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Normal modes and degrees of freedom in coupled oscillators

  1. Nov 16, 2015 #1
    Not a textbook/homework problem so I'm not using the format (hopefully that's ok).

    Can someone offer an explanation of normal modes and how to calculate the degrees of freedom in a system of coupled oscillators?

    From what I've seen the degrees of freedom seems to be equal to the number of oscillators in the system.
     
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  3. Nov 16, 2015 #2

    Geofleur

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    Consider two oscillators coupled together, for example the two masses of a double pendulum. Each mass is located via 3 coordinates, so a total of 6 variables is required to specify their positions in the absence of any constraints. However, the masses are connected to each other, and one of them is connected to the ceiling, which introduces 2 equations of constraint. The degrees of freedom are the number of variables in general minus the number of constraints, so 6 - 4 = 2 in this case. However, the number of degrees of freedom does not have to be equal to the number of oscillators. For example, consider a single pendulum that is free to swing in 3D, the so-called space pendulum. Then the number of variables in the absence of constraints is 3 and the number of constraints is 1, giving 2 degrees of freedom for the 1 oscillator.

    Normal modes are really great. We search for solutions where all the masses oscillate at a single frequency. Each such solution is called a normal mode. For the double pendulum, there are two such modes. In the first, both masses swing left and right together; in the second, one mass always swings in the opposite direction of the other. We can generate an arbitrary motion by combining these normal mode solutions! In fact, it really amounts to representing an arbitrary solution as a Fourier series. Wonderful stuff!
     
  4. Nov 17, 2015 #3
    Thanks for the reply! Your explanations were really helpful.

    Suppose I have a horizontal spring on a frictionless surface. It can only move in one direction (let's say the x-axis), so does that mean it only has 1 variable? 1 variable and 1 equation would mean 0 degrees of freedom?
     
  5. Nov 17, 2015 #4

    Geofleur

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    A spring, with nothing attached to it, would need 7 variables to specify completely its configuration in general: 3 for the x, y, and z coordinates of its center of mass, 3 Euler angles to describe the spring's orientation in space, and 1 more to describe the degree of elongation or compression. Saying that the spring can move only along the x-axis is really giving 4 equations of constraint: ## y = 0 ##, ## z = 0 ##, and 2 of the 3 Euler angles must be zero (I am assuming that we want the axis of the spring to coincide with the x-axis).

    With these constraints, there are only 7 - 4 = 3 possible motions left to the spring: it can move back and forth along the x-axis, it can be rotated about the x-axis, and it can be compressed/extended. So there are now 3 degrees of freedom. If we say that it cannot be rotated at all, then we are left with 2 degrees of freedom, motion along the x-axis and compression/extension.

    The spring would only have 0 degrees of freedom if it were unable to be rotated, unable to be compressed or elongated, and unable to move as a whole along any axis. Poor spring!
     
  6. Nov 17, 2015 #5
    I see, I didn't consider the compression/extension as a degree of freedom.

    Would motion along the x-axis be if the spring (the entire thing, not just one end) was being moved at some velocity along the x-axis?

    Also, how would it change if the spring had a mass on it? The mass would also need 3 coordinates, but since the mass is constrained to remain at the end of the spring, would the number of degrees of freedom be the same?
     
  7. Nov 17, 2015 #6

    Geofleur

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    Yes.

    Now that I think of it, I also assumed that the spring could not bend, which would enable a different kind of vibration (as opposed to elongating/compressing motions)! So let's say it can't bend, things are already complicated enough. The more complicated the system, the easier it is to miss some motions accessible to it.

    If the block could not be rotated relative to the spring, the degrees of freedom would be the same (I think!).
     
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