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Natural Frequency of a Spring System

  1. Sep 2, 2011 #1
    A mass m is confined to move in a channel in the x-direction and is connected to four identical springs with spring constant k, which are oriented at angles [itex]\phi[/itex] = 45° as shown, if the system is in static equilibrium.

    a) Ignoring friction, determine the natural frequency of vibration of the system.
    b) What is the natural frequency, if the springs are oriented at equal but arbitrary angles [itex]\phi[/itex]?

    [itex]\omega[/itex]n = [itex]\sqrt{\frac{k}{m}}[/itex]

    Not quite sure how to go about this one. I figure you need to take a small displacement of the system to the right or less then analyse the forces such that [itex]\Sigma[/itex]F = ma, and then derive the standard form of the spring equation from there.

    Any hints would be helpful!

    iwIpV.png
     
  2. jcsd
  3. Sep 2, 2011 #2
    I worked through it again and found the natural frequency to be:

    sqrt(4k/m)/(2 * Pi).

    I basically let the angles on both sides change by phi +/- dphi, and the same with the length L -/+ dL, and with x. If this is the correct answer then I suppose it worked out ok.
     
  4. Sep 3, 2011 #3

    vela

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    That's the answer I would expect if Φ=0. It seems odd to me that the frequency doesn't depend on the angle at all. Do you still think that expression is correct?
     
  5. Sep 3, 2011 #4
    You're right. As [itex]\phi[/itex] goes to 90 degrees, the natural frequency should decrease to 0, as there would be no component of force acting in the x-direction then. It appears as though multiplying by cosine would make this correct, but I don't know where that comes in in the working:

    GM6iG.png

    In the diagram, the angle on the left is phi + dphi, the angle on the right is phi - dphi.

    LYg2O.png
     
  6. Sep 9, 2011 #5

    Redbelly98

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    I have spotted two problems here:

    1. The problem statement says that x is the mass's displacement from equilibrium. Yet you use δx as the displacement, and x appears to be a constant equal to LcosΦ according to your diagram in Post #4. But wait, you are also using x as a variable, since you are trying to set up a differential equation for it. Use x as the displacement, as stated in the problem, and stay with that definition.

    2. A spring exerts a force -k·δL, not -k·(L±δL).
     
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