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CAF123
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Homework Statement
Consider the particle moving in the potential $$V = \frac{1}{2} mw^2(x^2-\ell^2)\,\,\text{for}\,\,|x| < \ell\,\,\text{and}\,\,0\,\,\text{for}\,\,|x| \geq \ell$$
The particle starts moving to the right at ##x = -\ell## with positive velocity ##v##. How long will it take the particle to reach x=l in the cases v=0 and w=0? For the general case, evaluate the period by using a substitution of the form x = Asinθ, for some suitably chosen A such that $$t = \frac{2}{w}\arcsin\left[\frac{\ell w}{(v^2 + w^2 \ell^2)^{1/2}}\right]$$ Recover the result for v=0.
The Attempt at a Solution
The potential is parabolic with a minimum of V and intersecting x at ±l. The period t derived was $$t = \left(\frac{m}{2}\right)^{1/2} \int_a^b \frac{dx}{(E-V)^{1/2}},$$ with a<b. If v=0, then the particle was released with no kinetic energy T and so its energy is all potential. Since F conservative, E=V always. This makes the denominator 0 and so t tends to infinity (i.e the particle never reaches l). If w=0, V(x) = 0 and so E=T which gives a period of t = ##\sqrt{2m/T}\ell##
In general, I reduced the expression for t down to $$t =\sqrt{2} \sqrt{\frac{m}{2}} \int_a^b \frac{dx}{\sqrt{2E - mw^2(x^2-\ell^2)}}$$ and then let ##x =(l/\sqrt{2}) \sin \theta##. This gives $$\sqrt{m} \int_{\theta_1}^{\theta_2} \frac{\ell \cos \theta d \theta}{\sqrt{2E + \frac{1}{2} m w^2\ell^2 \cos^2 \theta}}$$ but I can't see how to progress with this at the moment.
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