How do you solve this problem

$3\mathrm{.}24$: A machine resting on an elastic support can be modeled as a single$-$d

$$

$$ egree$-$of$-$freedom, spring$-$mass system arranged in the vertical direction. The ground is subject to a motion $y(t)$ of the form illustrated in figure P$3\mathrm{.}24.$ The machine has a mass of $5000$ kg $,$ and the support has stiffness $1\mathrm{.}5{10}^3\frac{N}{m}.$ Calculate the resulting vibration of the machine.

Figure P $3\mathrm{.}24$ : Triangular pulse input function.

$10$

Solution:

Griven, $m=5000kg,k=1\mathrm{.}5{10}^3\frac{N}{m},{}_n=\sqrt[\phantom{2}]{\frac{k}{m}}=0\mathrm{.}548ra\frac{d}{s}$ and that the ground motion is given by:

$y(t)=\{\begin{array}{cc}2\mathrm{.}5t,& 0t0\mathrm{.}2\\ 0\mathrm{.}75-1\mathrm{.}25t,& 0\mathrm{.}2t0\mathrm{.}6\\ 0,& t0\mathrm{.}6\end{array}$

The equation of motion is

$m{x}^{}+k(x-y)=0$

$=>m{x}^{}+kx=ky.$

$=>m{x}^{}+kx=F(t)$

The impulse response function computed from equn. $3\mathrm{.}12$ for an undamped system is

$h(t-)=\frac{1}{m{}_n}sin{}_n(t-)$

This gives the solution by integrating

$x(t)=\frac{1}{m{}_n}{\displaystyle \underset{0}{\overset{t}{}}}ky()sin{}_n(t-)d$

$={}_n{\displaystyle \underset{0}{\overset{t}{}}}f()sin{}_n(t-)d.$

for the interval $0t0\mathrm{.}2$ :

$x(t)={}_n{\displaystyle \underset{0}{\overset{t}{}}}2\mathrm{.}5sin{}_n(t-)d$

$=>x(t)=2\mathrm{.}5t-4\mathrm{.}56sin0\mathrm{.}548tmm$;$,0t0\mathrm{.}2$

$11$

for the interral $0\mathrm{.}2t0\mathrm{.}6$

$x(t)={}_n{\displaystyle \underset{0}{\overset{0\mathrm{.}2}{}}}2\mathrm{.}5sin{}_n(t-)d$

$+{}_n{\displaystyle \underset{0\mathrm{.}2}{\overset{t}{}}}(0\mathrm{.}75-1\mathrm{.}25)sin{}_n(t-)d$

$=0\mathrm{.}75-0\mathrm{.}5cos0\mathrm{.}548(t-0\mathrm{.}2)-1\mathrm{.}25t+2\mathrm{.}28sin0\mathrm{.}548(t-0\mathrm{.}2)$

Combining this with the solution from the first interval Jields:

$x(t)=0\mathrm{.}75+1\mathrm{.}25t-0\mathrm{.}5cos0\mathrm{.}548(t-0\mathrm{.}2)$

$+6\mathrm{.}48sin0\mathrm{.}548(t-0\mathrm{.}2)-4\mathrm{.}56sin0\mathrm{.}548(t-0\mathrm{.}2)mm$;

$0\mathrm{.}2t0\mathrm{.}6$

finally for the interval $t0\mathrm{.}6,$

$x(t)={}_n{\displaystyle \underset{0}{\overset{0\mathrm{.}2}{}}}2\mathrm{.}5tsin{}_n(t-)d$

$+{}_n{\displaystyle \underset{0\mathrm{.}2}{\overset{0\mathrm{.}6}{}}}(0\mathrm{.}75-1\mathrm{.}25t)sin{}_n(t-)d$

$+{}_n{\displaystyle \underset{0}{\overset{t}{}}}(0)sin{}_n(t-)d$

$=-0\mathrm{.}5cos0\mathrm{.}548(t-0\mathrm{.}2)-2\mathrm{.}28sin0\mathrm{.}548(t-0\mathrm{.}6)$

$+2\mathrm{.}28sin0\mathrm{.}548(t-0\mathrm{.}2)$