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Determine the steady state amplitude of the system shown.
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enter image description here

$m\ddot{x} = \ - k_b (x-y)$

$m\ddot{x} + k_b x = ky$

$m\ddot{x} + k_b x $ = k [0.005 sin 3st]

$m\ddot{x} + k_b x $ = 0.005 sin 3st

$K_{beam}\ = \ \frac{3 EI}{L^3}$

$K_{beam}\ = \frac{3 \times 210 \times 10^9 \times 4.1 \times 10^{-6}}{(1.8)^3}$

$K_{beam}\ = 442.90 \times 10^3$ N/m

$W_n = \sqrt{ \frac{K \ beam}{m}} = \sqrt{\frac{442.90 \times 10^3}{250}} = 42.09$ r/sec.

$r = \frac{w}{w_n} = \frac{35}{4209} = 0.831$

enter image description here

ANSLOM,

m$\ddot{x}$ = - k beam ( x – y)

$m\ddot{x} + k_b$x = k beam y

$m\ddot{x} + k_b$x = k (0.005. sin 35t)

$K_{eq}$ = $K_b$

$m_{eq}$ = m

Fo = (0.005) K

$\frac{X}{X_{st}} = \frac{1}{\sqrt{(1-r^2)^2 + (2 \xi r)^2}}$

$X_{st} = \frac{Fo}{k} = \frac{442.90 \times 10^3 \times 0.005}{442.90 \times 10^3}$

$X_{st} = 5 \times 10^{-3} \ m$

X = 0.016m

Case 2: Part 2: Problems in which wavelength is given that time:

  1. Amplitude ratio = $\frac{x}{y} = \frac{ \sqrt{ 1 + (2\zeta r)^2}}{\sqrt{ (1-r^2)^2 + (2 \xi r)^2}}$

  2. $r = \frac{w}{w_n},$ where, $w = 2 \pi f$ and $f = \frac{velocity}{\lambda \ (wave \ length)}$

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