Question

Solve the model for a driven spring/mass system with damping

$\mathbf{m}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{+}\mathbf{\beta }\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{+}\mathbf{kx}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

where the driving function f is as specie. Use a graphing utility to graph x(t) for the indicated values of t.$\mathbf{m}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{\beta }\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{k}\mathbf{=}\mathbf{1}$ 1, f is the square wave in Problem 54 with amplitude 5, and $\mathbf{a}\mathbf{=}\pi ,\mathbf{0}⩽\mathbf{t}⩽\mathbf{4}\pi$

Short answer

Laplace transform$x\left(t\right)=5\sum _{n=0}^{\infty }(-1{)}^n\left(1-e^{-(t-n\pi )}-(t-n\pi )e^{-(t-n\pi )}\right)U(t-n\pi )$

Step-by-step solution

Define Laplace transform:

The conversion of the function f (x) to the function $\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\int \infty \mathbf{0}\mathbf{e}\mathbf{-}\mathbf{xtf}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{dx}$is particularly useful in reducing the solution of a standard dividing line equation with constant coefficients in the polynomial equation solution

Find :xt

$x\text{'}\text{'}+2x\text{'}+x=f\left(t\right)$

From this we can come to know,

$\left\{\begin{array}{l}\mathcal{L}\left\{x\text{'}\right\}=s\mathcal{L}\left\{x\right\}-x\left(0\right)\\ \mathcal{L}\left\{x\text{'}\text{'}\right\}=s^2\mathcal{L}\left\{x\right\}-sx\left(0\right)-x\text{'}\left(0\right)\end{array}\right.$

Take Laplace transform on both side of equation,

$$\begin{gathered} \mathcal{L}\left\{f\right(t\left)\right\}=\mathcal{L}\left\{x\text{'}\text{'}+2x\text{'}+x\right\} \\ =\mathcal{L}\left\{x\text{'}\text{'}\right\}+2\mathcal{L}\left\{x\text{'}\right\}+\mathcal{L}\left\{x\right\} \end{gathered}$$

$\stackrel{\left(*\right)}{=}{s}^2\mathcal{L}\left\{x\right\}-sx\left(0\right)-x\text{'}\left(0\right)+2s\mathcal{L}\left\{x\right\}-2x\left(0\right)+\mathcal{L}\left\{x\right\}\left(x\left(0\right)=0,x\text{'}\left(0\right)=0\right)$

$=\left[s^2+2s+1\right]\mathcal{L}\left\{x\right\}$

Consider $\mathcal{L}\left\{x\right\}=X\left(s\right)$replace in above equation

$\mathcal{L}\left\{f\right(t\left)\right\}=\left[s^2+2s+1\right]X\left(s\right)$-----------------------------(1)

From solution 54,

$\mathcal{L}\left\{f\right(t\left)\right\}=\frac{5}{s\left(1+e^{-as}\right)}(a=\pi )$

$\mathcal{L}\left\{f\right(t\left)\right\}=\frac{5}{s\left(1+e^{-\pi s}\right)}$

Substitute this equation in equation (1)

$\left[s^2+2s+1\right]X\left(s\right)=\frac{5}{s\left(1+e^{-\pi s}\right)}$

$X\left(s\right)=\frac{5}{s\left(s^2+2s+1\right)\left(1+e^{-\pi s}\right)}$-----(2)

Evaluate the equation using partial fraction technique:

$\frac{5}{s\left(s^2+2s+1\right)}=\frac{A}{s}+\frac{Bs+C}{s^2+2s+1}$---------------------(3)

$=\frac{A\left(s^2+2s+1\right)+B{s}^2+Cs}{s\left(s^2+2s+1\right)}$

Since numerator and denominator of both sides are same,

$=\frac{A\left(s^2+2s+1\right)+B{s}^2+Cs}{s\left(s^2+2s+1\right)}$

$$\begin{gathered} 5=A{s}^2+2As+A+B{s}^2+Cs \\ 5=(A+B)s^2+(2A+C)s+A \end{gathered}$$

Compare coefficient on both sides

$$\begin{gathered} A+B=0 \\ 2A+C=0 \\ \underline{A}=5 \end{gathered}$$

$$\begin{gathered} B=-A \\ C=-2A \\ \underline{A=5} \end{gathered}$$

$$\begin{gathered} B=-5 \\ C=-10 \\ \underline{A-5} \end{gathered}$$

Substitute the value in equation (3)

$\frac{5}{s\left(s^2+2s+1\right)}=\frac{5}{s}-\frac{5s+10}{s^2+2s+1}$

Find: xs xs

$X\left(s\right)=\left[\frac{5}{s}-\frac{5s+10}{s^2+2s+1}\right]\frac{1}{\left(1+e^{-\pi s}\right)}$

$=\left[\frac{5}{s}-\frac{5s+10}{s^2+2s+1}\right]\left(1-e^{-\pi s}+e^{-2\pi s}-e^{-3\pi s}+\cdots \right)$

$=\sum _{n=0}^{\infty }(-1{)}^n\left[\frac{5}{s}-5\frac{s+1+1}{s^2+2s+1}\right]e^{-n\pi s}$

$=\sum _{n=0}^{\infty }(-1{)}^n\left[\frac{5}{s}-5\frac{s+1}{{(s+1)}^2}-5\frac{1}{{(s+1)}^2}\right]e^{-n\pi s}$

Inverse Laplace transform of both side

${\mathcal{L}}^{-1}\left\{X\right(s\left)\right\}=5\sum _{n=0}^{\infty }(-1{)}^n\left[{\mathcal{L}}^{-1}\left\{\frac{1}{s}{e}^{-n\pi s}\right\}-{\mathcal{L}}^{-1}\left\{\frac{1}{s+1}{e}^{-n\pi s}\right\}-{\mathcal{L}}^{-1}\left\{\frac{1}{{(s+1)}^2}{e}^{-n\pi s}\right\}\right]$

$\left\{\begin{array}{l}In{\mathcal{L}}^{-1}\left\{F\right(s\left)\right\}=f\left(t\right)anda>0thenf(t-a)U(t-a)={\mathcal{L}}^{-1}\left\{e^{-as}F\left(s\right)\right\}\\ {\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}=1,{\mathcal{L}}^{-1}\left\{\frac{1}{s-a}\right\}=e^{at},{\mathcal{L}}^{-1}\left\{\frac{1}{{(s-a)}^2}\right\}=t{e}^{at}\end{array}\right.$

$\stackrel{\left(**\right)}{=}5\sum _{n=0}^{\infty }\left(1-e^{-(t-n\pi )}-(t-n\pi )e^{-(t-n\pi )}\right)U(t-n\pi )$

Therefore, Laplace transform is

$x\left(t\right)={\mathcal{L}}^{-1}\left\{X\right(s\left)\right\}=5\sum _{n=0}^{\infty }(-1{)}^n\left(1-e^{-(t-n\pi )}-(t-n\pi )e^{-(t-n\pi )}\right)U(t-n\pi )$

Graph  :xt