Recent content by physgirl

right, so if the box isn't moving (yet), then for vertical, we just have normal force = gravity force (i think?), which is just mg=mg... then for horizontal, we have applied force (Fcos(theta))=frictional opposing force (u*F*cos(theta))... i think i get the confusion. so the normal force...

Homework Statement On a box (of mass "m"), force "F" is applied at "theta" degree angle above the horizontal. The box has static friction coefficient of "u." When this force is applied, and the box remains stationary, what expression describes the frictional force? Homework Equations...

i don't quite get Newton's third law... :( It sounds simple enough, F12 = -F21. However, I'm getting that mixed up with the second law now (ie. if there is a net F on the system, there will be acceleration). For instance, say that there's a box, and the "F" vector is pointing to the right...

anyone? :(

yes and yes. (1/sqrt(2pi))e^(i n phi) is the eigenfunction. and I really meant n^2 when I said m^2 previously... I just expressed my psi(phi, 0) in exponential terms. The psi I gave in the original post left out some constants, but when I do take into account those: psi(phi,0)=sqrt(1/4pi)...

Homework Statement I have to show that the p orbital wavefunctions are orthonormal to eath other in l=1 subspace. Homework Equations The Attempt at a Solution looking at my notes, I thought the expressions for p orb wavefunctions were: Psi_px=sqrt(3/4pi) cos(phi) sin(theta)...

I'm not sure how to express psi (phi, 0) as a sum of energy eigenfunctions in this case...

great, thanks!

oh, and then v is just g(x,y,z)...

As in... u=f(x,y,z)y dv=(dg(x,y,z)/dz) dxdydz So that... du = (df/dx)y + (df/dx)y + f + (df/dz)y v=...? Still not sure :(

Homework Statement If given, for instance, psi(phi, 0)=[1/sqrt(2pi)](cos^2(phi/2) + isin(phi)), which is the wavefunction at t=0, how do you go about finding the wavefunction at time t, psi(phi,t)?? Homework Equations The Attempt at a Solution Would it simply be psi(phi...

Homework Statement I have to show that in 3-d, Lx (angular momentum) is Hermitian. Homework Equations In order to be Hermitian: Integral (f Lx g) = Integral (g Lx* f) Where Lx=(hbar)/i (y d/dz - z d/dy) and f and g are both well behaved functions: f(x,y,z) and g(x,y,z) The Attempt...

I thought for a box particle, really the only significant boundary condition was that the wavefunction has to be continuous. Therefore the cosine term has to disappear to leave with just the sine term that WILL go to 0 when x=0... I'm looking at the solution for the oscillator SE and I just see...

So you would just use the operator H=-(hbar)/(2m)[d^2/dx^2]+(1/2)kx^2 Plug that into the Schrodinger equation... but when do you impose the new boundary condition...?

so it's a harmonic oscillator that has no access to x values less than 0?