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Two Pendulums connected by a spring

  1. Mar 18, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Identical pendulums of length ##L## are suspended from the ceiling a distance ##a## apart. Each of the pendulums carries a particle of mass ##m##. The two particles are connected horizontally by a spring of natural length ##a##. Consider only motion in the plane formed by the pendulums and the spring and assume that the displacments of the pendulums from the vertical, ##x_1, x_2## are small. (i.e the angles of deflection are small)

    1)Express the total energy as a sum of kinetic and potential terms using the variables ##x_1, x_2## and their derivatives (NB: Include the potential energy stored in a spring and gravity)

    2)Bring the energy into the form $$E = \frac{1}{2}\underline{\dot{x}^T} M \underline{\dot{x}} + \frac{1}{2}\underline{\dot{x}^T} K \underline{\dot{x}},$$ M, K matrices.

    3. The attempt at a solution

    1) I dealt with each pendulum separately. They are joined only via the spring and so this will come across in the changing elastic potential energy stored in the spring:
    Define coordinates where the pendulum is attached to the ceiling with positive x to left and positive y downwards. Let ##L = l##Then for this pendulum:
    $$V_1 = -mgy_1 = -mg(\sqrt{l^2 - x_1^2}),\,\,\,T_1 = \frac{m}{2}\left[\frac{\dot{x_1}^2 l^2}{l^2 - x_1^2}\right]$$

    For the other pendulum, again put x +ve to right and y postive down and origin at ceiling: Then $$V_2 = =mg\sqrt{l^2 - x_2^2}\,\,T_2 = \frac{m}{2}\left[\frac{\dot{x_2^2} l^2}{l^2 - x_2^2}\right]$$

    For the spring, I think the expression will change depending on whether the springs move in phase or not. If one is displaced to the right by x1 and the other to the left by x2 then ##U_S = \frac{1}{2}k(x_1 - x_2 - a)^2##

    I can then add all these together. I haven't used the fact that the displacements are small. I thought about neglecting sqaured terms etc.. but then all the kinetic terms would drop out. I am not sure about my expression for ##U_S## either since I can come up with another expression depending on the relative motion of the two pendula.

    Many thanks.
     
  2. jcsd
  3. Mar 18, 2013 #2
    It seems that the x1, x2 notation is not completely consistent. In the pendula' potential energy terms, x1 = 0 and x2 = 0 seem to denote the equilibrium position. But then the length of the string is x1 - x2 + a, yielding the extension as simply x1 - x2, not as x1 - x2 - a used in the spring's potential energy.

    That said, I am not even sure you can assume that the displacement is purely horizontal when deriving the spring's potential energy. You don't assume that for the pendula. Hmmm.
     
  4. Mar 18, 2013 #3

    CAF123

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    If x1 = x2 = 0 then the spring potential energy term should vanish surely?

    But I am always going to have some sort of extension from the natural length so if I denote the extension of the spring by Δx, then I have Δx + a (why is it +?) which gives a spring potential energy of ##\frac{1}{2}k(\Delta x + a)^2## If ##\Delta x## is zero, then I get a spring potential energy of ##\frac{1}{2}k(a)^2## which doesn't make sense I don't think. (since at equilibrium, the pendula are relaxed so there is no spring potential energy).

    Yes, I must have forgot to consider the y motion of the spring. Can you help me first getting the x component?
     
  5. Mar 18, 2013 #4
    It does not have to. E.g., the gravitational terms are not zero at equilibrium. Potential energy is defined up to an additive constant.

    Just to make sure we are on the same page. "Extension" means the increase in the length of the spring in comparison to its natural length. So if Δx is the extension, then Δx + a is the simply the length of the (extended) spring.

    Now the potential energy (as does the force) depends solely on the extension, not on the length, so it should be ##\frac{1}{2}k(\Delta x)^2##. In this particular case we do have zero for the minimal potential energy.

    The problem is there are no x and y components in potential energy. It is not a vector.
     
  6. Mar 18, 2013 #5

    CAF123

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    Ok, but when I was writing out the hookian force F = -kx, I always had something of the form F = -k(x-a) where x is the extension from natural length a. (e.g in my last problem about coupled masses with transverse oscillations). Why don't we have something of the form 1/2 k (x-a)^2 here?


    For the kinetic energy terms, I wrote ##T = \frac{m}{2}\left[\dot{x}^2 + \dot{y}^2\right]## When the spring is moved via the movement of the pendula , there will be some movement in x and in y direction. Picturing the situation, it seems rather difficult to come up with an expression for the potential energy of the spring.

    EDIT: Or could it be as simple as $$U = \frac{1}{2}k(x_1-x_2)^2 + \frac{1}{2}k(y_1 - y_2)^2?$$
     
    Last edited: Mar 18, 2013
  7. Mar 18, 2013 #6
    Because x1 and x2 are defined differently here. If you had at equilibrium x1 - x2 = a (or x2 - x1 = a) then (x1 - x2 - a) would denote the extension. But you have x1 = x2 = 0 at equilibrium, so (x1 - x2 - a) cannot possibly be the extension.

    It could be explained this way. x1 and x2 are defined relatively to the corresponding equilibrium points. We could define absolute coordinates X1 = x1, X2 = x2 - a. Then X1 - X2 = a at equilibrium, thus the extension is (X1 - X2 - a).

