Answer to Question #126851 in Mechanical Engineering for hamdety

Question

A compressor of 25 kg is mounted on isolator as shown in the figure.

The isolator is idealized by a spring and a damper of damping ration 𝜁= 0.3 . it is found that the compressor suffers from vibration due to unbalance of 0.001 sin50t. also, it is found that all the maximum impressed force is transmitted to the floor. What will be the TR and the maximum F_t  to the floor when the excitation force becomes 0.005sinwt ?

Answer 1

Here, in this question figure is missing so i m assuming my own figure and its related data.

mass of compressor =25 kg,here it is given idealized spring constant = $700\times 10^3$ N/m,

and damping ratio

,$\zeta=0.3$ , $F=F_osinwt$ ,$\omega_n=\sqrt{\frac{k}{m}}$

$\omega_n=\sqrt{\frac{700000}{25}}=167.33$ ,

$\omega=314.2 rad$ ,

$r=\frac{\omega}{\omega_n}=\frac{314.2}{167.33}=1.877$

$TR=\frac{1}{r^2-1}=\frac{1}{1.877^2-1}=0.396$

$\frac{F_T}{F_o}=\sqrt{\frac{1+(2\zeta r^2)}{(1-r^2)^2+(2\zeta r)^2}}$

$\frac{F_T}{0.001}=\sqrt{\frac{1+(2\times0.3\times 1.877^2)}{(1-1.877^2)^2+(2\times0.3\times 1.877)^2}}$

$F_T=0.000407$