Time dependent position given position dependent force

scoopaloop
Messages
12
Reaction score
0

Homework Statement


A particle of mass m is subject to a force F(x) = −kx^−2 (1) that attracts it toward the origin. (a) Determine the potential energy function U(x), defined by F(x) = − d U(x)/dx. (b) Assuming that the particle is released from rest at a position x0, show that the time t required for the particle to reach the origin is t = π sqrt(m/8k)(x_0)^(3/2)

Homework Equations


dt=dx/sqrt(2(E-U)/m)

The Attempt at a Solution


So, I found The potential energy to be k/x using that and the fact that at v=0 at x_0, I get dt=dx/sqrt(2((k/x_0)-k/x)/m). My only problem is integrating this I get a long nasty function that I feel I can't get x isolated. Maybe, I'm being lazy and need to gut through it, but is there an easier way to approach this?
 
Physics news on Phys.org
scoopaloop said:
My only problem is integrating this I get a long nasty function that I feel I can't get x isolated.
If we are to have any chance of helping you, you need to actually show us what you did.
 
I just want to know if this is the correct approach or if there is another way. I'm not asking you to do it for me. I'd imagine you either know how to do this problem or you don't, I don't see how writing some long pain in the butt equation will help you know how to do the problem.
 
Last edited:
The approach is fine, but you seem to be doing it wrong. You can easily factor out the dependence on ##x_0## by a change of variables and obtain a dimensionless integral.

You can also argue for the correct dependence of the time on ##x_0##, ##k##, and ##m## purely on dimensional grounds. There is only one combination of these parameters which results in a time. Of course, you still need to perform the integral to get the correct prefactor.
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top