Recent content by WJSwanson

That wouldn't be a typo. That's the result of evaluating the integral at x = lambda. :)

Okay, so part b.) was actually pretty easy and straightforward. I did it in like 3 minutes. I multiplied both sides by cos(\frac{2\pi m x}{\lambda} to get F(x) cos(\frac{2\pi m x}{\lambda}) = \Sigma_{n=0}^{\infty} A_m cos(\frac{2\pi m x}{\lambda}) cos(\frac{2\pi n x}{\lambda}) and from...

So that would mean I have F(x) = A_0 cos(\frac{2\pi n x}{\lambda}) = A_0 cos(0) = A_0 which in turn means that \int^{\lambda}{0} F(x)dx = \int^{\lambda}_0 A_0 dx = \lambda A_0 which then gives \lambda A_0 = \int^{\lambda}_0 F(x)dx \Rightarrow A_0 = \frac{1}{\lambda}...

How would the cosine terms be zero if they're cos(2πnx)? Wouldn't that require that x is always equal to 1/4 + some integer m, or something along those lines? I don't know how that would work unless there was some phase angle introduced to the argument, for n = 0. Right, sin(2πn) is 0 for...

I'm afraid I don't quite understand what you mean. I should replace F with an arbitrary function L?

Homework Statement For an even function, the Fourier series takes the form ^{\infty}_{n=0}\Sigma A_n cos(\frac{2\pi n x}{\lambda}) where \lambda is the wavelength of the function. In this problem you will see how to find the Fourier coefficients A_n. a.) Prove that A_0 =...

Homework Statement The RC circuit pictured below begins with capacitor C completely uncharged. I: What is the current at point A immediately after the switch closes? II:What is the current at point A 'at the end of the quaternary period' (I lol'd when he worded it like that) after the switch...

The short (and I'm sure dreadfully oversimplified) answer is basically just that they're called "fictitious" forces because they're the effects of reference frames that are moving/accelerating relative to each other.

Okay, I ended up figuring out part II. Since we know that \Delta\omega = 2\pi\Delta f = 6.28 * 10^{3} s^{-1} we infer that \Delta t \geq \frac{1}{2\Delta\omega} and since the probe sends the message at a rate of 2\Delta t = \frac{t}{2.0 * 10^{6}} we know that \Delta t =...

If your surface is simply defined as \theta = \pi/4 then all you need to do is solve your conversion factor from theta of \theta = cos^{-1}(\frac{z}{r}) = cos^{-1}(\frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}) Since your value of theta is a constant, you just have...

Homework Statement I: A telephone line can transmit a range of frequencies \Delta f = 2500 Hz. Roughly what is the duration of the shortest pulse that can be sent over this line? II: A space probe sends a picture containing 500 by 500 elements, each containing a brightness scale with 256...

So the moral of the story is that it's highly important to take into account your equilibrium conditions and how the interactions occur relative to the equilibrium conditions. It's also important for people who answer questions on PF not to get ahead of themselves and forget (like I did here) to...

Oh, here's the issue. Sorry I didn't check more closely before. I think your limits of integration were off. It should be W = \int^{\theta = 4\pi/9}_{\theta = \pi/6} -10\theta d\theta because the equilibrium position of the system is at position 1, where \theta = 0. You want to find the...

Looks right to me.

Essentially, yes. Basically, any non-ideal DC power source will operate similarly to an ideal DC power source except that it has an internal resistance. How might you apply that knowledge to the problem at hand?