}^{2}\lambda-\omega_{2}^{2}\lambda-\lambda^{2}-\omega_{2}^{4}=0 Which simplifies to...

Yeah, but it's just 2 X 1 anyway, so it's not *too* important, I don't think.

Homework Statement So I'm given two horizontal masses coupled by two springs; on the left there is a wall, then a spring with k_{1}, then a mass, then a spring with k_{2}, and finally another mass, not attached to anything on the right. The masses are equal and move to the right with x_{1}...

Ya that's what I said, but then the physics department at my university was all "meh meh meh, that's just how the sequence goes--take diffeqs and linear algebra next semester, after you've already taken waves and vibrations." Struggling super hard.

Sure, but that doesn't take into account the differing positions of the masses. I considered each mass to be a cylinder, along with the rod itself and used the parallel axis theorem to yield I=[1/12 m_1 [3(〖r_w〗^2+r_i^2 )+h^2 ]+mr_l1^2 ]+[1/12 m_2^2 [3(r_w^2+r_i^2 )+h^2 ]+mr_l2^2 ]+[1/12 m_r...

Hey, sorry for not using the exact template; I just have a general question about how to calculate the moment of inertia that I will have to apply to a number of instances. In this particular case, I need to calculate the moment of inertia for a rod pendulum. Of course, I could just use...