- #1

gionole

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- Homework Statement
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- Relevant Equations
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We have ##U(x) = U(a) + \frac{1}{2}U''(a)(x-a)^2## (by taylor series)

It's known that ##U'(a) = 0## and point ##a## is a turning point, hence at that point, Kinetic energy is 0, and ##E = U(a)##, hence:

We have ##U(x) = E + \frac{1}{2}U''(a)(x-a)^2##I need to get equation of motion and I want to use the following.

##t = \sqrt{\frac{1}{2}m} \int \frac{dx}{E - U(x)}##

##t = \sqrt{\frac{1}{2}m} \int \frac{dx}{E - E - \frac{1}{2}U''(a)(x-a)^2}##

##t = \sqrt{\frac{1}{2}m} \int \frac{dx}{ - \frac{1}{2}U''(a)(x-a)^2}##

I couldn't solve this. The book proceeds to mention that solution is:

##x(t) = a + se^{\pm \lambda t}## where ##s = x(0) - a## and ##\lambda^2 = -\frac{U''(a)}{m}##

Attaching the image.

It's known that ##U'(a) = 0## and point ##a## is a turning point, hence at that point, Kinetic energy is 0, and ##E = U(a)##, hence:

We have ##U(x) = E + \frac{1}{2}U''(a)(x-a)^2##I need to get equation of motion and I want to use the following.

##t = \sqrt{\frac{1}{2}m} \int \frac{dx}{E - U(x)}##

##t = \sqrt{\frac{1}{2}m} \int \frac{dx}{E - E - \frac{1}{2}U''(a)(x-a)^2}##

##t = \sqrt{\frac{1}{2}m} \int \frac{dx}{ - \frac{1}{2}U''(a)(x-a)^2}##

I couldn't solve this. The book proceeds to mention that solution is:

##x(t) = a + se^{\pm \lambda t}## where ##s = x(0) - a## and ##\lambda^2 = -\frac{U''(a)}{m}##

Attaching the image.