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Potential Energy of Anisotopic Oscillator

  1. Feb 24, 2008 #1
    1. The problem statement, all variables and given/known data
    1. The problem statement, all variables and given/known data[/b]
    There is a mass shown between two springs,
    ie. [--------O----------],
    on a horizontal frictionless table. Both springs are identical, with force constant k and unstretched length L. At equilibrium the mass rests at the origin, and the distances "a" (both are at an initial distance "a" from the mass) are not necessarily equal to L. (springs may initially be stretched or compressed). show that when mass goes to a position (x,y) with x and y small, the potential energy has the form:
    U = 0.5 * (k[tex]_{x}[/tex] * x[tex]^{2}[/tex] + k[tex]_{x}[/tex] * y[tex]^{2}[/tex]) (potential of an anisotropic oscillator)
    the k^x and k^y are supposed to be subscripts...

    show that if "a" < L the equilibrium is unstable and explain why


    2. Relevant equations
    U = 0.5 * (k[tex]_{x}[/tex] * x[tex]^{2}[/tex] + k[tex]_{x}[/tex] * y[tex]^{2}[/tex]) (potential of an anisotropic oscillator)
    U= -int (F dot dr)


    3. The attempt at a solution
    I wasn't able to make it very far in my attempt, but i am also not asking for a complete solution, just a push in the right direction:
    i figured at the beginning U=0, and Fnet =0.
    With Fspring1= -k(a-L), (spring on the right)
    and Fspring2 = -k(-(a-L))=k(a-L) (spring on the left)
    Fspring1=-Fspring2 initially

    if the spring were only stretched in the x direction a distance "b" to the right:
    Fspring1=-k(b-(a-L))
    since the spring was already stretched a distance a-L, (not sure if it should be b+(a-L))
    Fspring2=-k(b-(a-L))
    since the initial stretched distance is -(a-L), and adding a positive stretch of "b"
    therefore
    Fnet=Fspring1 + Fspring 2 = -2k*b
    therefore U = integral (-k*b) dx = -k*b[tex]^{2}[/tex] , which i figure must be wrong because shouldn't it be U = 0.5 * k * b[tex]^{2}[/tex] since y is zero in this case?

    then things get even more complicated when i try to go in the y direction, i figure gravity can be neglected due to the fact of "small y"
    i wasn't sure if somehow maybe i would be able to due the case in the x-direction and just add it to the y-direction solution by superposition, but i really doubt this is the case.

    If the mass was lifted directly upwards:
    the length of the spring would have a new length of "m" = [tex]\frac{a}{cos\phi}[/tex]
    where [tex]\phi[/tex] is the angle the spring with respect to the horizontal...

    i figured the forces would be:
    Fspring1 (x) = -k(m-L)*cos[tex]\phi[/tex]
    Fspring1 (y) =-k(m-L)*sin[tex]\phi[/tex]
    and the same for spring 2... as you can see i am just rambling with random thoughts i had trying to come up with a way to solve this. I am fairly lost, i don't need a complete solution but any helpful hints would be appreciated.
     
  2. jcsd
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