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Coupled Harmonic oscillator Setup

  1. Dec 7, 2011 #1
    I REALLY need help with this one guys! As of right now I believe I only need help with just the set up of the problem. The rest is just solving a differential equation and I assume the frequencies they want will just pop out.

    1. The problem statement, all variables and given/known data
    Two identical springs and two identical masses are attached to a wall. (It then shows a picture of two springs attached to a wall horizontally.) Find the normal modes and show that the frequencies can be written as [itex]\frac{1}{2}\sqrt{\frac{k}{m}}\left( \sqrt{5} \pm 1 \right)[/itex]


    2. Relevant equations
    That's the part I need help with!


    3. The attempt at a solution

    I get confused on these coupled harmonic oscillator problems because I do not fully understand how they set up newtons second law. This is my logic to the above problem.

    So when mass 1 (The left most mass) is displaced to the left from its equilibrium position, a restoring force of [itex]-kx_1[/itex] is created by the spring. At the same time, the other spring was also stretched in response to the first spring stretching and it exerts a force on mass 1 of [itex]-k(x_1+x_2)[/itex]. Where [itex]x_1[/itex] and [itex]x_2[/itex] are the distances from their respective equilibrium.
    So the DE for the first mass is..
    [tex]m\ddot{x}=-2kx_1-kx_2[/tex]
    I am confused on getting the second differential equation. I tried a few things but I kept getting solutions that weren't sinusoidal, but exponential since the roots weren't complex.

    Can anyone help?
     
  2. jcsd
  3. Dec 7, 2011 #2

    rude man

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    Could you show us the picture? How can anything be attached to a wall 'horizontally', when a wall is eo ipso vertical? :confused:
     
  4. Dec 7, 2011 #3

    rude man

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    And how are they coupled anyway?
     
  5. Dec 7, 2011 #4
    Sorry guys! Here is a really crappy picture I made in paint. I guess they would be sliding on a frictionless rod so gravity doesn't play any role or something.
    http://img853.imageshack.us/img853/6703/springyq.png [Broken]
     
    Last edited by a moderator: May 5, 2017
  6. Dec 8, 2011 #5
    Please guys, I really need to understand this. I don't need somebody to go though the whole problem I ONLY need help with the logic in setting up Newtons Second Law!
     
  7. Dec 8, 2011 #6

    vela

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    You have the basic idea, but you're getting the signs wrong. I prefer to displace the masses to the right so that, x1, for instance, is positive. It makes it easier to see what the signs on the terms should be in the equation. When you do that, both springs will push mass 1 to the left, the negative direction. If you displace mass 2 to the right, the force on mass 1 will be to the right, the positive direction. Therefore, you should get
    [tex]m \ddot{x}_1 = -kx_1 + k(x_2-x_1)[/tex]If you think about it, the force due to spring 2 should depend on [itex]x_2-x_1[/itex] because that quantity corresponds to the distance between the two masses, which is how much the second spring is stretched.

    What did you get for the equation for mass 2?
     
  8. Dec 8, 2011 #7
    A ha! I believe I am starting to understand!

    So if I displace mass 2 to the right, then a force from the second spring is pulling it to the left in the negative direction. [itex]-k(x_2-x_1)[/itex]. At the same time spring 1 is stretched and also pulling mass 1 (and consequently mass 2) to the left as well. [itex]-kx_1[/itex]. So that means the DE is..

    [tex]m \ddot{x_2}=-kx_1-k(x_2-x_1)[/tex]

    Would this be correct?
     
  9. Dec 8, 2011 #8

    vela

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    Not quite. Only the second spring exerts a force on mass 2, so only the second term should be there.

    The fact that displacing mass 2 to the right results in a force in the negative direction is reflected in the term -kx2. The fact that displacing mass 1 to the right will result in a force in the positive direction on mass 2 results in the term +kx1. When you combine those, you get -k(x2-x1), which you can interpret as the force of spring 2 on mass 2.
     
  10. Dec 8, 2011 #9
    Ohh okay! I believe I understand. I will now continue with the problem as from here on out it is simply solving the DE. (With some extra steps at the end.) I will post my results when finished.

    Thank you VERY much for your help. I was starting to lose faith that I would ever fully understand this. :\
     
  11. Dec 8, 2011 #10
    So I have gotten the previous problem and have moved on to the next problem. Now there are three masses confined to move on a ring. All the masses are connected to each other by springs. All masses are equal and all spring constants are equal. Would this be right for the DE's??

    [tex]m \ddot{x_1}=-k(x_1-x_2+x_3)[/tex]
    [tex]m \ddot{x_2}=-k(x_2-x_1-x_3)[/tex]
    [tex]m \ddot{x_3}=-k(x_1-x_2+x_3)[/tex]

    Somehow I do not think this is correct..
     
