I found the equations of motion as
##m\frac{\mathrm{d}^2x_1 }{\mathrm{d} t^2} = -\frac{mg}{l}x_1 + k(x_2-x_1)##
and
##m\frac{\mathrm{d}^2x_2 }{\mathrm{d} t^2} = -\frac{mg}{l}x_2 + k(x_1-x_2)##
I think the k matrix might be
##\begin{bmatrix}
mg/l + k & -k \\
-k & mg/l + k
\end{bmatrix}##...
Oh god of course, thank you.
So ##\frac{\partial r}{\partial x} = \frac{x}{r}##
And then from that I can show
##\frac{\partial f}{\partial x} = \frac{\partial f}{\partial r} \frac{\partial r}{\partial x} = \frac{df}{dr}\frac{x}{r}##
Im guessing I just use the chain rule for the next part?
Sorry...
Okay so I get
r=√x2+y2+z2r=x2+y2+z2
##\theta = \cos^{-1} (\frac {z} {\sqrt{x^2 + y^2 + z^2})##
ϕ=tan−1(yx)ϕ=tan−1(yx)
Am I right in thinking this makes
∂r∂x=sinθcosϕ∂r∂x=sinθcosϕ?
Ive found ##\delta x/\delta r## as ##sin\theta cos\phi##
##\delta r/\delta x## as ##csc\theta sec\phi##
But unsure how to do the second part? Chain rule seems to give r/x not x/r?
$$
Yes, Vela was right. Sorry I'm new here and had no idea I could use LaTeX.
To get the amplitude and phase would I have to move it into the form ##x = Cos(\omega t - \phi)##?
I found the steady state solution as
F_0(mw_0^2 - w^2m)Coswt/(mwy)^2 + (mw_0^2 -w^2m)^2
+ F_0mwySinwt/(mwy)^2 + (mw_0^2 -w^2m)^2
But I'm not sure how to sketch the amplitude and phase? Do I need any extra equations?
Right, so I thought I'd done this correctly but clearly not because my velocity is greater than the speed of light, where have I gone wrong?
P = (M, 0, 0, 0)
p1 = (E1, p1x, p1y, p1z)
p2 = (E2, p2x, p2y, p2z)
P = p1 + p2
p2 = P - p1
square each side
to get (p2)2 = P2 - 2Pp1 + p12
therefore
(m2)2...
Ah okay, I see, thank you! I've used four vectors and then the invariant to get E = 3.12GeV. I think I've got the second part too, using conservation of energy and time dilation. Thank you! :)