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Homework Help: Elastic pendulum - Lagrangian approach

  1. Sep 26, 2010 #1
    1. The problem statement, all variables and given/known data

    A spring of rest length [tex]L_0[/tex] (no tension) is connected to a support at one end and has a mass [tex]M[/tex] attached at the other. Neglect the mass of the spring, the dimension of the mass [tex]M[/tex], and assume that the motion is confined to a vertical plane. Also, assume that the spring only stretches without bending but it can swing in the plane.

    (a) Using the angular displacement of the mass from the vertical and the length that the spring has stretched from its rest length (hanging with the mass [tex]M[/tex]), solve Lagrange's equations for small stretching and angular displacement.

    (b) Solve Lagrange's equations to the next order in both stretching and angular displacement. This is still amenable to hand calculations. Using some reasonable assumptions about the spring constant, the mass, and the rest length, discuss the motion. Is resonance likely under the assumptions stated?

    2. Relevant equations

    Formula for the Lagrangian; Euler-Lagrange equations.

    3. The attempt at a solution

    The Lagrangian is:

    [tex]\mathcal{L} = T - V = \frac{M}{2}(\dot{r}^2 + r^2\dot{\phi}^2) - (\frac{k}{2} x^2 - Mgr\cos\phi)[/tex]

    where [tex]r = L_0 + x[/tex] (i.e. [tex]r[/tex] is the stretched length of the spring, and [tex]x[/tex] is the deviation from the rest length). [tex]\phi[/tex] is the angular displacement where zero is hanging straight down.

    The Euler-Lagrange equations in [tex]x[/tex] and [tex]\phi[/tex] are:

    [tex]\frac{\partial \mathcal{L}}{\partial x} - \frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{\partial \mathcal{L}}{\partial \dot{x}}\right) = M(L_0 + x)\dot{\phi}^2 - (kx - Mg\cos\phi) - M\ddot{x} = 0[/tex]

    [tex]\frac{\partial \mathcal{L}}{\partial \phi} - \frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{\partial \mathcal{L}}{\partial \dot{\phi}}\right) = - Mg(L_0 + x)\sin\phi - M((L_0 + x)^2\ddot{\phi} + 2(L_0 + x)\dot{x}\dot{\phi}) = 0[/tex]

    My guess at approximating for small angular displacements is to set [tex]\sin\phi = \phi[/tex] and [tex]\cos\phi = 1[/tex], but I don't know whether this should be before or after differentiating the Lagrangian, and I don't know the right way to approximate for small stretching.
     
    Last edited: Sep 27, 2010
  2. jcsd
  3. Sep 27, 2010 #2
    Small angle approximations must be applied after making derivatives of the lagrange equations. Remember that in this approx:

    [tex]
    \dot{\theta}=0
    [/tex]

    (In notation you have used, [tex]
    \dot{\phi}=0
    [/tex]
     
  4. Sep 27, 2010 #3
    But then must not [tex]\ddot{\phi} = 0[/tex]? If so, the second equation becomes [tex]Mg(L_0 + x)\phi = 0[/tex], which means [tex]\phi = 0[/tex]; a bit excessive for a small angle approximation.

    On the other hand, leaving [tex]\ddot{\phi}[/tex] alone, [tex]x[/tex] drops out completely from the second equation and (using [tex]x \ll L_0[/tex]) I get independent equations for [tex]x[/tex] and [tex]\phi[/tex] (equivalent to F=ma for a hanging spring and a swinging pendulum respectively).

    [tex]\ddot{x} = - g - \frac{k}{M}x[/tex]

    [tex]\ddot{\phi} = - \frac{g}{L_0}\phi[/tex]
     
    Last edited: Sep 27, 2010
  5. Sep 27, 2010 #4

    vela

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    Are you sure your derivative with respect to x is correct? I think you should get a term linear in x when you differentiate the kinetic energy.
     
  6. Sep 27, 2010 #5
    Aha you're right; I missed a power of 2. Silly.

    I'll edit the equations to make sense. Everything looks good now.
     
  7. Sep 28, 2010 #6
    After checking my books and exercises I think the best way is:

    First of all calculate the equilibrium angle ([tex] \ddot{\phi}=\dot{\phi}=0} [/tex] You'll find an angle [tex] \phi_{0}[/tex] (remember that you'll get a constant.

    Then you can find the small angle frecuency by doing an approximation:

    [tex] \phi=\phi_{0}+\epsilon=0} [/tex] where [tex]\epsilon[/tex] is the angle for small vibrations.

    Use this in the equation for the angle in order to find an equation for [tex]\epsilon[/tex].
    You can also check Marion's texbook (chapter 7).
     
    Last edited: Sep 28, 2010
  8. Sep 28, 2010 #7
    I'm not sure I understand. From your equation, [tex]\phi = \epsilon = 0[/tex] (since [tex]\phi_0 = 0[/tex] clearly satisfies [tex] \ddot{\phi}=\dot{\phi}=0} [/tex]).

    Is something wrong with the result I got above?
     
  9. Sep 28, 2010 #8

    vela

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    It's been a long time since I took classical mechanics, so this might be a misguided suggestion. Are you familiar with diagonalizing the Lagrangian? If the technique applies, I seem to recall it made solving this type of problem pretty straightforward.
     
  10. Sep 30, 2010 #9
    Be careful: Your [tex] L_0[/tex] is not the [tex] L_0[/tex] given in the question.
    You have to use the rest length with a mass hanging in a gravitational field which will be slightly larger.
     
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