Recent content by Phyrrus

Ok, thanks for all your help. How do I find them?

So once I find the prefactors, I am given a 3 set of relations from which I can solve for the 3 constants?

Ok, so then v_{j}=cos(\omega_{0}t) and -v_{i}=sin(\omega_{0}t) And then the constants -\omega_{0}^{2}\rho_{0}=(q/m)*B_{0}

I'm still not entirely sure what I am supposed to do. Is this right? d^2r/dt^2 = -\omega_{0}^{2}\rho_{0}(cos\omega_{0}t i + sin\omega_{0} j) Now we can say that, the above acceleration expression is equal to the Lorentz acceleration of (q/m)*(v_{0}\timesB_{0}? But now how do I split...

I'm not going to lie, I couldn't do it.

I'm sorry, I still don't quite get it

Ok, but how do I calculate v_{0}\timesB_{0} when v_{0}=(w_{0},u_{0}) ? Don't I need to separate w_{0} into x and y components?

The \vartheta was the angle used to separate w_{0} into x and y components. But I'm sorry, can you please elaborate more? Do you mean all I need to do is differentiate r(t) twice and equate coefficients?

Homework Statement Assume the Earth's magnetic field is almost homogeneous with direction along the z-axis, with a small inhomogeneous modification which make the field lines converge towards the z-axis. Also ignore relativistic and gravitational effects. . First assume the magnetic field, B...

1. Homework Statement [/b] Consider the dynamical system \frac{dx}{dt}=rx-\frac{x}{1+x} where r>0 Draw the bifurcation diagram for this system. Homework Equations The Attempt at a Solution Well fixed points occur at x=0,\frac{1}{r}-1 and x=0 is stable for 0<r<1 and unstable for...

thanks mate

Homework Statement Find and compare the transformations of the angle of the velocity vector of a particle and the angle of an inclined stick. The relationship between the two frames is as usual. In frame S' a stick makes an angle of θ' with the x' axis. What is the angle θ measured in the S...

Ahhhh yes, got it, it all works out now. Thanks a lot mate. But just how did you get that? I presume it's quite similar to the first question, chain rule again?

Thanks mate, finally got it in the end. Just a little bit stuck on part b of the question though, have to prove that: \nabla^2f = \frac{2}{r}\frac{df}{dr}+\frac{d^2f}{dr^2} I've been using the vector identities so that I get: = \frac{1}{r}\frac{df}{dr}(\nabla\bulletr)...

Sorry, I made a mistake in the OP, it's supposed to be \nablaf = \frac{1}{r}\frac{df}{dr}r Thanks for your reply, but I'm not really sure where the r=√(r^2) come into it? How do you arrive at the 1/r out the front? The only thing I can think of is that it's supposed to reduced it to a...