How much force is required to excite a vibrating machine with a load?

[slobberingant]

TL;DR Summary

Need help understanding a vibrating force formula that was given to me at work.

I have been given a formula at work to use for calculating how much force is required to excite a vibrating machine with load. Only a proportion of material load on a vibrating machine is considered.

The formula is
F = 2*S / (K*M)
Where:
S = stroke (mm)
K = (140/w)^2
w = angular velocity (rad/s)
M = (M1 + (M1 + M2)) / (M1 * (M1 + M2)) (1/kg)
M1 = machine weight (kg)
M2 = material weight (kg)

I've manage to derive that the formula is based on Newton's second law and the acceleration of simple harmonic motion.
F = ma
a = Aw^2 - simple harmonic acceleration formula
Where:
A = amplitude (mm) => Stroke = 2*A

Therefore
F = mAw^2

My formula above is different in the following ways.
- w^2 ((rad/s)^2) becomes K - (140/w)^2 (1/((rad/s)^2)) but is divided rather than multiplied.
- m (kg) becomes M (1/kg) and is also divided rather than multiplied.

What does the M variable in my formula above do? - It seems like a ratio between machine and material mass. However, in isolation, the values it produces can be lower than the machine weight which would result in incorrect force values.

Where does the 140 constant in K come from?

 

[jack action]

- w^2 ((rad/s)^2) becomes K - (140/w)^2 (1/((rad/s)^2)) but is divided rather than multiplied.

No. w^2 becomes 1/K

- m (kg) becomes M (1/kg) and is also divided rather than multiplied.

No. m becomes 1/M

What does the M variable in my formula above do? - It seems like a ratio between machine and material mass. However, in isolation, the values it produces can be lower than the machine weight which would result in incorrect force values.

Actually, if M1 >> M2, then 1/M tends toward 0.5*M1, which multiplied by the 2 in the equation really gives F = M1 * [acceleration].

If M2 >> M1, then 1/M tends toward M1, which gives F = 2 * M1 * [acceleration]. I'm not sure what it means though.

Where does the 140 constant in K come from?

Not sure, but I think it might have to do with some pre-calculated damping for the machine.

More info: https://www.brown.edu/Departments/E...Notes/vibrations_forced/vibrations_forced.htm

 

[slobberingant]

Thanks Jack.
Fantastic insight.
Regarding M, if there is little load relative to mass then use the mass of machine only. This makes sense as larger machines are less effected by material load. This logic works with M2 >> M1. This approach helps greatly.
I will do some more reading on damping.
Thank you.