Question

If the impressed force $f\left(t\right)5sint$acts on the system for $0⩽t<2\pi$and is then removed.

Short answer

The inverse Laplace transform equation is,

$x\left(t\right)=\frac{1}{15}Sint-\frac{1}{60}Sin\left(4t\right)-\frac{1}{15}Sin(t-2\pi )U(t-2\pi )+\frac{1}{60}Sin\left(4\right(t-2\pi \left)\right)U(t-2\pi )$

Step-by-step solution

Definition of "inverse" Transform Laplace

The transformation of a Laplace transform into a function of time is known as the inverse Laplace transform.

Determine the inverse Laplace transform.

Weight $\left(w\right)=32\text{lb}$

$32\text{lb}$weight stretches the spring $2\text{ft}$.

$$\begin{gathered} 32=k·2k \\ =16\text{lb}/\text{ft} \end{gathered}$$

We know that $m=\frac{w}{g}$ Where, $g=32\text{ft}/{\text{Sec}}^2$ Is the acceleration due to gravity.

Substituting $w=32\text{lb}$ and $g=32\text{ft}/{\text{Sec}}^2$ , we get

$$\begin{gathered} m=\frac{w}{g} \\ =\frac{32}{32} \\ =1\text{(slugs)} \end{gathered}$$

The weight is released from the equilibrium point, which it represents $x\left(0\right)=0$. This is an initial condition.

Also, the weight is released from rest, and it represents $x^{\text{'}}\left(0\right)=0$ .

Determine the Laplace transform equation.

Now the force function is,

$$\begin{gathered} f\left(t\right)=\left\{\begin{array}{ll}Sint,& 0⩽t<2\pi \\ 0,& t⩾2\pi \end{array}\right. \\ =\sin t-\sin tU(t-2\pi ) \end{gathered}$$

The general equation of motion for a spring mass system with an external force is:

$m{x}^{\text{'}\text{'}}+kx=f\left(t\right)$

Substituting $m=1,k=16$ and

in the above equation, we get.

$f\left(t\right)=\sin t-\sin tU(t-2\pi )$

Take Laplace transform of both side,

$$\begin{gathered} \mathcal{L}\{\sin t-\sin tU(t-2\pi \left)\right\}=\mathcal{L}\left\{x^{\text{'}\text{'}}+16x\right\} \\ =s^2\mathcal{L}\left\{x\right\}-sx\left(0\right)-x^{\text{'}}\left(0\right)+16\mathcal{L}\left\{x\right\} \\ =\left(s^2+16\right)\mathcal{L}\left\{x\right\}\left(x\left(0\right)=0,x^{\text{'}}\left(0\right)=0\right) \\ \mathcal{L}\left\{x^{\text{'}\text{'}}\right\}=s^2\mathcal{L}\left\{x\right\}-sx\left(0\right)-x^{\text{'}}\left(0\right) \end{gathered}$$

Denote $L\left\{x\right\}=X\left(s\right)$ and replace in the above equation

$\mathcal{L}\{\sin t-\sin tU(t-2\pi \left)\right\}=\left(s^2+16\right)X\left(s\right)$

Find the value

Now find $L\{\sin t-\sin tU(t-2\pi \left)\right\}$ in the following way

$$\begin{gathered} \mathcal{L}\{\sin t-\sin tU(t-2\pi \left)\right\}=\mathcal{L}\left\{\sin t\right\}-\mathcal{L}\left\{\sin tU\right(t-2\pi \left)\right\} \\ =\mathcal{L}\left\{\sin t\right\}-\mathcal{L}\left\{\sin \right(t-2\pi \left)U\right(t-2\pi \left)\right\} \\ =\frac{1}{s^2+1}-\frac{1}{s^2+1}{e}^{-2\pi s} \end{gathered}$$

$\left\{\begin{array}{l}\text{If}F\left(s\right)=\mathcal{C}\left\{f\right(t\left)\right\}\text{and}a>0\text{, then}\mathcal{C}\left\{f\right(t-a\left)JJ\right(t-a\left)\right\}=e^{-as}F\left(s\right)\\ \sin kt=\mathcal{C}\left\{\frac{k}{s^2+k^2}\right\}\end{array}\right.$

Substitute the known values.

