How Does Adding a Mass Impact Vibration Amplitude in an Undamped System?

[jasonbot]

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.

omega=1800*(2π/60)=188,5rad/s
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 :|

 

[LawrenceC]

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.

 

[jasonbot]

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=-0.000109mOh and thanks! <--- Sorry I forgot to add that :|

 

[LawrenceC]

That looks good to me. I went through your arithmetic...