Question

A vibratory body of mass 150 kg supported on springs of total stiffness 1050 kN/m has rotating unbalance force of 525 N at speed of 6000 rpm. If the damping factor is 0.3.

Phase angle of vibration is

• A. ${157}^{\circ }$
• B. ${60}^{\circ }$
• C. ${120}^{\circ }$
• D. ${175}^{\circ }$

Answer 1

The correct option is D ${175}^{\circ }$

$\varphi ={\tan }^{-1}\left(\frac{2\pi r}{1-r^2}\right)$

$={\tan }^{-1}\left(\frac{2\times 0.3\times 7.5}{1-(7.5{)}^2}\right)$

$\varphi ={175.2}^{\circ }$