Substitute (2) in (1)

$$\begin{gathered} \left(s^2+16\right)X\left(s\right)=\frac{1}{s^2+1}-\frac{1}{s^2+1}{e}^{-2\pi s} \\ X\left(s\right)=\frac{1}{\left(s^2+1\right)\left(s^2+16\right)}-\frac{1}{\left(s^2+1\right)\left(s^2+16\right)}{e}^{-2\pi s} \end{gathered}$$

Decompose the fraction $\frac{1}{\left(s^2+1\right)\left(s^2+16\right)}$. Using the partial fraction technique.

$\frac{\left(s^2+1\right)\left(s^2+16\right)}{\left(s^2+1\right.}+\frac{As+D}{s^2+16}=\frac{(As+B)\left(s^2+16\right)+(Cs+D)\left(s^2+1\right)}{\left(s^2+1\right)\left(s^2+16\right)}$

Since the fractions in above equation have the same denominators,

It takes their numerators to be equal.

$$\begin{gathered} 1=A{s}^3+16As+B{s}^2+16B+C{s}^3+Cs+D{s}^2+D \\ 1=(A+C)s^3+(B+D)s^2+(16A+C)s+16B+D \end{gathered}$$

Find the Inverse Laplace transform

Comparing coefficient both side on we get

$$\begin{gathered} A+C=0 \\ 16A+C=0 \\ B+D=0 \\ 16B+D=1 \end{gathered}$$

Based on the previous two systems, we get $A=0,B=\frac{1}{15},C=0$ and $D=-\frac{1}{15}$ Substituting $A,B,CandD$in the equation (3), we get.

$\frac{1}{\left(s^2+1\right)\left(s^2+16\right)}=\frac{1}{15\left(s^2+1\right)}-\frac{1}{15\left(s^2+16\right)}$

Based on, the equation (*) becomes.

$X\left(s\right)=\frac{1}{15\left(s^2+1\right)}-\frac{1}{15\left(s^2+16\right)}-e^{-2\pi s}\left(\frac{1}{15\left(s^2+1\right)}-\frac{1}{15\left(s^2+16\right)}\right)$

Take inverse Laplace transform of both side.

$$\begin{gathered} {\mathcal{L}}^{-1}\left\{X\right(s\left)\right\}=\frac{1}{15}{\mathcal{L}}^{-1}\left\{\frac{1}{s^2+1^2}\right\}-\frac{1}{15}\frac{1}{4}{\mathcal{L}}^{-1}\left\{\frac{4}{s^2+4^2}\right\}-\frac{1}{15}{\mathcal{L}}^{-1}\left\{\frac{e^{-2\pi s}}{s^2+1^2}\right\}+\frac{1}{15}\frac{1}{4}{\mathcal{L}}^{-1}\left\{\frac{e^{-2\pi s}4}{s^2+4^2}\right\} \\ =\frac{1}{15}Sint-\frac{1}{60}Sin\left(4t\right)-\frac{1}{15}Sin(t-2\pi )U(t-2\pi )+\frac{1}{60}Sin\left(4\right(t-2\pi \left)\right)U(t-2\pi ) \\ x\left(t\right)=\frac{1}{15}Sint-\frac{1}{60}Sin\left(4t\right)-\frac{1}{15}Sin(t-2\pi )U(t-2\pi )+\frac{1}{60}Sin\left(4\right(t-2\pi \left)\right)U(t-2\pi ) \end{gathered}$$

Hence, the inverse Laplace transform equation is,

$x\left(t\right)=\frac{1}{15}Sint-\frac{1}{60}Sin\left(4t\right)-\frac{1}{15}Sin(t-2\pi )U(t-2\pi )+\frac{1}{60}Sin\left(4\right(t-2\pi \left)\right)U(t-2\pi )$