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Effect of mass and springs on the damping of mass spring system

  1. Dec 17, 2008 #1
    If I have mass-spring system with certain damping factor, what is the expected reduction in the damping if the mass is reduced by 25% and the spring stiffness by 50%? It is clear that the damping will decrease but how much? can anyone helps me in that?
  2. jcsd
  3. Dec 22, 2008 #2
    No body helped in that. I really need a help

    Does using kinetic and potential energies help in that? Any idea?
  4. Dec 23, 2008 #3
    The critical damping ratio of a system is often used to compare system damping to that which would result in a critically damped case (i.e. quickest settling time, no overshoot). This is given by:

    [tex]\zeta = \frac{c}{2 \sqrt{k m}}[/tex]

    where stiffness is k, mass is m and damping constant is c.

    You can see the effect that this has on oscillation amplitude as follows. Consider the variation of amplitude of an underdamped single degree of freedom mass-spring-dashpot system (bit of a mouthful) with time:

    x(t) = [tex]e^{-\zeta \omega _{n} t} ( A cos(\omega _{d} t) + B sin(\omega _{d} t))[/tex]


    [tex]\omega _{d} = \omega _{n} \sqrt{1-\zeta^{2}}[/tex]
    [tex]A = x(0)[/tex]
    [tex]B = \frac{1}{\omega _{d}}(\zeta \omega _{n} x(0) + x^{.}(0))[/tex]
    (last term is supposed to contain first derivative w.r.t. time)

    A good resource to show how this varies with different values of mass, stiffness and damping constant can be found here.
    Last edited: Dec 23, 2008
  5. Dec 23, 2008 #4
    Thank you Timmay

    What I have is reduction in mass and stiffness but the damping c will decreases accordingly. So zeta and c are not fixed and unknown.

    Zeta can be found if the initial displacement, the final displacement and the time difference. Therefore for both cases if we started with same displacement, the final displacement in the second case (reduction 50% in stiffness and 25% in mass) will be larger as the reduction in mass is less than the reduction in stiffness. if we can find the increase in the final displacement of the second case in compared with the final displacement in the first (original)case, the problem is solved. Potential and kinetic energies may help, guess. I need furtrher help in that from you Timmay and other colligues.
  6. Dec 23, 2008 #5
    If you can measure the variation of displacement with time of your system for unforced damped vibrations (again I'm assuming underdamped) then you could use the log decrement method.

    If the system parameters do not change within a single test, then the ratio between successive peaks or troughs (local maxima and minima) will remain constant. It can be shown that for any two adjacent local maxima:

    [tex]\frac{x (t _{m})}{x (t _{m+1})} = e^{ \frac {2 \zeta \pi}{\sqrt {1- \zeta^{2}}}[/tex]

    The log decrement ( [tex]\delta[/tex] ) is then equal to:

    [tex]\delta = ln\frac {x (t _{m})}{x (t _{m+1})} = \frac {2 \zeta \pi}{\sqrt{1- \zeta^{2}}}[/tex]

    You can calculate and rearrange to find [tex]\zeta[/tex] for each test, and then express in terms of the first equation in the original post to show the effect on damping constant for each set of conditions. I guess from your last post that one of these two steps should satisfy you.
  7. Dec 24, 2008 #6
    Thank you timmay

    I've done the two tests experimentally and I got reduction in zeta to bout 50%. However I'm looking for a mathematical proof explaining this reduction.
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