Question

Figure (a) shows a mechanical vibratory system. When 2 N of force (step input) is applied to system, the mass Oscillates, as shown in Figure (b). Assuming that the mechanical system show here is at rest before excitation force f(t) = 2 N is given. The displacement x is measured from the equilibrium position. f(t) = 2 N 0.0095m (m) (sech chile (a) . . (B) How to Sety 3-1. Draw the free-body-diagram and drive the mathematical model of the system. 3-2. Find the transfer function of the system using 's' (complex variable). (3-3. Determine m, b, and k of the system from the response curve in Figure (b).

Please solve 3-1, 3-2, 3-3.

I am most interested in the solution for 3-3 however.

Please show all steps clearly

Thank you

Answer 1

1)

$m\ddot x = f(t) - F_s-F_D$

$\Rightarrow m\ddot x = f(t) - kx - b\dot x$

$\Rightarrow m\ddot x+ kx + b\dot x = f(t)$

2)

Take Laplace transform of the above equation

$L[m\ddot x+ kx + b\dot x] =L[ f(t)]$

assuming zero initial conditions

$(ms^2+bs+k)X(s)=F(s)$

$\Rightarrow \frac{X(s)}{F(s)} = \frac{1}{(ms^2+bs+k)}$

3)

f(t) = 2N. So, F(s) = 2/s

$\Rightarrow X(s) = \frac{1}{(ms^2+bs+k)}*\frac{2}{s}$

For the given response.

steady-state value is 0.1

$\Rightarrow X(s) _{ss}= \lim_{s\rightarrow 0}s*\frac{1}{(ms^2+bs+k)}*\frac{2}{s} = 0.1$

$\Rightarrow \frac{2}{k} = 0.1\Rightarrow k = 20N/m$

percentage overshoot:(standard second-order system)

$\large \%OS = \frac{0.0095}{0.1} *100= 9.5 = e^{\frac{-\zeta \pi}{\sqrt{1-\zeta^2}}}*100$

$\large \Rightarrow e^{\frac{-\zeta \pi}{\sqrt{1-\zeta^2}}} = 0.095$

$\large \Rightarrow\frac{-\zeta \pi}{\sqrt{1-\zeta^2}} = ln(0.095)$

$\large \Rightarrow \zeta = 0.5996$

settling time is 4 seconds:(standard second-order system)

$t_s(2\%) = \frac{4}{\zeta w_n} = 4sec\Rightarrow \zeta w_n = 1$

$\Rightarrow w_n = \frac{1}{0.5596} = 1.6678$

For a spring-mass system, we know that:

$\Rightarrow w_n =\sqrt{ \frac{k}{m} }= 1.6678$

but k = 20

$\Rightarrow m = 7.19kg$

Also,

$\frac{b}{m} = 2\zeta w_n = 2*1 = 2$

$\Rightarrow b = 2*7.19 = 14.38Ns/m$

Ans: m = 7.19;k = 20;b = 14.38