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Multi-DOF vibration systems

  1. Jan 10, 2012 #1
    Hi all,

    I have a question about Multi DOF vibrating systems. For free vibration of undamped MDOF systems, we have the equations of motion as :

    [itex]M[/itex] [itex]\ddot{q}[/itex] + [itex]K[/itex] [itex]{q}[/itex] = [itex]{0}[/itex] (1)

    Where,
    [itex]M[/itex] - n x n mass matrix
    [itex]K[/itex] - n x n stiffness matrix
    [itex]{q}[/itex] - n x 1 vector of generalized coordinates

    Most vibrations book try to obtain the eigenvalue problem by assuming the solution to (1) as
    [itex]{q}[/itex] = [itex]{Q}[/itex] [itex]e^{j \omega t}[/itex] (2)

    For a 2-DOF system, (2) is
    [itex]q_{1}[/itex] = [itex]Q_{1}[/itex] [itex]e^{j \omega t}[/itex] (scalar eqn, 1st component of [itex]{q}[/itex])
    [itex]q_{2}[/itex] = [itex]Q_{2}[/itex] [itex]e^{j \omega t}[/itex] (scalar eqn, 2nd component of [itex]{q}[/itex])

    My question is why do we assume the same exponent for all components of [itex]{q}[/itex]? Why not assume (for a 2 DOF system) the solution as
    [itex]q_{1}[/itex] = [itex]Q_{1}[/itex] [itex]e^{j \omega_{1} t}[/itex]
    [itex]q_{1}[/itex] = [itex]Q_{1}[/itex] [itex]e^{j \omega_{2} t}[/itex]

    ie, [itex]{q}[/itex] = [itex]W[/itex] [itex]Q[/itex], where [itex]W[/itex] is a diagonal matrix containing the exponent terms?

    Thanks,
    yogesh
     
  2. jcsd
  3. Jan 10, 2012 #2

    AlephZero

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    The reason is that the idea of "vibration" means that the whole object keeps repeating exactly the same motion in each cycle of vibration. If the frequencies of different parts of the object were different, that would not happen.

    Of course the actual motion of the system can be a linear combination of all the possible modes of vibration, but that would mean you were assuming a motion like
    [itex]q_1 = Q_{11}e^{j\omega_1 t} + Q_{12}e^{j\omega_2 t}[/itex]
    [itex]q_2 = Q_{21}e^{j\omega_1 t} + Q_{22}e^{j\omega_2 t}[/itex]
    which isn't very useful if you want to find [itex]\omega_1[/itex] and [itex]\omega_2[/itex].
     
  4. Jan 11, 2012 #3
    ^AlephZero,
    Thanks for the reply. Does that mean if I give random initial displacements (and/or velocities) to both the masses and let go, and trace out [itex]q_{1}[/itex] and [itex]q_{2}[/itex], we will see periodic variations in the response?
     
  5. Jan 11, 2012 #4

    AlephZero

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    The motion would be a combination of both vibration modes. Unless the two frequencies are in a simple ratio like 2:1 or 3:2 the motion will not look "periodic". For example a graph like [itex]\cos(t) + 0.5 \cos(\sqrt2 t)[/itex] shows the sort of motion you would get in the general case.
     
  6. Jan 11, 2012 #5
    Agree with what you said. Let's say we didn't know about modes and we want to solve the EOM for a 2-DOF system. If we want to assume a solution to the equations, how would you proceed? What would prompt us to use the same frequency for both the masses?
     
  7. Jan 11, 2012 #6

    AlephZero

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    If you forget about physics and just consider this as a math problem, the EOM is a linear 2nd order differential equation.

    If the EOM had 1 DOF, the standard solution method is to assume a solution [itex]x = A e^{pt}[/itex], substitute into the equation, and get a quadratic equation for p. You then find the corresponding values of A from the initial conditions [itex]x(0)[/itex] and [itex]\dot x(0)[/itex].

    For a matrix equation the same method works, if you take x and A as vectors. The equation for p is now an eigenvalue equation [itex](K + p^2M)A = 0[/itex]. For each eigenvalue [itex]p_i[/itex] there is a corresponding eigvenvector [itex]A_i[/itex] (multiplied by an arbitrary scaliing factor). The general solution is then
    [tex]\sum_i c_i A_i e^{p_i t}[/tex]
    where as before the constants [itex]c_i[/itex] are determined by the intial conditions.

    Each separate solution [itex]A_i e^{p_i t}[/itex] looks like a "vibrating mode shape" of the structure.

    Of course all the p's are pure imaginary numbers, so we write [itex]p_i = j \omega_i[/itex] and the eigenvalue equation is usually written as [itex](K - \omega^2M)A = 0[/itex], but I'm are pretending I don't know any physics here!

    It is possible to prove mathematically that this really is the most general solution of the EOM.
     
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