Hi,
sorry for the delay.
In q.1 - to get the units, go back to the basics. For example, energy is force times distance, and force is acceleration times mass. From that you can get to the mass. hbar is energy times time, or distance times momentum - you can use those. To get the units of K...
That's right - intuitively we can expect that the time will go to infinity as we take the velocity to zero, and it seems that this is what happens.
But since the question is from a Putnam final contest, I guess that the person who wrote it knew what he was doing.
What is also hard for me...
Hi,
If I understand what you mean by your question (to take a small displacement approximation and linearize the problem), please note that the pendulum starts from *upward* position. This isn't a small amplitude oscillation problem.
If I didn't do it wrong, I think it's supposed to be -2j.
What this means is that this field was created by a current density which is uniform all throughout space its direction is the y axis (or the -y) and its magnitude is 2T_0.
Obviously this kind of current density doesn't exist in...
part 1: This is technical.
In part 2, what is u?
In part 3, think about how to use the uncertainty principle. In the ground state, the kinetic energy and the potential energy can be assumed to be approximately equal. You can also assume that the uncertainty relation holds and that you have...
Well, to be honest I'm not sure how I should use this assumption.
I thought about including it by adding a constant term to the energy, which changes the integral to having:
a + 2g(1-cos\theta)
instead of
2g(1-cos\theta)
in the denominator. But this evaluates to a hypergeometric function...
The question is about the *instant* the switch is opened/closed.
At the instant the switch is closed, the current on the S coil is zero (why)? Therefore, the current on the P coil is zero as well. At the instant the switch is opened, there is a current on the P coil, so it's bigger.
Well, if something is not a magnetic field, it's not a magnetostatic field either. And yes, that's the correct equation. In the magnetostatic case there are no time-varying fields, so you just need to find j.
Hint: A magnetostatic field is a solution to one of the Maxwell equations in the case where everything is not dependent on time. Which equation is it?
Also, obviously, in order for something to be magnetostatic, it has to first be a magnetic field.
Your arguments about determining which...
Homework Statement
Hi,
this is supposed to be an easy question, but for some reason I can't get it to work. The question is:
A weightless rod is hinged at O so that it can rotate without friction in a vertical plane. A mass m is attached to the end of the rod A, which is balanced...
the invariant quantities in Galilean relativity are quantities that don't change when viewed from frames moving at constant velocities relative to each other, where the transformation rules are the ones given by the Galileo transformation. Try performing a general transformation on each of the...
just before the motor is turned off, the energy in the system is the sum of: elevator K, counterweight K, elevator U, counterweight U, and the rotational kinetic energy of the pulley. After the elevator comes to rest, the energy in the system is the sum of elevator U and counterweight U after...
The assumption is that the line generates a certain potential (which you should have/be able to calculate) throughout space and that the proton moves inside that potential. You are not supposed to take into account the potential that the proton creates when you work out its equations of motion...
I think that by scattering state they simply mean that it's a state that isn't bound - for instance, if E>V_F (the potential at F). Because then the idea is that the particle is scattered by the potential.
Thanks, and I'm sorry if I wrote more than I should have, but what I'm saying is that I have a counterexample to what he's trying to show, and that seems really weird to me. What am I missing here?
umm... It doesn't seem like your solution satisfies the initial condition: y(0)=0. Other than that, It's just a parabola, so there's no asymptotic behavior - it simply diverges...
My attempt was: We know that: p(A = a1) = |<a1 | \psi>|^2 and p(B = b2) = |<b2 | \psi>|^2. Also, the probability of measuring a1 and then b2 is: p(a1,b2) = |<b2 | a1><a1 | \psi>|^2, and similarly: p(b2,a1) = |<a1 | b2><b2 | \psi>|^2. However, what I don't understand is what does the fact that...
Magnetic charges have never been found, and for all we know do not exist. (Although there have been many attempts to find them.) This is stipulated in one of the Maxwell equations: divB = 0.
There is nothing to calculate. Since the law says that: \vec{dB} = C\vec{dl} \times \hat{r} , where C is a constant that contains all the relevant factors, we can see that dB will always be perpnedicular to both dl and r. In this case they both lie on the plane P, so dB is always normal to P.
1. Homework Statement [/b]
There are N 3-dimensional quantum harmonic oscillators, so the energy for each one is:
E_i = \hbar \omega (\frac{1}{2} + n_x^i + n_y^i + n_z^i). What is the total number of states from energy E_0 to E, and what is the density of states for E?
The Attempt at a...