March 17th 1997

Analysis of the Damped Mass-Spring Differential Equation

Contents:

Terms:

    Term
    M
    R
    K

    Q


    Fm
    x
    X
    Y
    Z
    t

    j

    Units
    kg
    kg/s
    N/m


    1/s

    N
    m
    m
    m/s
    m/s^2
    s

    Description
    Mass
    Mechanical resistance
    Spring constant

    Quality Factor
    Natural Frequency

    Maximum external force applied to mass
    Mass displacement (+x is outward)
    Maximum mass displacement
    Maximum mass velocity
    Maximum mass acceleration
    time

    The imaginary constant square root of -1

    There are three intuitive properties of the damped mass-spring system, they are defined as follows:

    Mass:
    I will assume that the mass is constant for all my analysis. The application of a force (F) a free mass (M) causes it to accelerate according to:

          2
          d x
    F = M ---
            2
          dt 

    Mechanical Resistance:
    When the mass moves at a velocity dx/dt it experiences a force in the opposite direction due to resistance. This resistive force is a linear function of velocity:

            dx
    F = - R --
            dt

    Spring Constant:
    A restoring force proportional to the driver position obeys the following:

    F = - K x

Damped Mass-Spring Differential Equation:

    Using the above equations we can relate all the forces on the mass to get a second order differential equation. The position and velocity of the mass generates forces that accelerate it. Therefore we get the following:

             2
            d x       dx
      F = M --- = - R -- - K x
              2       dt
            dt

    But this equation will only describe the damped mass-spring system without any outside interaction. We add an extra term, Fm*exp(jt), as an external force on the cone. We have a modified equation to reflect this added force.

             2
            d x       dx             jt
      F = M --- = - R -- - K x + Fm e                        (1)
              2       dt
            dt

    There are many well defined ways to solve this problem. The approach we will take will be the simplest of all; assuming the solution for x is of the form x=X*exp(jt), where X is a complex number indicating the maximum displacement and phase with respect to the force (Fm).

         2   jt            jt  
        d X e          d X e            jt       jt
      M -------- = - R --------- - K X e    + Fm e           (2)
             2            dt
           dt

              2    jt              jt        jt       jt
    =>   - M   X e    = - R  j X e    - K X e    + Fm e       (3)

              2               Fm
    =>   - M   + R  j + K = --                                (4)
                               X

    By making the assumption that force is sinusoidal, we have transformed our equation from the time domain to the frequency domain. X can be solved given any single frequency (). With the application of Fourier theory we can intuitively see how equation 4 describes the complete behavior of the damped mass-spring system given any periodic force. Unfortunately the three properties, M, R and K, are no longer "intuitive" in this frequency domain. We will have to define new quantities that help us see what this equation really means.

Quality Factor and Natural Frequency:

    I now introduce natural frequency (). It is the frequency that the mass-spring system would alternate at if mechanical resistance was zero. It can also be defined as the frequency of greatest mass velocity. Quality factor (Q) can best be described as efficiency at natural frequency. We will define Q and  and see that these definitions agree with their intuitive concept. Their definitions are:

        sqrt(M K)                        / K \
    Q = ---------               = sqrt |---|
            R                            \ M /

    We can substitute these two terms into (4) to get a simpler result:

            2   R       K   Fm
    =>   -   + -  j + - = ---                            (5)
                M       M   M*X

            2           2   Fm
    =>   -   + --  j +  = ---                          (6)
                Q             M*X

    Lets now solve this equation finding all the important points: maximum displacement, maximum velocity, and maximum acceleration. We will also see the effects of Q on these values so that we may get a better understanding of what this value is.

Displacement, Velocity and Acceleration:

    For sake of simplicity we will introduce two more terms; maximum velocity (Y) and maximum acceleration (Z). We can derive their relationships to X by recalling that

                        jt
               dx   d Xe        jt
    Velocity = -- = ----- = jXe                       (7)
               dt     dt

                    2     2  jt
                   d x   d Xe        2   jt
    Acceleration = --- = ------- = -  Xe              (8)
                     2      2
                   dt     dt

Therefore, we can define Maximum velocity (Y) is the coefficient of velocity, and maximum acceleration (Z) as the coefficient of acceleration:

                                        2
         Y = jX       and       Z = -   X                 (9)

We can substitute these two into (6) to find the expressions for X,Y and Z.

