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.
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 above...
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...
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...
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) +...
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...
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...
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
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...
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} ...
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)
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...