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Non Similar Coupled Pendulum

  1. Oct 10, 2013 #1
    1. The problem statement, all variables and given/known data
    (The picture is having trouble being directed so I attached it)

    Hey guys, I'm trying to figure out the normal modes of a pendulum with two different masses. The lengths of the strings are the same and the weight of the strings are negligible.

    I can easily find everything if the masses are equal but when the masses are different it kind of screws me up.

    2. Relevant equations

    F=ma
    Fs=k(x1-x2)=kΔx
    m1*a1=-m1*g*(x1/l)-kΔx
    m2*a2=-m2*g*(x2/l)-kΔx

    3. The attempt at a solution
    x(double dot)n=an

    Working it all out I get:

    a1+a2+(g/l)*(x1+x2)+Δx((k/m1)-(k/m2))

    This doesn't help me get a simple solution like x(t)=A*cos(ωt+[itex]\phi[/itex]) because of the addition of the Δx((k/m1)-(k/m2))
     

    Attached Files:

  2. jcsd
  3. Oct 11, 2013 #2

    TSny

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    Hello.

    OK, so you have two coupled differential equations:

    m1*a1=-m1*g*(x1/l)-kΔx
    m2*a2=-m2*g*(x2/l)-kΔx

    I think you have one sign error in these equations if Δx stands for the same quantity in both equations. (You didn't state what Δx means. Correction: You did state that Δx is x1-x2. Can you see which of the equations has a sign error?) If the spring exerts a negative force on one of the masses, will it also exert a negative force on the other mass?

    You're going to try to find normal mode solutions in the form

    x1 = A1cos(ωt + ##\phi##)
    x2 = A2cos(ωt + ##\phi##)

    Rather than add the two differential equations, try solving them simultaneously for A1 and A2 and see what happens.
     
    Last edited: Oct 11, 2013
  4. Oct 11, 2013 #3
    Oops, I did miss a negative. So it'd be
    m1*a1=-m1*g*(x1/l)-kΔx
    m2*a2=-m2*g*(x2/l)+kΔx

    Ploughing through it and setting some constant (I'm gonna pick [itex]\eta[/itex]) [itex]\eta[/itex]=mn*xn

    And using [itex]\omega[/itex]g,n2=(g*mn)/l
    So then:
    [itex]\eta[/itex]1+[itex]\omega[/itex]g,12*x1
    [itex]\eta[/itex]2+[itex]\omega[/itex]g,22*x2

    Then solving them I get (I think):
    x1=A1cos([itex]\omega[/itex]g,1+[itex]\phi[/itex]1)
    x2=A2cos([itex]\omega[/itex]g,2+[itex]\phi[/itex]2)
     
  5. Oct 11, 2013 #4

    TSny

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    That looks good.

    I don't follow what you did there.

    In a normal mode, both particles will have the same frequency. That is, you should have

    x1=A1cos([itex]\omega[/itex]t+[itex]\phi[/itex]1)
    x2=A2cos([itex]\omega[/itex]t+[itex]\phi[/itex]2)

    with the same value of ##\omega## for both particles. You need to substitute these expressions into your differential equations.
     
    Last edited: Oct 11, 2013
  6. Oct 11, 2013 #5
    Maybe I'm missing the concept then. I don't really understand why they'd have the same frequency. If one mass is heavier than the other wouldn't it have a lower frequency?
     
  7. Oct 11, 2013 #6

    TSny

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    It's essentially part of the definition of "normal mode" that each particle has the same frequency. See:
    http://en.wikipedia.org/wiki/Normal_mode.

    Also, in a normal mode, the particles move either "in phase" with each other or "out of phase" with each other. This mean that you can assume that the phase constant ##\phi## is the same for each particle. "Out of phase" then means that A1 and A2 will have opposite signs. In finding the normal modes, you can actually just let ##\phi = 0##.

    So, you can make the assumption that the motion of each particle is described by
    x1=A1cos([itex]\omega[/itex]t)
    x2=A2cos([itex]\omega[/itex]t)

    If you sub these into your two equations of motion, you will be able to determine the possible normal mode frequencies and whether the particles move in phase or out of phase in each mode.

    [EDIT: You can explore the normal modes of a similar system here: http://www.falstad.com/coupled/
    You can change to 2 masses and you can vary the individual masses (click on "mouse:pull string"). You can select either normal mode or a combination of the modes (slide the two dots near the middle left side of the screen).]
     
    Last edited: Oct 11, 2013
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