# Oscillation Period

## Homework Statement

Assume that the potential is symmetric with respect to zero and the system has amplitude ##a##, show that the period is given by : ##T=\sqrt{8m}\int^a_0\frac{dx}{\sqrt{V(a)-V(x)}}.##

## Homework Equations

##E = \frac12 m(\frac{dx}{dt})^2+V(x)##

## The Attempt at a Solution

For a particle, I know that at ##t=0## if we release it from rest at position ##x=a## we then have ##\frac{dx}{dt}=0## at ##t=0## and thus ##E=V(a)##. So when the particle reaches the origin for the first time it has gone through one quarter of a period of the oscillator. Thus, I have to integrate with respect to t from ##0## to ##\frac{T}{4}## and rearrange the equation ##E## for ##\frac{dx}{dt}##. But from here I am not sure how to set it up properly to get ##T=\sqrt{8m}\int^a_0\frac{dx}{\sqrt{V(a)-V(x)}}.##

If you substitute your value of E in Hamilton's equation, can you separate the variables (functions of x on one side, functions of t on the other)?

If you substitute your value of E in Hamilton's equation, can you separate the variables (functions of x on one side, functions of t on the other)?

I have ##V(a) -V(x) = \frac12m(\frac{dx}{dt}2) \implies 2\sqrt{V(a)-V(x)} = m\frac{dx}{dt} \implies \implies \sqrt{\frac{2}{m}}dt = \frac{dx}{\sqrt{V(a)-V(x)}}## but I am confused on how they to get ##T##.

What about integrating both sides? You know the limits of the integrals (for t and x) from your first post, so what do you get?

When I am integrate I am not sure how they got the ##\sqrt{8m}##.

What do you get when you integrate dt from 0 to T/4?

What do you get when you integrate dt from 0 to T/4?

We will get ##T/4## if we integrate dt from 0 to T/4.

Ok, then what is (T/4)⋅(2/m)1/2?
This is just algebra...

Ok, then what is (T/4)⋅(2/m)1/2?
This is just algebra...
Oh wow.. how dumb am I. Thank you very much!

juliocezario30
Can you show step by step procedure? I am confused

haruspex