Recent content by physconomics

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    Two pendula connected by a spring - normal modes

    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}##...
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    Equation of Motion of a Particle acted on by a retarding force

    I really can't figure out where to even start on this question
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    Is the Chain Rule Applied to Spherical Polar Coordinates Different?

    Is ##\frac{\partial}{\partial x} \frac{x}{r} = \frac{r - \frac{x^2}{r}}{r^2}##
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    Is the Chain Rule Applied to Spherical Polar Coordinates Different?

    Right so I'm a bit stuck, I'm getting ##\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}(\frac{df}{dr}\frac{x}{r}) = \frac{\partial}{\partial x}(\frac{df}{dr}) \frac{x}{r} + \frac{df}{dr}\frac{1}{r} ## where ##\frac{\partial}{\partial x}(\frac{df}{dr}) = \frac{\partial}{\partial...
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    Is the Chain Rule Applied to Spherical Polar Coordinates Different?

    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...
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    Is the Chain Rule Applied to Spherical Polar Coordinates Different?

    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=sin⁡θcos⁡ϕ?
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    Is the Chain Rule Applied to Spherical Polar Coordinates Different?

    Okay, so how else would I get ##\partial x/\partial r##?
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    Is the Chain Rule Applied to Spherical Polar Coordinates Different?

    Sorry, I meant ##\frac{\partial x}{\partial r}##
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    Is the Chain Rule Applied to Spherical Polar Coordinates Different?

    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?
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    Damped Oscillator and Oscillatory Driving Force

    $$ 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)##?
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    Damped Oscillator and Oscillatory Driving Force

    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?
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    Particle decaying into two daughter particles - special relativity

    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...
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    Electron and positron collision producing a b0 meson pair

    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! :)
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    Electron and positron collision producing a b0 meson pair

    I'm confused doesn't the question say they're created at the threshold energy?
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