                                    -1
          Fm  /    2           2 \ 
      X = --- | -   + --  j +  |                    (10)
           M  \        Q           / 

              /               2 \ -1
          Fm  |             | 
      Y = --- | j  + -- - j -- |                       (11)
           M  \       Q        / 

              /               2 \ -1
          Fm  |             | 
      Z = --- | 1 - j ---- - -- |                       (12)
           M  |       Q      2 |
              \                /

Maximum Displacement:

    |X| is at its maximum (or minimum) if it's derivative is zero. Take (10), find |X| and take the derivative:

                                        -1
              Fm  /    2           2 \ 
    =>    X = --- | -   + --  j +  |                    (10)
               M  \        Q           / 

                                               -1/2
                   /            2           2 \
                Fm | /  2    2 \    /    \  |
    =>    |X| = -- | |  -   |  + | --  |  |             (13)
                M  \ \         /    \ Q    /  /

                        /  2    2 \           /    \ 
                      2 |  -   | (-2) + 2 | --  | --
                Fm 1    \         /           \ Q    / Q
    =>    0 = - -- -  -----------------------------------   (14)
                M  2                              -3/2
                      /            2           2 \
                      | /  2    2 \    /    \  |
                      | |  -   |  + | --  |  |
                      \ \         /    \ Q    /  /

    The denominator is irrelevant to finding our zero solution,

                        2                   /         \ 1/2 
               2   2                      |      1  |
    =>    0 =  -  + ---      =>     = | 1 - --- |     (15)
                         2                  |       2 |
                       2Q                   \     2Q  /

    If we look at  we see that it is very close to  when the Q is very high, or the mechanical resistance is low. Also notice that for Q < 1/sqrt(2),  is purely imaginary; the maximum point is at  = 0, with a value of |X| = Fm/K. For the case of Q >= 1/sqrt(2), the point of maximum displacement exists, lets find the value. Substitute (15) into (10) to find it.

                                -1/2
               Fm 1  / 1     1  \
    =>   |X| = -- -- | -- - --- |      Q > 1/sqrt(2)        (16)
               M   2 |  2     4 |
                   \ Q    4Q  /

    Here is a logarithmic graph of |X|, normalized to Fm/(M*^2) = Fm/K on the vertical axis, and =1 Hz on the horizontal. The green lines are the maximum points for Q=2 and Q=1. There would be a line for Q=10, but it is so close to  that it can not be seen. Also notice that for a Q>2 the maximum displacement will be about 'Q' times greater than Fm/K. For Q=1/sqrt(2)=.707, and Q=.5 the point of maximum displacement is at zero Hertz, and therefore not shown. A feature that should be noticed is that at , |X| = Q*Fm/K. Another point: above resonance, |X| falls off at a rate of 6dB per octave; below it is flat.

Maximum Velocity:

    As before, we have to find  that maximizes |Y|. Use equation (11), find |Y|, then set the derivative of |Y| to zero.

                  /               2 \ -1
              Fm  |             | 
    =>    Y = --- | j  + -- - j -- |                       (11)
               M  \       Q        / 

                    / /      2 \2   2 \ -1/2
                Fm  | |      |    | 
    =>    |Y| = --- | |  - -- | + -- |                     (17)
                 M  | |       |    2 |
                    \ \        /   Q  / 

                   /      2 \ /  1  \
                   |      | | --- | 
                 2 |  - -- | |   2 |
              Fm   \       / \    /
    =>    0 = -- ------------------------                   (18)
              M  / /      2 \2   2 \ -3/2
                 | |      |    | 
                 | |  - -- | + -- |
                 | |       |    2 |
                 \ \        /   Q  / 

    Similarly, the denominator is irrelevant to finding our zero solution.

