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2.9 HW9

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2.9.1 Problem 1

Since this is an undamped system, the equation of motion is

$$m\ddot{x}+kx=F\left ( t\right )$$ Where $F\left ( t\right ) =Ap\left ( t\right )$ and $p\left ( t\right )$ is the pressure. Therefore $$F\left ( t\right ) =\left ( 50\times 10^{3}\right ) A\left ( 1-e^{-3t}\right )$$ The term $50\times 10^{3}$ was added above because the units were given in $kPa$ and need to convert them to $Pa.$ The equation of motion becomes $$\begin{align*} m\ddot{x}+kx & =\left ( 50\times 10^{3}\right ) A\left ( 1-e^{-3t}\right ) \\ & =\left ( 50\times 10^{3}\right ) A-\left ( 50\times 10^{3}\right ) Ae^{-3t} \end{align*}$$

To simplify notations, let $\beta =\left ( 50\times 10^{3}\right ) A$. The above now becomes

$$\begin{equation} m\ddot{x}+kx=\beta -\beta e^{-3t} \tag{1} \end{equation}$$ The solution to the above can be found by adding the two particular solutions of $$\begin{equation} m\ddot{x}+kx=\beta \tag{2} \end{equation}$$ And $$\begin{equation} m\ddot{x}+kx=-\beta e^{-3t} \tag{3} \end{equation}$$ To the homogeneous solution of $m\ddot{x}+kx=0.$ This can be done since the ODE is linear. The particular solution to (2) is found by assuming $x_{p}\left ( t\right ) =C_{1}$ where $C_{1}$ is some constant and substituting this into (1) and solving for $C_{1}$ gives $kC_{1}=\beta$ or $C_{1}=\frac{\beta }{k}$, hence $$\begin{equation} x_{p,1}\left ( t\right ) =\frac{\beta }{k} \tag{4A} \end{equation}$$ The particular solution to (2) is now found. From the lookup table, assuming $x_{p}\left ( t\right ) =C_{1}e^{-3t}$ and substituting this into (2), and since $\dot{x}_{p}=-3C_{1}e^{-3t}$ and $\ddot{x}_{p}=9C_{1}e^{-3t}$ gives $$\begin{align*} 9mC_{1}e^{-3t}+kC_{1}e^{-3t} & =-\beta e^{-3t}\\ 9mC_{1}+kC_{1} & =-\beta \\ C_{1} & =\frac{-\beta }{9m+k} \end{align*}$$

Therefore

$$\begin{equation} x_{p,2}\left ( t\right ) =\frac{-\beta }{9m+k}e^{-3t} \tag{4B} \end{equation}$$ Now that the particular solutions are known (4A,4B), they are added to the homogeneous solution (which is known) and the complete solution for (1) is $$\begin{align} x\left ( t\right ) & =\overset{x_{h}\left ( t\right ) }{\overbrace{A\cos \omega _{n}t+B\sin \omega _{n}t}}+\overset{x_{p}\left ( t\right ) }{\overbrace{x_{p,1}\left ( t\right ) +x_{p,2}\left ( t\right ) }}\nonumber \\ & =A\cos \omega _{n}t+B\sin \omega _{n}t+\frac{\beta }{k}-\frac{\beta }{9m+k}e^{-3t} \tag{5} \end{align}$$

Initial conditions are now applied to determine $A,B$. Since $x\left ( 0\right ) =0$ the above becomes

$$\begin{align*} 0 & =A+\frac{\beta }{k}-\frac{\beta }{9m+k}\\ A & =\frac{\beta }{9m+k}-\frac{\beta }{k} \end{align*}$$

The solution (5) becomes

$$\begin{equation} x\left ( t\right ) =\left ( \frac{\beta }{9m+k}-\frac{\beta }{k}\right ) \cos \omega _{n}t+B\sin \omega _{n}t+\frac{\beta }{k}-\frac{\beta }{9m+k}e^{-3t} \tag{6} \end{equation}$$ Taking derivative of the above $$\dot{x}\left ( t\right ) =-\omega _{n}\left ( \frac{\beta }{9m+k}-\frac{\beta }{k}\right ) \sin \omega _{n}t+\omega _{n}B\cos \omega _{n}t+3\frac{\beta }{9m+k}e^{-3t}$$ Since $\dot{x}\left ( 0\right ) =0$ then $$\begin{align*} 0 & =\omega _{n}B+3\frac{\beta }{9m+k}\\ B & =\frac{-3\beta }{\left ( 9m+k\right ) \omega _{n}} \end{align*}$$

