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Coil Spring exhibiting slightly different stiffness when compressing than when expanding

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f_c

Nuclear
Mar 12, 2022
1
Hi,

I came across to this problem for research and educational reasons.

Could someone suggest how to best describe the oscillation motion of a Coil Spring exhibiting slightly different stiffness when compressing than whn expanding?

Mathematically it consists in "just" solving the standard harmonic oscillator, but with the additional (maybe subtle) condition that the "K" constant of the spring changes across the equilibrium position
of the coil spring at x = 0. This can be done for example solving the associated differential equations using commercially available software. Before starting to work on that I would like to ask if someone is aware about work of this type already done in some engineering textbook or software tutorials (?)

thanks,

Best Regards,

F_C
 
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I usually solve them numerically, to be honest. The extreme case is when the spring loses contact completely, as in the jounce bumper of a car, which has a non linear rate of its own. This introduces odd order harmonics to the vibration spectrum, as will your case, if you are driving with a pure sine wave of force, or vice versa if you are driving with a pure displacement sine wave.

If you are thinking about a spring mass system then I'm gently wondering in your case whether you can do SHM for each half of the cycle and then make sure the velocities and displacements are equal at x=0.

So for the shm, at maximum extension energy =1/2*k1*r1^2, and minimum 1/2*k2*r2^2, and obviously they are equal. The velocity at x=0 is 1/2*m*v^2, and again is equal to the max extension energy. In each half of the oscillation the angular frequency w=v/r, so it takes pi/w seconds to perform its half of the oscillation, add the times up and invert and there's your fundamental frequency. I think. If I've got the derivation right then w_combined=1/(pi*sqrt(m))*sqrt(k1*k2)/(sqrt(k1)+sqrt(k2)) which looks reasonable but needs checking. I am slightly mystified why SHM, a very linear analysis, is able to cope with a non linear spring.


<Note above has been edited>

A common way to get this effect in practice is to have two concentric springs of different lengths, so one only comes into play in compression. Oh, here we go




Cheers

Greg Locock


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