# Recent content by rmfw

1. ### Intensity of natural light after polarization

Thanks, this made it clear.
2. ### Intensity of natural light after polarization

I read that if a beam of unpolarized light goes through a polarizer, the intensity of the polarized beam is equal to half the intensity of the original beam. Can someone explain me why? I thought the intensity would be the same.
3. ### Theory of small oscilations

Homework Statement (see pic) \frac{l_{2}}{l_{1}} = \frac{1}{4} \frac{g}{l_{1}} = 1 I need to find the normal modes of oscillation. (for small oscillations) The Attempt at a Solution I solved the problem using the matricial way and got the following matrix: (I simplified it using the...
4. ### Classical mechanics

Ayo everybody, I'm doing a problem about theory of small oscilatons (see pic) and I got the following for potential energy: V= mg(\frac{l_{2}}{2} +\frac{l_{1}}{2} \theta^{2}_{1} + \frac{l_{2}}{4} \theta^{2}_{2}) (after the aproximation cos \theta ~ 1 - \frac{\theta^{2}}{2} Knowing that V...
5. ### Transmission of light and velocity of light

Hi ! I think part of the wave energy is transferred to the air particles and as Compton showed the wave length of the wave is reduced after that energy is transferred, and by the formula v= λ * ƒ it's clear why the velocity is lower than in vacuum. I'm not sure if this is the right...
6. ### Determine the wave function of the particle

Hi Simon Bridge, thank you for your reply. I did write psiII as a sum of complex exponentials and I got the following: \frac{B}{A} = \frac{\frac{k}{l}e^(ika-ila)-\frac{k}{l}e^(-ika + ila) - e^(-ika -ila) - e^(-ika + ila)}{\frac{k}{l}e^(ika-ila) - \frac{k}{l}e^(ika+ila) +e^(ika+ila) +...
7. ### Determine the wave function of the particle

Homework Statement A particle coming from +∞ with energy E colides with a potential of the form: V = ∞ , x<0 (III) V = -V0 , 0<x<a (II) V = 0, x>a (I) a) Determine the wave function of the particle considering that the amplitude of the incident wave is A. Writting the amplitude of the...
8. ### Infinite potential

Potential and wave function If a particle comes from +infinite and collides with a potential of the form : V = ∞ , x < 0 (I) -V0 , 0<x<a (II) 0 , x≥a (III) Is the wave function for region (I) = 0? And for region (II) = A sin(kx) with k constant? Really need to...
9. ### Cosmic simulation software

Thanks, I'v installed Universe Sandbox and it's really cool .
10. ### Cosmic simulation software

Hey, does anyone know what software is used on the astrophysics videos made by NASA? Or similar modeling software where I can play around a bit creating solar systems and stuff like that? Thanks
11. ### Bar suspended by rope kinetic energy

Ok I understood that, but my question still remains, (see pic attached in this post) in this case, I know the answer for kinetic energy is T = \frac{1}{2} \dot{\theta}^2 \frac{m L^2}{3} = \dot{\theta}^2 \frac{m L^2}{6} But, according to this statement "The total kinetic energy is given by...
12. ### Bar suspended by rope kinetic energy

Ok I already know the answer is I = \frac{m L ^2}{12} . But there's something else bothering me, if there were no rope, only the bar fixed at its beginning (origin) and oscillating in a 2D plane, then the kinetic energy would be only in rotational motion calculated using I = \frac{m L ^2}{3} ...
13. ### Bar suspended by rope kinetic energy

I have added a picture to the original post. Ok so T = \frac{ m v_{centerofmass}^2}{2} + \frac{I w^2}{2} So now I still don't know if I should use I = \frac {m L^2}{12} (rotation about the center of mass) or I = \frac {m L^2}{3} (rotation at the beginning of the bar)
14. ### Bar suspended by rope kinetic energy

Also sorry for that doublepost, didn't meant to.
15. ### Bar suspended by rope kinetic energy

If I needed to calculate the kinetic energy of a bar suspended by a rope at one of its ends, in other words, a bar pendulum, do I need to calculate the kinetic energy of the center of mass plus the kinetic energy relatively to the center of mass or do I need to calculate the kinetic energy of...