Recent content by piano.lisa

I don't know. That's why I'm asking.

Homework Statement Consider the infinite square well described by V = 0 if 0<x<a and v = infinity otherwise. At t=0, the particle is definitely in the left half of the well, and described by the wave function, \psi (x,0) = \frac{2}{\sqrt{a}}sin\left \frac{2 \pi x}{a} \right if 0 < x <...

Thank you very much. I ended up solving this earlier today. Thank you for all your help.

I still haven't reached any solution to my problem. Any help is appreciated.

Thank you, The first part I understand. However, the driving force can be at an arbitrary point, which confuses me. The driving force I'm used to dealing with is krcos(\omega t), or more simply, I think cos(\omega t) will suffice. Since it could be at an arbitrary point, I do not...

Just to let everyone know, I still don't have solutions for my above questions. Any help is appreciated, thank you.

Thank you for that part as well. Another student also recognized that dimensional inconsistency, and has asked the professor, but is still waiting for a response.

Thank you so much. We do have the same expansion of k^2. Except, at the point where 0 = \beta u - \beta^2 - \alpha^2, I solved for \beta using the quadratic formula, and you used a more efficient method.

This was my method: k^2 = \frac{\omega^2}{v^2} = \frac{(\alpha + i\beta)^2}{(u - \beta +i\alpha)^2} Then, I expand the brackets, and rationalize the denominator. k^2 = \frac{(\alpha^2 - \beta^2 + 2i\alpha\beta)[(u^2 + \beta^2 - \alpha^2 - 2u\beta) - i(2u\alpha - 2\beta\alpha)]}{[(u^2 +...

Taking into account what you said, I simplified further using the fact that w is complex, and I obtained: \beta = \frac{u \pm \sqrt{u^2 - 4\alpha^2}}{2} Substituting \alpha = \omega - i\beta into the above, I obtained: \beta = \frac{\omega^2}{u + 2i\omega} Which is slightly the same as...

Before I can discuss the driven part... how do I do the first part with the normal modes? At t=0, the mode is n=3... how about some other time? How can I find it?

I think that's what I did... Here, I'll show you: k^2 = \omega^2 / v^2 = \frac{(\alpha + i\beta)^2}{(u + i\omega)^2} = \frac{(\alpha^2 - \beta^2 + 2i\alpha \beta)(u^2 - \omega^2 - 2iu\omega)}{(u^2 - \omega^2 + 2iu\omega)(u^2 - \omega^2 - 2iu\omega)} k^2 = \frac{(u^2 - \omega^2)(\alpha^2 -...

Using (v)(v) to solve, I obtained: \beta = \frac{\alpha}{2u\omega}[(\omega^2 - u^2) \pm (u^2 + \omega^2)] However, \beta is still in terms of \alpha, so I'm not sure what I'm doing wrong. Therefore, I obtained \beta = \frac{\alpha \omega}{u} OR \beta = -\frac{\alpha u}{\omega} Likewise...

Why is it tedious to square the complex numbers? Do I have to do, for example, (v)(v*) ?

Homework Statement Consider an infinitely long continuous string with tension \tau. A mass M is attached to the string at x=0. If a wave train with velocity \frac{\omega}{k} is incident from the left, show that reflection and transmission occur at x=0 and that the coefficients R and T are...