Question

A machine of 250 kg mass is supported on springs of total stiffness of 100 kN/m. Machine has an unbalanced rotating force of 350 N at speed of 3600 rpm. Assuming a damping factor of 0.15, the value of transmissibility ratio is

• A. 0.0531
• B. 0.9922
• C. 0.0162
• D. 0.0028

Answer 1

The correct option is C 0.0162

$\omega =\frac{2\pi \times 3600}{60}=377\text{rad/s}$

Natural frequency,

${\omega }_n=\sqrt{\frac{k}{m}}=\sqrt{\frac{100\times 1000}{250}}=20\text{rad/s}$

$r=\frac{\omega }{{\omega }_n}=\frac{377}{20}=18.85$

Transmissibility ratio,

$TR=\frac{\sqrt{1+(2\xi r{)}^2}}{\sqrt{(1-r^2{)}^2+(2\xi r{)}^2}}$

$=\frac{\sqrt{1+(2\times 0.15\times 18.85{)}^2}}{\sqrt{(1-(18.85{)}^2{)}^2+(2\times 0.15\times 18.85{)}^2}}$

$=0.0162$