Question

A seismic instrument has natural frequency of 40 Hz and a damping ratio of 0.66. If the system is excited by a frequency of 60 Hz, then the error due to proximity of excited frequency with natural frequency of the instrument is

• A. -12.3%
• B. -9.6%
• C. - 3.9%
• D. -4.8%

Answer 1

The correct option is C - 3.9%

The ratio of output displacement to input displacement is given by

$\frac{(x_2-x_1)M}{x_{M_1}}=\frac{u^2}{\sqrt{(1-u^2{)}^2+(2\xi u{)}^2}}$

Now, normalized frequency, $u=\frac{60}{40}=1.5$

$\frac{(x^2-x_1)M}{x_{M_1}}=\frac{(1.5{)}^2}{\sqrt{(1-{1.5}^2{)}^2+(2\times 0.66\times 1.5{)}^2}}=0.961$

$Error=(0.961-1)\times 100=-3.9\%$