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Using the given values, determine the steady state amplitude of the block.
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$KE \ = \ (KE)_m \ + \ (KE) \ pulley$

$\frac{1}{2} \ mx^2 \ + \ \frac{1}{2} \ I \dot{θ}^2$

$= \frac{1}{2} \ m \ [r \theta]^2 \ + \ \frac{1}{2} \ I. \ \dot{θ}^2$

$= \frac{1}{2} \ [mr^2 + I] \ \dot{θ}^2$

$I_{eq} \ = \ mr^2 \ + \ I$

$= 10 \times \ (0.1)^2 \ + \ 0.1$

$= 0.2 \ kg-m^2$

$PE \ = \ \frac{1}{2} \ k(2r \theta)^2 \ + \ \frac{1}{2} \ k (r \theta)^2$

$= \frac{1}{2} \ [4Kr^2] \ \theta^2 \ + \ \frac{1}{2} \ kr^2 \ \theta^2$

$= \frac{1}{2} \ [4Kr^2 + kr^2] \ \theta^2$

$Keq \ = \ 4kr^2 \ + \ kr^2$

$= 5 \times \ 1.6 \times \ 10^5 \times \ (0.1)^2$

= 8000 N/m

$W_n \ = \ \sqrt{ \frac{Keq}{Ieq}} \ = \ \sqrt{ \frac{8000}{0.2}} \ = \ 200$ rad/sec

$W. D \ = \ - \int \ c \dot{y} \ dy$

$y \ = \ r \ \theta$

$\dot{y} \ = \ r \dot{θ}$

$C_{eq} \ = \ Cr^2$

$= 640 \ \times \ (0.1)^2$

$= 6.4 \ \frac{Ns}{m}$

$r = \frac{w}{w_n} = \frac{180}{200} = 0.9$

$\xi = \frac{C_{eq}}{C ceq} = \frac{6.4}{2 \ \times \ 0.2 \ \times \ 200} = 0.08$

$\theta_{st} = \frac{M_o}{K_{eq}} = \frac{100}{8000} = 0.0125$ rad/s

$\frac{\theta}{0.0125} = \frac{1}{\sqrt{ (1-(0.9)^2]^2 + (2 \ \times \ 0.08 \ \times 0.9)^2}}$

$\theta = 0.052$rad