Determine the necessary stiffness and the damping constant of a accelerator if the maximum error is to be limited to 3 percent for measurement in the frequency range of 0 to 100Hz. Assume that the suspended mass is 0.05kg.

For maximum error.

$(\frac{\ddot{Z}}{\ddot{Y}})_{max} \ = \ P_{max} \ = \ \frac{1}{2 \xi\sqrt{1 - \xi^2}}$

In accelerometer always select less values of ratio.

i.e. $\frac{\ddot{Z}}{\ddot{Y}} = 1.03$ OR 0.97.

$\therefore$ 0.97 is correct value.

$0.97 = \frac{1}{2 \xi\sqrt{1 - \xi^2}}$

$0.97 \times \xi \sqrt{1 - \xi^2} = 1$

$1.94 \xi \sqrt{1 - \xi^2} =1$

$\xi \sqrt{1 - \xi^2} = 0.5154$

$\xi^2 (1- \xi^2) = 0.265$

$\xi^2 - \xi^4 – 0.265 = 0$

$- \xi^4 + \xi^2 – 0.265 = 0$

$\xi = 0.707$

But $\xi \lt \frac{1}{\sqrt{2}} = 0.707$

Go for $\frac{\ddot{Z}}{\ddot{Y}} = 1.03$

$1.03 = \frac{1}{2 \xi \sqrt{ 1 - \xi^2}}$

$1.03 \times 2 \xi \sqrt{1 - \xi^2} = 1$

$2.06 \xi \sqrt{1 - \xi^2} = 1$

$\xi \sqrt{1 - \xi^2} = 0.485$

$\xi (1 - \xi^2) = 0.235$

$\xi^4 - \xi^2 + 0.235 = 0$

$\xi^2 = 0.622$ $\xi^2 = 0.377$

$\xi = 0.788$ $\xi = 0.614$

$r_p = \frac{w_p}{w_n}$

$r_p = \sqrt{1 – 2 \xi^2}$ - - - - standard for peak.

$= \sqrt{ 1 – 2 \times 0.614^2}$

$r_p$ = 0.495

$w_p = 2\pi 100$

= 628.31 rad/sec

$r = \frac{w_p}{w_n}$

$w_n = \frac{w_p}{r} = \frac{628.31}{0.495} = 1269.33$

$w_n = \sqrt{\frac{k}{m}}$

$k = 80.55 \times 10^3 \frac{N}{m}$