Substituting this in (6) gives the final solution

$$\begin{equation} x\left ( t\right ) =\left ( \frac{\beta }{9m+k}-\frac{\beta }{k}\right ) \cos \omega _{n}t-\frac{3\beta }{\left ( 9m+k\right ) \omega _{n}}\sin \omega _{n}t+\frac{\beta }{k}-\frac{\beta }{9m+k}e^{-3t} \tag{7} \end{equation}$$ Since $$\omega _{n}=\sqrt{\frac{k}{m}}=\sqrt{\frac{1000}{10}}=10$$ And $$\begin{align*} \beta & =\left ( 50\times 10^{3}\right ) A\\ & =\left ( 50\times 10^{3}\right ) \pi \left ( \frac{0.1}{2}\right ) ^{2}\\ & =392.70 \end{align*}$$

Then numerically, the solution (7) is

$$\begin{align*} x\left ( t\right ) & =\left ( \frac{392.70}{90+1000}-\frac{392.70}{1000}\right ) \cos 10t-\frac{3\left ( 392.70\right ) }{\left ( 90+1000\right ) 10}\sin 10t+\frac{392.70}{1000}-\frac{392.70}{90+1000}e^{-3t}\\ & =-0.032\cos 10t-0.108\sin 10t+0.393-0.360e^{-3t} \end{align*}$$

Below is a plot of the above to illustrate the solution for some arbitrary time $t$.

2.9.2 Problem 2

The force on the piston is

$$F\left ( t\right ) =Ap\left ( t\right )$$ Where $A$ is the area of the piston which is $A=\pi \left ( \frac{d}{2}\right ) ^{2}$. Since this is undamped system, the equation of motion is $$m\ddot{x}+kx=F\left ( t\right )$$ To solve using Duhamel integration, the impulse response $g\left ( t\right ) =\frac{1}{m\omega _{n}}\sin \left ( \omega _{n}t\right )$ is used. The integration is done using the two options.

2.9.2.1 Option 1

$$\begin{align} x_{conv}\left ( t\right ) & =\int _{0}^{t}F\left ( \tau \right ) g\left ( t-\tau \right ) d\tau \nonumber \\ & =\frac{A}{m\omega _{n}}\int _{0}^{t}p\left ( t\right ) \sin \left ( \omega _{n}\left ( t-\tau \right ) \right ) d\tau \nonumber \\ & =\frac{A}{m\omega _{n}}\int _{0}^{t}50\left ( 1000\right ) \left ( 1-e^{-3\tau }\right ) \sin \left ( \omega _{n}\left ( t-\tau \right ) \right ) d\tau \nonumber \end{align}$$

Where $50\left ( 1000\right )$ is used since the units are in $kPa$. The above becomes

$$\begin{align} x_{conv}\left ( t\right ) & =\left ( 5\times 10^{4}\right ) \frac{A}{m\omega _{n}}\int _{0}^{t}\left ( 1-e^{-3\tau }\right ) \sin \left ( \omega _{n}\left ( t-\tau \right ) \right ) d\tau \nonumber \\ & =\left ( 5\times 10^{4}\right ) \frac{A}{m\omega _{n}}\left ( \int _{0}^{t}\sin \left ( \omega _{n}\left ( t-\tau \right ) \right ) d\tau -\int _{0}^{t}e^{-3\tau }\sin \left ( \omega _{n}\left ( t-\tau \right ) \right ) d\tau \right ) \tag{1} \end{align}$$

