# Vibration Absorber question

• jasonbot
You even remembered to add the 12 kg to the mass of the compressor. I don't see any mistakes. The only thing I might suggest is that you state that you are solving for the changed displacement amplitude in your summary. So it would be "In summary, the changed displacement amplitude for the overall system is -0.000109m." But overall, you did a great job summarizing the content.

## Homework Statement

An air compressor of mass m1 = 150 kg is mounted on a spring of stiffness k1= 3000 N/m. The measured
displacement amplitude X1 = 0.0002 m at the operating speed of N = 1800 r/min. If an undamped isolation
system with mass m2=12 kg and a natural frequency w2 = 105 rad/s is fixed to the compressor, calculate the
changed displacement amplitude X1 for the overall system

## Homework Equations

As far as I understand this is the relevant equation: X1/delta_st=(1-(omega/omega2)^2)/((1+k2/k1-(omega/omega_1)^2)(1-(omega/omega_2)^2)-k2/k1)

## The Attempt at a Solution

As far as I understand its a vibration absorber problem.

omega1=(k1/m1)^.5=4.47
k2=omega_2^2*m2=105^2*12=132300N/m

X1/delta_st=(1-(omega/omega2)^2)/((1+k2/k1-(omega/omega_1)^2)(1-(omega/omega_2)^2)-k2/k1)

X1/delta_st=-5,836x10^-4

delta_st=-0,342m

I don't think that's right :|

It is a dynamic absorber problem. While I have not gone through your arithmetic, I see that delta_st is static deflection which is how much the weight of the compressor deflects the spring under static conditions.

You should be solving for X1. Here is your equation: X1/delta_st=-5,836x10^-4
Solve it for X1 by plugging in delta_st.

That makes sense actually.

So assuming that I go

delta_st=F0/k=(150+12)*9.81/3000=0.52974

X1=delta_st*-5,836x10^-4=-0.000309

so then I'm assuming the changed displacement amplitude will be the difference between measured and X1 from calculations.

so X1'=-0.000309+0.0002=-0.000109m

Oh and thanks! <--- Sorry I forgot to add that :|

Last edited:
That looks good to me. I went through your arithmetic...