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.

The amplitude of vibration is

• A. 0.009 mm
• B. 0.006 mm
• C. 0.003 mm
• D. 0.01 mm

Answer 1

The correct option is A 0.009 mm

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

$\xi =0.3,$

$F=525N,$

$\omega =\frac{2\pi N}{60}=20\pi \text{rad/s}$

${\omega }_n=\sqrt{\frac{k}{m}}=\sqrt{\frac{1050\times 1000}{150}}=83.66\text{rad/s}$

$r=\frac{200\pi }{83.66}=7.5$

$A=\frac{\frac{525}{1050000}}{\sqrt{(1-(7.5{)}^2{)}^2+(2\times 0.3\times 7.5{)}^2}}$

$\xi =0.3,$

$=9\times {10}^{-6}\text{m}=0.009\text{mm}$