Chapter 4, Problem 4.51P

To determine

(a)

The value of $x\left(t\right)$ at $c=680$.

Answer to Problem 4.51P

The value of $x\left(t\right)$ at $c=680$ is $x\left(t\right)=4.167-4.81e^{-2t}+0.641e^{-15t}$.

Explanation of Solution

Given information:

The value of step function is $5000$ and the equation of motion is $40\ddot{x}+c\dot{x}+12000x=f\left(t\right)$ the value of $c$ is $680$.

Write the Equation of motion.

$40\ddot{x}+c\dot{x}+12000x=f\left(t\right)$ ....... (I)

Here, acceleration is $\ddot{x}$, constant is $c$, velocity is $\dot{x}$, the step function is $f\left(t\right)$ and the displacement is $x$.

Calculation:

Substitute $680$ for $c$ and $5000$ for $c$ in Equation (I).

$\begin{array}{l}40\ddot{x}+680\dot{x}+12000x=5000\\ \ddot{x}+17\dot{x}+30x=125\end{array}$ ...... (II)

Take Laplace transform in equation (II)

$\begin{array}{l}L\left(\ddot{x}\right)+17L\left(\dot{x}\right)+30L\left(x\right)=L\left(125\right)\\ s^{2}X\left(s\right)-sx\left(0\right)-\dot{x}\left(0\right)+17\left[sX\left(s\right)-x\left(0\right)\right]+30X\left(s\right)=\frac{125}{s}\end{array}$ ........ (III)

Substitute $0$ for $x\left(0\right)$ and $0$ for $\dot{x}\left(0\right)$ in Equation (III).

$\begin{array}{l}{s}^{2}X\left(s\right)-s\times 0-0+17\left[sX\left(s\right)-0\right]+30X\left(s\right)=\frac{125}{s}\\ s^{2}X\left(s\right)+17\left[sX\left(s\right)\right]+30X\left(s\right)=\frac{125}{s}\\ X\left(s\right)=\frac{125}{s\left(s+2\right)\left(s+15\right)}\end{array}$ ........ (IV)

Apply partial fraction on Equation (IV).

$\begin{array}{l}\frac{125}{s\left(s+2\right)\left(s+15\right)}=\frac{A}{s}+\frac{B}{s+2}+\frac{C}{s+15}\\ 125=A\left(s+2\right)\left(s+15\right)+Bs\left(s+15\right)+Cs\left(s+2\right)\end{array}$ ...... (V)

Equate constant term of equation (V).

$\begin{array}{l}125=30A\\ A=4.167\end{array}$

Equate coefficient of $s$ of Equation (V) and substitute $4.167$ for $A$.

$\begin{array}{l}0=17(4.167)+15B+2C\\ 15B+2C=-70.839\end{array}$ ....... (VI)

Equate coefficient of $s^2$ of Equation (V) and substitute $4.167$ for $A$.

$\begin{array}{l}0=4.167+B+C\\ B+C=-4.167\end{array}$ ...... (VII)

Solve equation (VI) and (VII).

$\begin{array}{l}B=-4.81\\ C=0.641\end{array}$

Substitute $-4.81$ for $B$, $0.641$ for $C$ and $4.167$ for $A$ in Equation (V).

$\frac{125}{s\left(s+2\right)\left(s+15\right)}=\frac{4.167}{s}+\frac{-4.81}{s+2}+\frac{0.641}{s+15}$

Substitute $\frac{4.167}{s}+\frac{-4.81}{s+2}+\frac{0.641}{s+15}$ for $\frac{125}{s\left(s+2\right)\left(s+15\right)}$ in equation (IV).

$X\left(s\right)=\frac{4.167}{s}+\frac{-4.81}{s+2}+\frac{0.641}{s+15}$ ....... (VIII)

Take inverse Laplace of Equation (VIII).

$\begin{array}{l}{L}^{-1}X\left(s\right)=L^{-1}\left[\frac{4.167}{s}+\frac{-4.81}{s+2}+\frac{0.641}{s+15}\right]\\ L^{-1}X\left(s\right)=L^{-1}\left(\frac{4.167}{s}\right)+L^{-1}\left(\frac{-4.81}{s+2}\right)+L^{-1}\left(\frac{0.641}{s+15}\right)\\ x\left(t\right)=4.167-4.81e^{-2t}+0.641e^{-15t}\end{array}$.

Conclusion:

The value of $x\left(t\right)$ at $c=680$ is $x\left(t\right)=4.167-4.81e^{-2t}+0.641e^{-15t}$.

To determine

(b)

The value of $x\left(t\right)$ at $c=400$.

Answer to Problem 4.51P

The value of $x\left(t\right)$

$c=400$ is $x\left(t\right)=4.167-4.167e^{-5t}\cos \sqrt{5}t-18.635\sin \sqrt{5}t$.

