Question

$\text{A 35-kg flow monitoring device is placed on a table in a laboratory. A pad of stiffness}2\times {10}^5\text{N/m and damping ratio 0.08 is placed between the apparatus and the table. The table is bolted to the laboratory floor. Measurements indicate that the floor has a steady-state vibration amplitude of 0.5 mm at a frequency of 30 Hz. What is the amplitude of acceleration of the flow monitoring device?}$

• A. $1.22{\text{m/s}}^2$
• B. $2.44{\text{m/s}}^2$
• C. $3.66{\text{m/s}}^2$
• D. $4.88{\text{m/s}}^2$

Answer 1

The correct option is C $3.66{\text{m/s}}^2$

The natual frequency and frequency ratio are,

$\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{2\times {10}^5}{35}}=75.6\frac{\text{rad}}{\text{s}}$

$r=\frac{\omega }{{\omega }_n}=\frac{2\pi \times 30}{75.6}=2.49$

$\text{The amplitude of absolute displacement of the flow measuring devices is calculated using equation,}X=Y\sqrt{\frac{1+(2\xi r{)}^2}{(1-r^2{)}^2+(2\xi r{)}^2}}$

$=0.0005\times \sqrt{\frac{1+[2\times 0.08\times 2.49{]}^2}{[1-(2.49{)}^2{]}^2+[2\times 0.08\times 2.49{]}^2}}$

$=1.03\times {10}^{-4}m$

The acceleration amplitude is

$A={\omega }^{2}X$

$[2\pi \times 30{]}^2(1.03\times {10}^{-4})$

$=3.66{\text{m/s}}^2$