Question 6.7: An accelerometer is designed to have a natural frequency of ...
An accelerometer is designed to have a natural frequency of 30 \mathrm{rad} / \mathrm{s} and a damping ratio of 0.7. It is calibrated to read correctly the input acceleration at very small values of the exciting frequency.
(a) What would be the percentage error in the instrument reading if the harmonic motion being measured has a frequency of 20 \mathrm{rad} / \mathrm{s} ?
(b) The displacement input imparted to the frame of the instrument is given by u_{g}=500 \sin 10 t+400 \sin 20 t. Plot the variation of the input acceleration with time. Obtain the curve of instrument reading versus time and compare it with the input.
(a) From Equation 6.89a,
\rho =\frac{m A}{k} \frac{1}{\sqrt{\left(1-\beta^2\right)^2+(2 \xi \beta)^2}} (6.89a)
\frac{\rho \omega^{2}}{A}=\frac{1}{\sqrt{\left(1-\beta^{2}\right)^{2}+(2 \xi \beta)^{2}}}
where A is the amplitude of input acceleration. For \beta=0, \rho \omega^{2} / A=1. The instrument is calibrated so that its reading is equal to \omega^{2} times \rho. This ensures that at low frequencies the instrument will read the amplitude of input acceleration correctly. For \beta=20 / 30=0.67 and \xi=0.7
\frac{\rho \omega^{2}}{A}=0.921
This implies that for an input frequency of 20 \mathrm{~Hz}, the instrument reading \rho \omega^{2} is 92.1 \% of A or 7.9 \% too low.
(b) The input acceleration is given by
\ddot{u}_{g}=500 \sin 10 t+400 \sin 20 t (a)
The two harmonic components of the input acceleration are plotted in Figure E6.6a and b, respectively. The instrument reading for the input motion with a frequency of 10 \mathrm{rad} / \mathrm{s} is given by
a_{1}=500 \times \frac{1}{\sqrt{\left(1-\beta_{1}\right)^{2}+\left(2 \xi \beta_{1}\right)^{2}}} \sin \left(10 t-\phi_{1}\right)
where \beta_{1}=10 / 30=1 / 3, \xi=0.7, and \tan \phi_{1}=2 \xi \beta_{1} /\left(1-\beta_{1}^{2}\right). Substitution of the given values yields
a_{1}=0.996 \times 500 \sin (10 t-0.483) (b)
In a similar manner, the instrument reading for the input motion with a frequency of 20 \mathrm{rad} / \mathrm{s} is given by
a_{2}=0.921 \times 400 \sin (20 t-1.034) (c)
Relationships given by Equations \mathrm{b} and \mathrm{c} are shown on Figure \mathrm{E} 6.7 \mathrm{a} and \mathrm{b} by dashed lines. It is observed that the curve for a_{1} is shifted by 0.483 / 10=0.0483 \mathrm{~s} with respect to the input acceleration, while the curve for a_{2} is shifted by 1.034 / 20=0.0517 \mathrm{~s}.
Finally, the total input acceleration is compared with the resultant instrument reading in Figure E6.7c. Because the shift for the two components is slightly different, the output signal is a bit distorted, but the distortion is fairly small.
