Question

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

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

where the driving function f is as specied. Use a graphing utility to graph x(t) for the indicated values of t$\mathbf{m}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,}\mathbf{\beta }\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{k}\mathbf{=}\mathbf{5}\mathbf{,}\mathbf{f}$. is the meander function in Problem 53 with amplitude 10, and$\mathbf{a}\mathbf{=}\pi ,\mathbf{0}⩽\mathbf{t}⩽\mathbf{2}\pi$

Short answer

Laplace transform $x\left(t\right)=2\left(1-e^{-t}\cos 3t-\frac{1}{3}{e}^{-t}\sin 3t\right)$

$+4\sum _{n=0}^{\infty }(-1{)}^n\left(1-e^{-(t-n\pi )}\cos 3(t-n\pi )-\frac{1}{3}{e}^{-(t-n\pi )}\sin 3(t-n\pi )\right)U(t-n\pi )$

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  Xs :

$\frac{1}{2}x\text{'}\text{'}+x\text{'}+5x=f\left(t\right)$

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,

$$\begin{gathered} \mathcal{L}\left\{f\right(t\left)\right\}=\mathcal{L}\left\{\frac{1}{2}x\text{'}\text{'}+x\text{'}+5x\right\} \\ =\frac{1}{2}\mathcal{L}\left\{x\text{'}\text{'}\right\}+\mathcal{L}\left\{x\text{'}\right\}+5\mathcal{L}\left\{x\right\} \end{gathered}$$

$\stackrel{\left(*\right)}{=}\frac{1}{2}\left[s^2\mathcal{L}\left\{x\right\}-sx\left(0\right)-x\text{'}\left(0\right)\right]+s\mathcal{L}\left\{x\right\}-x\left(0\right)+5\mathcal{L}\left\{x\right\}\left(x\left(0\right)=0,x\text{'}\left(0\right)=0\right)$

$=\left[\frac{1}{2}{s}^2+s+5\right]\mathcal{L}\left\{x\right\}$

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

$\mathcal{L}\left\{f\right(t\left)\right\}=\left[\frac{1}{2}{s}^2+s+5\right]X\left(s\right)$--------------------(1)

From solution 53,

$\mathcal{L}\left\{f\right(t\left)\right\}=\frac{10\left(1-e^{-as}\right)}{s\left(1+e^{-as}\right)}(a=\pi )$

$\mathcal{L}\left\{f\right(t\left)\right\}=\frac{10\left(1-e^{-\pi s}\right)}{s\left(1+e^{-\pi s}\right)}$

Substitute this equation in equation (1)

$$\begin{gathered} \left[\frac{1}{2}{s}^2+s+5\right]X\left(s\right)=\frac{10\left(1-e^{-\pi s}\right)}{s\left(1+e^{-\pi s}\right)} \\ \left[s^2+2s+10\right]X\left(s\right)=\frac{20\left(1-e^{-\pi s}\right)}{s\left(1+e^{-\pi s}\right)} \end{gathered}$$

Evaluate the equation using partial fraction technique:

$\frac{20}{s\left(s^2+2s+10\right)}=\frac{A}{s}+\frac{Bs+C}{s^2+2s+10}$

$=\frac{A\left(s^2+2s+10\right)+B{s}^2+Cs}{s\left(s^2+2s+10\right)}$

Have same numerator and denominator

$$\begin{gathered} 20=A{s}^2+2As+10A+B{s}^2+Cs \\ 20=(A+B)s^2+(2A+C)s+10A \end{gathered}$$

Compare coefficient on both sides,

$$\begin{gathered} A+B=0 \\ 2A+C=0 \\ 10A=20 \end{gathered}$$

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

$$\begin{gathered} B=-2 \\ C=-4 \\ A=2 \end{gathered}$$

Substitute the value in equation (3)

$\frac{20}{s\left(s^2+2s+10\right)}=\frac{2}{s}-\frac{2s+4}{s^2+2s+10}$

Find :Xs

$$\begin{gathered} X\left(s\right)=\left(1-e^{-\pi s}\right)\left[\frac{2}{s}-\frac{2s+4}{s^2+2s+10}\right]\frac{1}{\left(1+e^{-\pi s}\right)} \\ =\left(1-e^{-\pi s}\right)\left[\frac{2}{s}-\frac{2s+4}{s^2+2s+10}\right]\left(1-e^{-\pi s}+e^{-2\pi s}-e^{-3\pi s}+\cdots \right) \end{gathered}$$

$$\begin{gathered} =\left[\frac{2}{s}-\frac{2s+4}{s^2+2s+10}\right]\left(1-2e^{-\pi s}+2e^{-2\pi s}-2e^{-3\pi s}+\cdots \right) \\ =\frac{2}{s}-\frac{2(s+1)+2}{{(s+1)}^2+9}+2{\sum _{n=1}^{\infty }(-1)}^n\left[\frac{2}{s}-\frac{2s+4}{s^2+2s+10}\right]e^{-n\pi s} \end{gathered}$$

$=\frac{2}{s}-\frac{2(s+1)}{{(s+1)}^2+9}-\frac{2}{{(s+1)}^2+9}+4\sum _{n=1}^{\infty }(-1{)}^n\left[\frac{1}{s}-\frac{s+1}{{(s+1)}^2+9}-\frac{1}{{(s+1)}^2+9}\right]e^{-n\pi s}$

Inverse Laplace transform of both side

${\mathcal{L}}^{-1}\left\{X\right(s\left)\right\}=2{\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}-2{\mathcal{L}}^{-1}\left\{\frac{s+1}{{(s+1)}^2+3^2}\right\}-\frac{2}{3}{\mathcal{L}}^{-1}\left\{\frac{3}{{(s+1)}^2+3^2}\right\}$

$+4\sum _{n=1}^{\infty }(-1{)}^n\left[{\mathcal{L}}^{-1}\left\{\frac{1}{s}{e}^{-n\pi s}\right\}-{\mathcal{L}}^{-1}\left\{\frac{s+1}{{(s+1)}^2+3^2}{e}^{-n\pi s}\right\}-\frac{1}{3}{\mathcal{L}}^{-1}\left\{\frac{3}{{(s+1)}^2+3^2}{e}^{-n\pi s}\right\}\right]$

role="math" localid="1664029728797" $$\begin{gathered} \stackrel{\left(**\right)}{=}2\left(1-e^{-t}\cos 3t-\frac{1}{3}{e}^{-t}\sin 3t\right) \\ \sum _{n=1}^{\infty }{(-1)}^n\left(1-e^{-(t-n\pi )}\cos 3(t-n\pi )-\frac{1}{3}{e}^{-(t-n\pi )}\sin 3(t-n\pi )\right)U(t-n\pi ) \end{gathered}$$

Therefore, Laplace transform is

$x\left(t\right)={\mathcal{L}}^{-1}\left\{X\right(s\left)\right\}=2\left(1-e^{-t}\cos 3t-\frac{1}{3}{e}^{-t}\sin 3t\right)$

$+4\sum _{n=1}^{\infty }(-1{)}^n\left(1-e^{-(t-n\pi )}\cos 3(t-n\pi )-\frac{1}{3}{e}^{-(t-n\pi )}\sin 3(t-n\pi )\right)U(t-n\pi )$

Graph  xt :