Question

A machine of 250 kg mass is supported on springs of total stiffness 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:
1. 0.0531
2. 0.9922
3. 0.0162
4. 0.0028

Answer 1

Correct Answer - Option 3 : 0.0162

\(\omega = \frac{{2\pi \times 3600}}{{60}} = 377rad/s\)

Natural frequency, \({\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{{100 \times 1000}}{{250}}} = 20\ rad/s\)

Now, \(r = \frac{\omega }{{{\omega _n}}} = \frac{{377}}{{20}} = 18.85\)

Transmissibility ratio \(\left( {TR} \right) = \frac{{\sqrt {1 + {{\left( {2\xi r} \right)}^2}} }}{{\sqrt {{{\left( {1 - {r^2}} \right)}^2} + {{\left( {2\xi r } \right)}^2}} }}\)

\(= \frac{{\sqrt {1 + {{\left( {2 \times 0.15 \times 18.85} \right)}^2}} }}{{\sqrt {\left[ {1 - {{\left( {18.85} \right)}^2}} \right]^2 + {{\left( {2\times0.15 \times 18.85} \right)}^2}} }}\)

= 0.0162