                    2
                   
    =>     0 =  - ---        =>       =                 (19)
                    

    Notice that the point of maximum velocity is also the natural frequency of the mass spring system. Sub (19) into (11) to get the amplitude at this frequency:

               Fm Q    Fm
    =>     Y = -- -- = --                                   (20)
               M     R

    Lucky for us that Y can be solved to a simple form. We can analyze it with greater ease. We notice that Y is in phase with the force applied. In the time domain we see the velocity only depends on the force applied and the mechanical resistance in the system. With frequency domain variables we see that Y is directly proportional to Q. This is a graph to show the relationship between Q and velocity |Y|. It is normalized to Fm/(*M) = Fm/R on the vertical axis, and =1 on the horizontal axis. The amplitude, as  recedes from , falls at a rate of 3dB per octave.

Maximum Acceleration:

    Finally for the calculation of maximum acceleration. It will prove the most useful in analyzing driver parameters. Follow the same method as in the last 2 cases:

                  /               2 \ -1
              Fm  |             | 
    =>    Z = --- | 1 - j ---- - -- |                       (12)
               M  |       Q      2 |
                  \                / 

                    / /     2 \2    2  \ -1/2
                Fm  | |     |      | 
    =>    |Z| = --- | |1 - -- | + ---- |                    (21)
                 M  | |     2 |    2 2 |
                    \ \      /    Q  /

                   /      2 \ /    \      2
                   |      | | -2 |   2 
                 2 | 1 - -- | | -- | - ----
                   |      2 | |  3 |    3 2
              Fm   \       / \   /    Q
    =>    0 = -- --------------------------                 (22)
              M  / /      2 \2    2  \ -3/2
                 | |      |      | 
                 | | 1 - -- | + ---- |
                 | |      2 |    2 2 |
                 \ \       /    Q  /

                   2      2                  /          \-1/2
                                         |      1   |
    =>    0 = 1 - --  - ----    =>     =  | 1 - ---- |   (23)
                   2       2                 |        2 |
                       2 Q                  \     2 Q  /

    As with the case of displacement, (23) is undefined for Q <= 1/sqrt(2); the acceleration is strictly increasing, with respect to , in this region. For the values that (23) is defined, substitute (23) into (12) to find amplitude of this acceleration:

                              -1/2
               Fm / 1     1  \
    =>   |Z| = -- | -- - --- |     Q > 1/sqrt(2)            (24)
               M  |  2     4 |
                  \ Q    4Q  /

    The maximum amplitude is independent of the natural resonant frequency.  Here is a logarithmic graph of |Z|, normalized to Fm/M on the vertical axis, and =1 Hz on the horizontal. The green lines indicate the points of maximum amplitude for Q=2 and Q=1. Also notice that for a Q>2 the maximum displacement will be 'Q' times greater than Fm/M. For Q=1/sqrt(2)=.707, and Q=.5 the point of maximum displacement is at infinite Hertz, and therefore not shown. All graphs show that |X| falls off at a rate of 6dB per octave below resonance, and above it is flat. Another feature that should be noticed is that at , |X| = Q*Fm/M. If you suspected that this graph has an uncanny similarity to the displacement graph, you are correct. They are mirror images of each other.

Conclusions:

    Displacement has a near flat response below resonance. Acceleration has near flat response above resonance. We see in these two cases that the Q determines the ratio of 'amplitude at ' to 'amplitude in the flat region of response'. For values of Q<1/sqrt(2) we know that there is no maximum point, so knowing the amplitude at  is useful to visualizing the graph.

    Although the maximum amplitude is never realized at , it is good enough to approximate, and say ' is the point of maximum amplitude' if Q>=2. These two generalizations help us visualize the effects of Q. Unfortunately we can not make the same sort of generalization for .707 < Q < 2. Only experience can help us here.

    The velocity graph has a simple mathematical form, but it has few analytical points we can use to remember its shape. Q has the effect of reducing the peak at resonance. Other than that, no other information can be gleaned unless it is compared to a graph with a different Q.