  12. Dec 8, 2011 #11

    rude man

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    At first I was worried about the fact that centripetal as well as longitudinal accelerations are involved. But on reflection then centripetal fiorce does no work on the masses since it's always orthogonal to the motion around the ring. So we van treat this problem similarly to the previous one:

    Using s instead of x to indicate arc motion, and counting s positive for counterclockwise direction, I would write

    [tex]m_1 \ddot{s_1} = k[s_2 - s_1) + (s_1 - s_3)[/tex], or
    [tex]m_1 \ddot{s_1} = k(s_2 - s_3)[/tex]

    etc. for m2 and m3. Of course, m1 = m2 = m3 but I wanted to indicate how I arrived at my equations.

    Vela?
     
  13. Dec 8, 2011 #12
    But why? My biggest problem is understanding the logic behind the formulation :[
     
  14. Dec 8, 2011 #13

    rude man

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    Hang on. I'm working on an explanation. BTW there has to be the assumption that the springs are all compressed or just barely neutral, they can't slosh around the ring....
     
  15. Dec 9, 2011 #14

    vela

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    With the masses on the ring, you need to be a bit more careful because of the periodicity of the physical position, e.g., [itex]\theta=0[/itex] and [itex]\theta=2\pi[/itex] correspond to the same point.

    Do the springs lie along chords or do they lie along the circumference of the circle?
     
  16. Dec 9, 2011 #15
    It's a rigid frictionless circle. My professor did an example in class with TWO masses and came up with these as the DE.
    [tex]m \ddot{x_1}=-2k(x_1-x_2)[/tex]
    [tex]m \ddot{x_2}=-2k(x_2-x_1)[/tex]

    I can KIND OF understand these equations, but I am definitely not 100% :\
     
  17. Dec 9, 2011 #16

    vela

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    Forget my last post. You don't have to worry about periodicity if you're only looking at small oscillations.

    You get the equations essentially the same way you did before.

    If you displace mass 1 a bit to the right, the force from the spring between masses 1 and 2 will point to the left. Similarly, the force from the spring between masses 1 and 3 will also point to the left. So a displacement from equilibrium will result in a force in the opposite direction, so [itex]\ddot{x}[/itex] should be proportional to [itex]-x_1[/itex] for both terms. Therefore, the equation of motion must be
    [tex]m \ddot{x}_1 = -k(x_1-x_3) - k(x_1-x_2)[/tex]I didn't bother with looking at the effect of displacing the other masses on mass 1 because we know the forces have to be proportional to [itex]x_1-x_2[/itex] and [itex]x_1-x_3[/itex]. It's just a matter of choosing the right sign for each term to get the direction right.

    Can you get the other two equations now? And does the equation from the two-masses-on-a-ring system make sense now?
     
  18. Dec 9, 2011 #17
    I think I finally understand! (I really mean it this time ;) )
    So would the other two equations of motion be this?
    [tex]m \ddot{x_2}=-k(x_2-x_1)-k(x_2-x_3)[/tex]
    [tex]m \ddot{x_2}=-k(x_3-x_1)-k(x_3-x_2)[/tex]
     
  19. Dec 9, 2011 #18

    vela

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    Yes! You can figure that out by doing the similar analysis for each mass or by appealing to the symmetry of the problem.
     
  20. Dec 9, 2011 #19
    Awesome! I went back to the two mass problem on a circle and it all makes sense now!

    Thank you so much! I very much appreciate it. :D!

    Another instance where these forums save my ***!
     
  21. Dec 10, 2011 #20
    Man I REALLY hate to revive this topic but I am trying to solve the three equations with 4 unknowns that arise from the DE's and it is absolutely insane. Are their any tricks I can do to make this easier? This is what I am doing..
    -Assume solution is of the form..
    [tex]x_1=c_1e^{pt},x_2=c_2e^{pt},x_3=c_4e^{pt}[/tex]
    -Solve for [itex]c1/c2[/itex] in two equations
    -Set these two equations equal to each other and solve for [itex]c_3/c_2[/itex]
    -Plug expression obtained from [itex]c_3/c_2[/itex] into both equations for [itex]c_1/c_2[/itex].
    -Set both [itex]c_1/c_2[/itex] equations equal to each other and solve for p.

    That last step is ridiculous, as I cannot see a straight forward way to solve it.

    The one part I do not understand is when I plug my assumptions in for [itex]x_1,x_2 and x_3[/itex] I get a 3x3 matrix. I then put this matrix in reduced row echelon form and I get 0,0,0. How can it be 0,0,0?
     
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