Explanation of Solution

Given information:

The value of step function is $5000$ and the equation of motion is $40\ddot{x}+c\dot{x}+12000x=f\left(t\right)$ the value of $c$ is $400$.

Write the Equation of motion.

$40\ddot{x}+c\dot{x}+12000x=f\left(t\right)$ ......... (IX)

Here, acceleration is $\ddot{x}$, constant is $c$, velocity is $\dot{x}$, the step function is $f\left(t\right)$ and the displacement is $x$.

Calculation:

Substitute $400$ for $c$ and $5000$ for $c$ in Equation (IX).

$40\ddot{x}+400\dot{x}+12000x=5000$ ........ (X)

Take Laplace transform in equation (X)

$\begin{array}{l}L\left(40\ddot{x}\right)+400L\left(\dot{x}\right)+1200L\left(x\right)=L\left(5000\right)\\ 40s^{2}X\left(s\right)-sx\left(0\right)-\dot{x}\left(0\right)+400\left[sX\left(s\right)-x\left(0\right)\right]+1200X\left(s\right)=\frac{5000}{s}\end{array}$ ........ (XI)

Substitute $0$ for $x\left(0\right)$ and $0$ for $\dot{x}\left(0\right)$ in Equation (XI).

$\begin{array}{l}40s^{2}X\left(s\right)-s\times 0-0+400\left[sX\left(s\right)-0\right]+1200X\left(s\right)=\frac{5000}{s}\\ s^{2}X\left(s\right)+10\left[sX\left(s\right)\right]+30X\left(s\right)=\frac{125}{s}\\ X\left(s\right)=\frac{125}{s\left(s^2+10s+30\right)}\end{array}$ ........ (XII)

Apply partial fraction on Equation (XII).

$\begin{array}{l}\frac{125}{s\left(s^2+10s+30\right)}=\frac{A}{s}+\frac{Bs+C}{s^2+10s+30}\\ 125=A\left(s^2+10s+30\right)+s\left(Bs+C\right)\end{array}$ ...... (XIII)

Equate constant term of equation (XIII).

$\begin{array}{l}125=30A\\ A=4.167\end{array}$

Equate coefficient of $s$ of Equation (XIII) and substitute $4.167$ for $A$.

$\begin{array}{l}0+10A+C\\ C=-10\left(4.167\right)\\ C=-41.67\end{array}$ ....... (XIV)

Equate coefficient of $s^2$ of Equation (XIII) and substitute $4.167$ for $A$.

$\begin{array}{l}0=A+B\\ B=-4.167\end{array}$ ...... (XV)

Substitute $-4.167$ for $B$, $-41.67$ for $C$ and $4.167$ for $A$ in Equation (XIII).

$\frac{125}{s\left(s^2+10s+30\right)}=\frac{4.167}{s}-\frac{4.167s+41.67}{s^2+10s+30}$

Substitute $\frac{4.167}{s}-\frac{4.167s+41.67}{s^2+10s+30}$ for $\frac{125}{s\left(s^2+10s+30\right)}$ in equation (XIII).

$X\left(s\right)=\frac{4.167}{s}-\frac{4.167s+41.67}{s^2+10s+30}$ ....... (XVI)

Take inverse Laplace of Equation (XVI).

$\begin{array}{l}{L}^{-1}X\left(s\right)=L^{-1}\left[\frac{4.167}{s}-\frac{4.167s+41.67}{s^2+10s+30}\right]\\ L^{-1}X\left(s\right)=L^{-1}\left(\frac{4.167}{s}\right)-L^{-1}\left(\frac{4.167s+41.67}{s^2+10s+30}\right)\\ L^{-1}X\left(s\right)=4.167-L^{-1}\left(\frac{4.167s+41.67}{s^2+10s+30+25-25}\right)\\ L^{-1}X\left(s\right)=4.167-L^{-1}\left(\frac{4.167s+41.67}{{\left(s+5\right)}^2+5}\right)\end{array}$

Further simplifying

$\begin{array}{l}{L}^{-1}X\left(s\right)=4.167-L^{-1}\left(\frac{4.167s+41.67}{{\left(s+5\right)}^2+5}\right)\\ x\left(t\right)=4.167-L^{-1}\left(\frac{4.167s}{{\left(s+5\right)}^2+5}\right)-L^{-1}\left(\frac{41.67}{{\left(s+5\right)}^2+5}\right)\\ x\left(t\right)=4.167-4.167e^{-5t}\cos \sqrt{5}t-\frac{41.67}{\sqrt{5}}\sin \sqrt{5}t\\ x\left(t\right)=4.167-4.167e^{-5t}\cos \sqrt{5}t-18.635\sin \sqrt{5}t\end{array}$.

Conclusion:

The value of $x\left(t\right)$$c=400$ is $x\left(t\right)=4.167-4.167e^{-5t}\cos \sqrt{5}t-18.635\sin \sqrt{5}t$.