The first integral in (1) becomes

$$\begin{align} \int _{0}^{t}\sin \left ( \omega _{n}\left ( t-\tau \right ) \right ) d\tau & =-\left ( \frac{\cos \left ( \omega _{n}\left ( t-\tau \right ) \right ) }{-\omega _{n}}\right ) _{0}^{t}\nonumber \\ & =\frac{1}{\omega _{n}}\left ( \cos \left ( \omega _{n}\left ( t-\tau \right ) \right ) \right ) _{0}^{t}\nonumber \\ & =\frac{1}{\omega _{n}}\left ( \cos \left ( \omega _{n}\left ( t-t\right ) \right ) -\cos \left ( \omega _{n}t\right ) \right ) \nonumber \\ & =\frac{1}{\omega _{n}}\left ( 1-\cos \left ( \omega _{n}t\right ) \right ) \tag{2} \end{align}$$

The second integral in (1) is found using the handout integration tables

$$\int e^{ax}\sin \left ( b+cx\right ) dx=\frac{ae^{ax}\sin \left ( b+cx\right ) }{a^{2}+c^{2}}-\frac{ce^{ax}\cos \left ( b+cx\right ) }{a^{2}+c^{2}}$$ In this case $a=-3$ and $b=\omega _{n}t$ and $c=-\omega _{n}$. The above becomes after substitution $$\begin{align} \int _{0}^{t}e^{-3\tau }\sin \left ( \omega _{n}\left ( t-\tau \right ) \right ) d\tau & =\left ( \frac{-3e^{-3\tau }\sin \left ( \omega _{n}\left ( t-\tau \right ) \right ) }{9+\omega _{n}^{2}}-\frac{-\omega _{n}e^{-3\tau }\cos \left ( \omega _{n}\left ( t-\tau \right ) \right ) }{9+\omega _{n}^{2}}\right ) _{0}^{t}\nonumber \\ & =\frac{1}{9+\omega _{n}^{2}}\left ( -3e^{-3\tau }\sin \left ( \omega _{n}\left ( t-\tau \right ) \right ) +\omega _{n}e^{-3\tau }\cos \left ( \omega _{n}\left ( t-\tau \right ) \right ) \right ) _{0}^{t}\nonumber \\ & =\frac{1}{9+\omega _{n}^{2}}\left ( \omega _{n}e^{-3t}-\left ( -3\sin \left ( \omega _{n}t\right ) +\omega _{n}\cos \left ( \omega _{n}t\right ) \right ) \right ) \nonumber \\ & =\frac{\omega _{n}e^{-3t}+3\sin \left ( \omega _{n}t\right ) -\omega _{n}\cos \left ( \omega _{n}t\right ) }{9+\omega _{n}^{2}} \tag{3} \end{align}$$

Substituting (2,3) into (1) gives the final result

$$\begin{equation} x_{conv}\left ( t\right ) =\left ( 5\times 10^{4}\right ) \frac{A}{m\omega _{n}}\left ( \frac{1}{\omega _{n}}\left ( 1-\cos \left ( \omega _{n}t\right ) \right ) -\frac{\omega _{n}e^{-3t}+3\sin \left ( \omega _{n}t\right ) -\omega _{n}\cos \left ( \omega _{n}t\right ) }{9+\omega _{n}^{2}}\right ) \tag{4} \end{equation}$$ Because initial conditions are zero the solution is $$\begin{align*} x\left ( t\right ) & =x_{h}\left ( t\right ) +x_{cov\left ( t\right ) }\\ & =x_{cov\left ( t\right ) } \end{align*}$$

Substituting all the numerical values, and since $\omega _{n}=\sqrt{\frac{k}{m}}=\sqrt{\frac{1000}{10}}=10$ then (4) becomes

