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An accelerometer has a suspended mass of 0.01 kg with a damped natural frequency of vibration 150 Hz
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When mounted on an engine undergoing an acceleration of 9.81 $m/s^2$ by the instrument. Find the damping cmst and the spring stiffness of the accelerometer. Take recorded acceleration $\alpha. 6m/s^2$ when m/c runs at 5000 rpm.

m = 0.01 kg

$F_d$ = 160 Hz

$W_d$ = $2 \pi \times 160$

= 1005.30 Hz

$\ddot{z} = 9.6 \ m/s^2$

$\ddot{y} = 9.8 \ m/s^2$

$N = 5000 \ rpm$

$w = \frac{2 \pi \times 5000}{60} = 523.59 \ Hz$

$\rightarrow$ $w_d = w_n \sqrt{1 - \xi^2}$

$1005.30 = w_n \sqrt{1 - \xi^2}$

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

$w_n = \frac{523.598}{r}$

$\therefore$ $\frac{1005.30 \times r}{523.598} = \sqrt{1 - \xi^2}$

$1.92 \times r = \sqrt{1 - \xi^2}$

$3.69 r^2 = 1 - \xi^2$

$\xi^2 = 1- 3.69 r^2$

$\frac{\ddot{z}}{\ddot{y}} = \frac{1}{\sqrt{ (1-r^2)^2 + (2 \xi r)^2}}$

$\frac{9.6}{9.81} = \frac{1}{\sqrt{(1- r^2)^2 + 4 \times (1- 3.69r^2)}}$

$0.978 = \frac{1}{\sqrt{ (1-r^2)^2 + 4 r^2 (1- 3.69 r^2)}}$

$\sqrt{(1-r^2)^2 + 4 r^2 (1- 3.69 r^2)} = 1.0224$

$1 – 2r^2 + r^4 + 4 r^2 – 14.76 r^4 = 1.0453$

$r^4 – 14.76 r^4 + 4 r^2 – 2 r^2 + 1 – 1.0453 = 0$

$- 13.76 r^4 + 2 r^2 – 0.0453 = 0$

$r ^2 = 0.028$ R = 0.1673

OR $r^2 = 0.1172$ R = 0.3423

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

$0.3423 \times w_n = 523.598$

$w_n = 1529.446$

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

$K = 23.29 \ \frac{kN}{m}$