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For what excitation frequencies will the steady state amp. Of the machine of fig. be less than 1.5mm.
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For what excitation frequencies will the steady state amp. Of the machine of fig. be less than 1.5mm.

$w_n = \sqrt{ \frac{k}{m}} = \sqrt{ \frac{3.1 \times 10^5}{125}}$

$\frac{x}{Xst} = \frac{1}{\sqrt{ (1-r^2)^2 + (2 \xi r)^2}}$

$Xst = \frac{fo}{k} = \frac{400}{3.1 \times 10^5} = 1.29 \times 10^{-3} \ m$

$\xi = \frac{c}{c_c} = \frac{925}{2 \times 125 \times 49.79}$

$\xi$ = 0.074

$\frac{x}{Xst} = \frac{1.5 \times 10^{-3}}{1.29 \times 10^{-3}} = \frac{1}{\sqrt{ (1-r^2)^2 + (2 \times 0.074 \times r)^2}}$

$1.162 = \frac{1}{\sqrt{(1-r^2)^2 + (0.0219r^2)}}$

$1.3502 = \frac{1}{(10r^2)^2 + (0.0219r^2)}$

$1 – 2 r^2 + r^4 + 0.219 r^2 = 0.7406$

$0.2593 = 2r^2 – r^4 – 0.0219 r^2$

$r^4 – 1.9781 r^2 + 0.2593 = 0$

Assume $r^2 = x$

$x^2 – 1.9781x + 0.2593 = 0$

$x_1 = 1.836$

$x_2 = 0.141$

$r_1 = \sqrt{x_1} = \sqrt{1.836} = 1.35 \ m$

$r_2 = \sqrt{x_2} = \sqrt{0.141} = 0.37 \ m$

$r_1 = \frac{w_1}{w_n}$

$w_1 = r_1 \times w_n$

$= 1.34 \times 49.79$

$r_2 = \frac{w_2}{w_n}$

$w_2 = r_2 . w_n$

$= 0.37 \times 49.79$

= 18.42 r/sec

Case 2: Forced vibration with rotating unbalance.

List of formulae

$m\ddot{x} + c\dot{x} + kx = mo.e.w^2. \ sin \ wt$

1. $Fo = m0.e.w^2$

2. $X = \frac{fo}{\frac{k}{\sqrt{(1-r^2)^2 + (2 \xi r)^2}}}$

$X = \frac{ \frac{mo.e.w^2}{k} \times \frac{m}{m}}{\sqrt{(1-r^2)^2 + (2 \xi r)^2}}$

$\frac{X}{\frac{mo.e}{m}} = \frac{r^2}{\sqrt{(1-r^2)^2 + (2 \xi r)^2}}$

3. Unbalance = $mo .e = \frac{Fo}{w^2}$