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robertjordan
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Homework Statement
A particle oscillates with amplitude A in a one-dimensional potential that is symmetric about x=0. Meaning U(x)=U(-x)
First find velocity at displacement x in terms of U(A), U(x), and m.
Then show that the period is given by ##4\sqrt{\frac{m}{2U(A)}}\int_0^A \frac{1}{1-\frac{U(x)}{U(A)}}dx##
(hint: ##dt=\frac{dx}{v}##)
Homework Equations
Total energy = ##U(x) + \frac{1}{2}mv^2##
Period = time it takes to complete an oscillation
I found velocity by noting that at x=A, all the energy is potential, so total energy of the system (which doesn't change) is U(A), so we can set up the equation:
##U(A)=U(x)+\frac{1}{2}mv^2## and solving for v we get:
##v=x'=\sqrt{\frac{2(U(A)-U(x))}{m}}##.
The Attempt at a Solution
Now I was thinking we could multiply the velocity at x=0 by Δt, then do v(0)Δt+v(Δt)Δt, and then v(0)Δt+v(Δt)Δt+v(2Δt)Δt and so on and so forth... this will help us find how far the particle has traveled at any time t. But obviously the right way to do the above process is by integrating our ##v## equation with respect to time, which will give us distance traveled over that time which will let us know how long it takes to travel A distance (1/4 of an oscillation).
##\int_0^t v dt = \int_0^t \sqrt{\frac{2(U(A)-U(x))}{m}}dt##
But we know ##\frac{dx}{dt}=v## so we can write ##dt=\frac{dx}{v}##...
umm... I'm stuck...
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