Accelerometer constant displacement

Sep 10, 2008 2 Replies

I have a basic problem I hope that someone wouldn't mind helping me with:



I'm using a shaker table and want to test at several frequencies, but keep the same displacement.



Am I correct in thinking that I can say ma=kx and f=sqrt(k/m) (where m is mass, a acceleration, k spring constant, f frequency and x displacement) and so all I have to do to keep displacement constant is keep (a/f^2) constant?



For a mass bobbing on a spring, it is appropriate to equate the inertial force (ma) to the spring deflection force (kx).

Referring to a wiki entry:

you can see that acceleration when given as a function of time, a(t) =3D -A omega ^2 cos (omega t + phi) so for constant amplitude A

-a(t) / omega^2 cos (omega t + phi) is constant

i.e amplitude is proportional to -a / f^2 as you say.

And frequency is PROPORTIONAL to sqrt(k/m)

Regards

BrianW

On Thu, 11 Sep 2008 06:59:31 -0700 (PDT), Brian meant to write:

dunno where the extra garbage after the equals sign sprang from. Mr Gates, I suppose?

Brian W

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