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

by Zaid
Tags: damping, effect, mass, spring, springs
 P: 122 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: $$\zeta = \frac{c}{2 \sqrt{k m}}$$ 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) = $$e^{-\zeta \omega _{n} t} ( A cos(\omega _{d} t) + B sin(\omega _{d} t))$$ where: $$\omega _{d} = \omega _{n} \sqrt{1-\zeta^{2}}$$ $$A = x(0)$$ $$B = \frac{1}{\omega _{d}}(\zeta \omega _{n} x(0) + x^{.}(0))$$ (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.
 P: 122 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: $$\frac{x (t _{m})}{x (t _{m+1})} = e^{ \frac {2 \zeta \pi}{\sqrt {1- \zeta^{2}}}$$ The log decrement ( $$\delta$$ ) is then equal to: $$\delta = ln\frac {x (t _{m})}{x (t _{m+1})} = \frac {2 \zeta \pi}{\sqrt{1- \zeta^{2}}}$$ You can calculate and rearrange to find $$\zeta$$ 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.