GIVEN:
Amplitude, A = 0.1mm
Force, F =1 N
mass of motor, m = 120 kg
operating speed, N = 720 rpm
[tex]\frac{A}{F}[/tex] = [tex]\frac{0.1\times 10^{-3}}{1} = 0.1\times 10^{-3}[/tex]
Formula Used:
[tex]A = \frac{F}{\sqrt{(K_{t} - m\omega ^{2}) +(\zeta \omega ^{2})}}[/tex]
Solution:
Let Stiffness be denoted by 'K' for each mounting, then for 4 mountings it is 4K
We know that:
[tex]\omega = \frac{2 \pi\times N}{60}[/tex]
so,
[tex]\omega = \frac{2 \pi\times 720}{60}[/tex] = 75.39 rad/s
Using the given formula:
Damping is negligible, so, [tex]\zeta = 0[/tex]
[tex]\frac{A}{F}[/tex] will give the tranfer function
Therefore,
[tex]\frac{A}{F}[/tex] = [tex] \frac{1}{\sqrt{(4K - 120\ ^{2})}}[/tex]
[tex]0.1\times 10^{-3}[/tex] = [tex] \frac{1}{\sqrt{(4K - 120\ ^{2})}}[/tex]
Required stiffness coefficient, K = 173009 N/m = 173.01 N/mm