Recent content by whozum

So instead of one person relatively secure and one in trouble, we have, two in trouble? I think that's fairly counterproductive..

No, I'm just using this as a tex notepad since no one was helping when I needed it. I'm good with all these. v = \lambda f \lambda = 4L v = 4L f f = \frac{v}{4L}

Prove \left(\frac{ia - 1}{ia+1}\right)^{ib} = exp(-2b cot^{-1} a) and \sum_{n=1}^{N} \cos(2n-1) x = \frac{\sin 2n x}{2\sin x}}

f = \frac{1}{2L}\sqrt{\frac{mg}{\mu}}

\alpha ^{x+n} 2 dy + \left( \alpha ^3 x^2 - 3 \alpha ^n y) dx = 0

iii) (challenging) now consider a wave made from a superposition of modes: \sum_{n=1}^{\infty} a_n\cos(\omega_n t)\sin(\frac{2\pi n}{L} x) calculate the total energy of this wave and comment on your answer. I get the KE and PE each to be 0 by the expression in post 3. We have...

\frac{T}{2} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} a_n a_m\cos(\omega _n t) \cos(\omega_m t) {\left(\frac{2\pi}{L} x\right)}^2 n m \int_0^L \cos \left(\frac{2\pi n}{L} x\right) \cos\left(\frac{2\pi m}{L} x\right) dx

The differential equation modelling this system is given by: m\Ddot{x} + 2\beta \dot{x} + \omega^2 x = 0 With \beta \ and \ \omega defined as \beta = \frac{b}{2m} \ and \ \omega = \sqrt{\frac{k}{m}} with m being mass, k the characteristic constant of the physical system, and b...

Using Newton's 2nd Law for a damped oscillator: ma = -kx - \alpha x which is a second order linear DE. To solve it we use the trial integrating factor e^{\lambda x} [/tex] to come to the root equation mx^2 + \alpha x + k = 0 where we can find our two solutions to be r_{1} \ and...

hmm..67890

You don't have to be mean.

She was pretty upset about that. It's not that bad out. Trib, tonight was your night but you lost it haha. We went to the drive in and almost watched a couple movies. Feel free to take us all out to dinner though o:) Smurf, pretty sure your hand doesn't have an STD. Both hands..

Shes in a hotel for the week, its working out pretty well.. and I say that in the least suggestive way possible. I'm on my way down there in a bit but don't think I'll be able to keep her company all day. We'll see. Any messages for her?

She's loving it, I unfortunately have to deal with it during the summer so I know how treturous it can be. I know that's spelled wrong.

It's a hookah bar/arabic restaraunt. She posted that at about 11:45 so it's not so late, the place is open til about 2ish but most restaurants are open til about midnight or so, unless its really chic.