    Using the absolute coordinates defined above, the length of the spring is ## \sqrt {(X_1 - X_2)^2 + (y_1 - y_2)^2} ##, and so its extension is ## \sqrt {(X_1 - X_2)^2 + (y_1 - y_2)^2} - a ##, and the potential energy is ## \frac 1 2 k (\sqrt {(X_1 - X_2)^2 + (y_1 - y_2)^2} - a)^2 ##.
     
  8. Mar 18, 2013 #7

    CAF123

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    Or equivalently, ##U = \frac{1}{2}k (\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2})^2##?
    The expression I had in my edit (I am not sure if you caught it initally) summed together the potential energy in the x direction and the potential energy in the y direction. Is that also a satisfactory equation? I believe the one derived above takes some extension in the x direction and some extension in the y direction and computes the resulting extension via Pythagoras. Whereas the one I derived initially considers the cases separetely and I then add them together.

    EDIT: I only just reliased now that both are exactly the same.
     
    Last edited: Mar 18, 2013
  9. Mar 18, 2013 #8
    I do not think it is equivalent to that. The first one has a clear geometrical and physical meaning, this one is not obvious at all.

    It is actually equivalent to the sum of the potential energies you had, which does not seem justified, either.

    Anyway, I think this is becoming too complicated so you could probably just ignore the vertical displacement for the spring and use the potential energy term you derived previously.
     
  10. Mar 18, 2013 #9

    CAF123

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    So given the below as my energy:
    $$E = -mg \sqrt{l^2 - x_1^2} + \frac{m}{2}\left[\frac{\dot{x_1}^2l^2}{l^2 - x_1^2}\right] -mg\sqrt{l^2 - x_2^2} + \frac{m}{2}\left[\frac{\dot{x_2}^2l^2}{l^2 - x_2^2}\right] + \frac{1}{2}k(x_1 - x_2)^2 + \frac{1}{2}k(y_1 - y_2)^2$$

    Before I try to put this into the matrix form, I need to somehow use the small angle approx/small displacement. At the moment I have squared terms in ##x_1,x_2,y_1,y_2## and I think to put in matrix form they should be linear.
     
  11. Mar 18, 2013 #10

    CAF123

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    I derived it by considering the separate extensions in x and y and then summed them together. E.g some x1 -x2 in x direction and y1 - y2 in y direction. Why is this not correct?

    When you say ignore the vertical displacement, do you mean since we have small displacements the spring stays near enough horizontal and so the potential is only ##1/2 k (x_1 - x_2)^2##
     
  12. Mar 18, 2013 #11
    I do not understand why you are required to have two matrices. The expressions around M and K are identical, so you could just use N = M + K. Are you sure this what the problem is about? Perhaps one of the terms should be quadratic in displacements, not in velocities?
     
  13. Mar 18, 2013 #12

    CAF123

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    There is an error in the problem statement. How could I be so careless.
    It should read:
    Bring the energy into the matrix form: $$\frac{1}{2}\underline{\dot{x}}^T M \underline{\dot{x}} + \frac{1}{2} \underline{x}^T K \underline{x}$$
     
  14. Mar 18, 2013 #13
    Because extensions do not follow the Pythagoras scheme.

    Assume you have right triangle with sides a and b. Its hypotenuse is ## \sqrt {a^2 + b^2} ##. Now, let the sides have some extensions A and B. Then the hypotenuse is ## \sqrt {(a + A)^2 + (B + b)^2} ##, and the extension of the hypotenuse is ## \sqrt {(a + A)^2 + (B + b)^2} - \sqrt {a^2 + b^2} \ne \sqrt {A^2 + B^2} ##.

    Yes, that is what I mean. I know I said otherwise earlier :)
     
  15. Mar 18, 2013 #14
    This is better. So you have to retain the terms quadratic in velocities and displacements.
     
  16. Mar 18, 2013 #15

    CAF123

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    But I also have squared roots etc.. so how I can separate say ##x_1^2## from ##\sqrt{l^2 - x_1^2}?##
     
  17. Mar 18, 2013 #16
    Taylor series. Retain the terms up to the power of two.
     
  18. Mar 18, 2013 #17

    CAF123

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    Yes, so my expression becomes after simplification: $$-2mgl + mgx_1^2/2l + \frac{m}{2}\left[\frac{\dot{x_1}^2l^2}{l^2 - x_1^2}\right]+ mgx_2^2/2l + \frac{m}{2}\left[\frac{\dot{x_2}^2l^2}{l^2 - x_2^2}\right] + 1/2 kx_1^2 - kx_1x_2 + 1/2 k x_2^2$$

    Set that first constant term = 0.
     
  19. Mar 18, 2013 #18
    You need to take care of the denominator in the kinetic energy terms.
     
  20. Mar 18, 2013 #19

    CAF123

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    So I need to Taylor expand ##l^2(l^2 - x_1^2)^{-1}##. I get then ##1 + \frac{x_1^2}{l^2}##.

    When I multiply this back onto ##\frac{m}{2} \dot{x_1}^2,## should I neglect the term that contains the coupled term ##\dot{x_1}^2 x_1^2##. (It will be sufficiently small anyway).
     
  21. Mar 18, 2013 #20
    Yes, I think this is what you are supposed to do.
     
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