Potential energy equilibrium positions

AI Thread Summary
The discussion revolves around finding equilibrium positions for a particle with the potential energy function U(x) = x + sin((2 rad/m) x) within the range of 0 to π meters. The first derivative of U(x) was calculated and set to zero, leading to the solutions x = π/3 and x = 2π/3, with the latter being overlooked initially. The second derivative test indicated that x = π/3 corresponds to an unstable equilibrium due to a negative value, while the stability of x = 2π/3 was not initially assessed. Participants emphasized the importance of considering all possible solutions to the cosine equation to identify all equilibrium points. The discussion concluded with a reminder to check for additional values within the specified range.
frostking
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Homework Statement


A particle has potential energy

U(x) = x + sin ((2 rad/m) x)

over the range of x greater or equal to 0 meters and less than or equal to pi meters
Where are the equilibrium positions in this range and for each is it a point of stable or unstable equilibrium?

Homework Equations


derv of u(x) set = to 0 and then the second derv of u(x)


The Attempt at a Solution




I solved derv of u(x) = 1 + 2 cos(2x) then set = to 0

so 2 cos(2x) = -1 divide by 2 and cos(2x) = -1/2
2x = arch cos of (-1/2) = 2pi/3 rad

x = 1 pi/3 rad

I get this part but the answer says that x can = 2pi/3 rad as well and I do not understand why.

To determine if equilibrium is unstable or stable I took the second derv and at x = pi/3 and second derv of - 4 sin(2x) I got less than 0 so a maximum and therefore unstable equilibrium

Can someone please help me understand why I should have known to consider 2pi /3? Thanks for your efforts, Frostking
 
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There are an infinite number of solutions to \cos(2x) = -1/2. You just found one - there's another solution which will give you 2\pi/3 as the final answer.
 
Thanks, yes I should have realized I needed to check for other values less than or equal to pi!
 
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