If the system is initially at rest and a velocity of 10cm $s^{-1}$ is imparted to the mass. Determine

1.the displacement and velocity of the mass as a lime function.

2.The displacement at t = 1 sec. Now, if an excitation force of (24 sin 15t) N is applied to the mass, find the steady state response of the system.

m = 20kg

Fo = 24N

W = 15 r/sec

K = $8 \ \times \ 10^3 \ N/m$

C = 130 Ns/m

$\dot{x}$ = $10 \ \times \ 10^{-2} \ m/sec$

1] Expression for x and x in terms of t.

2] Find x at t = 1 sec

3] SSA = x = ?

1] $\xi \ = \ \frac{c}{c_e} = \frac{130}{2 \times 20 \times \sqrt{ \frac{8 \times 10^3}{20}}}$

$\xi \ = \ 0.1625 \ \lt \ 1$

$\therefore$ system is under damped.

$\therefore$ displacement for $\xi \lt 1$ is given by

$x = x [ \frac{e^{-\xi w_nt}}{u} \ sin \ (wt + \phi)]$ - - - - [1]

Boundary conditions are

1]T = 0, x = 0

2] T = 0, $\dot{x}$ = $10 \times \ 10^{-2}$ m/sec

Using first boundary condition,

$0 \ = \ X \ sin \ \phi$

As $X \ = \ \neq \ 0$

$Sin \ \phi \ = \ 0$

$\phi \ = \ 0 $

$\dot{x} = \ x [sin \ (w_d t) (- \xi . w_n).e^-\xi w_nt + e^{-\xi w_nt} cos (w_d. t) . w_d]$ - - - [2]

$10 \times 10^{-2} = x [w_d]$

$10 \times 10^{-2} = x$

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

$\frac{10 \times 10^{-2}}{20 \sqrt{ 1 – (0.1625)^2}} = X$

$X = 5.067 \times 10^{-3} m$

Substitute $\phi$ and X in equation [1]

$x = 5067 \times 10^{-3} [ e^{-325t} \ sin \ (19.7st)]$ - - - - [3]

From equation [2]

$\dot{x} = 5.067 \times 10^{-3} [ sin \ (19.73t)e^{-3.25t} \times (-3.25) + e^{-3.25t} \times cos(19.73t) \times 19.73]$

2] From equation 3, at t = 1 sec

$x = 5.067 \times 10^{-3} [0.038 \times sin \ 19.73]$

$x = 6.63 \times 10^{-3} m$

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

$r = \frac{w}{w_n} = \frac{15}{20} = 0.75$

$X_{st} = \frac{f_o}{k} = \frac{24}{8 \times 10^3} = 3 \times 10^{-3} \ m$

$\frac{X}{3 \times 10^{-3}} = \frac{1}{\sqrt{ (1 – (0.75)^2 + (2 \times 0.1625 \times 0.75)^2}}$

$= 5.99 \times 10^{-3} \ m$