$$\begin{align*} x\left ( t\right ) & =\left ( 5\times 10^{4}\right ) \frac{\pi \left ( \frac{0.1}{2}\right ) ^{2}}{\left ( 10\right ) \left ( 10\right ) }\left ( \frac{1}{10}\left ( 1-\cos \left ( 10t\right ) \right ) -\frac{10e^{-3t}+3\sin \left ( 10t\right ) -10\cos \left ( 10t\right ) }{109}\right ) \\ & =3.927\left ( \frac{1}{10}\left ( 1-\cos \left ( 10t\right ) \right ) -\frac{10e^{-3t}+3\sin \left ( 10t\right ) -10\cos \left ( 10t\right ) }{109}\right ) \\ & =3.927\left ( \frac{1}{10}\left ( 1-\cos \left ( 10t\right ) \right ) +\frac{10}{109}\cos \left ( 10t\right ) -\frac{10}{109}e^{-3t}-\frac{3}{109}\sin \left ( 10t\right ) \right ) \\ & =3.927\left ( \frac{1}{10}-\frac{10}{109}e^{-3t}-\frac{3}{109}\sin \left ( 10t\right ) -\frac{9}{1090}\cos \left ( 10t\right ) \right ) \end{align*}$$

This is a plot of the above, which agrees with plot from the direct integration method. This verifies the above result

2.9.2.2 Option 2

$$\begin{align} x_{conv}\left ( t\right ) & =\int _{0}^{t}F\left ( t-\tau \right ) g\left ( \tau \right ) d\tau \nonumber \\ & =\frac{A}{m\omega _{n}}\int _{0}^{t}p\left ( t-\tau \right ) \sin \left ( \omega _{n}\tau \right ) d\tau \nonumber \\ & =\frac{A}{m\omega _{n}}\int _{0}^{t}50\left ( 1000\right ) \left ( 1-e^{-3\left ( t-\tau \right ) }\right ) \sin \left ( \omega _{n}\tau \right ) d\tau \nonumber \end{align}$$

Where $50\left ( 1000\right )$ is used, since the units are in $kPa$. The above becomes

$$\begin{align} x_{conv}\left ( t\right ) & =\left ( 5\times 10^{4}\right ) \frac{A}{m\omega _{n}}\int _{0}^{t}\left ( 1-e^{-3\left ( t-\tau \right ) }\right ) \sin \left ( \omega _{n}\tau \right ) d\tau \nonumber \\ & =\left ( 5\times 10^{4}\right ) \frac{A}{m\omega _{n}}\left ( \int _{0}^{t}\sin \left ( \omega _{n}\tau \right ) d\tau -\int _{0}^{t}e^{-3\left ( t-\tau \right ) }\sin \left ( \omega _{n}\tau \right ) d\tau \right ) \tag{1} \end{align}$$

The first integral in (1) is now evaluated

$$\begin{align} \int _{0}^{t}\sin \left ( \omega _{n}\tau \right ) d\tau & =-\frac{1}{\omega _{n}}\left ( \cos \left ( \omega _{n}\tau \right ) \right ) _{0}^{t}\nonumber \\ & =\frac{-1}{\omega _{n}}\left ( \cos \left ( \omega _{n}t\right ) -1\right ) \nonumber \\ & =\frac{1}{\omega _{n}}\left ( 1-\cos \left ( \omega _{n}t\right ) \right ) \tag{2} \end{align}$$

The second integral in (1) is

$$\begin{align} \int _{0}^{t}e^{-3\left ( t-\tau \right ) }\sin \left ( \omega _{n}\tau \right ) d\tau & =\int _{0}^{t}e^{-3t+3\tau }\sin \left ( \omega _{n}\tau \right ) d\tau \nonumber \\ & =\int _{0}^{t}e^{-3t}e^{3\tau }\sin \left ( \omega _{n}\tau \right ) d\tau \nonumber \\ & =e^{-3t}\int _{0}^{t}e^{3\tau }\sin \left ( \omega _{n}\tau \right ) d\tau \tag{3} \end{align}$$

This integral is found using tables

$$\int e^{ax}\sin \left ( bx\right ) dx=\frac{e^{ax}\left ( a\sin \left ( bx\right ) -b\cos \left ( bx\right ) \right ) }{a^{2}+b^{2}}$$ Where in this case $a=3$ and $b=\omega _{n}$ Therefore (3) becomes $$\begin{align} e^{-3t}\int _{0}^{t}e^{3\tau }\sin \left ( \omega _{n}\tau \right ) d\tau & =e^{-3t}\left ( \frac{e^{3\tau }\left ( 3\sin \left ( \omega _{n}\tau \right ) -\omega _{n}\cos \left ( \omega _{n}\tau \right ) \right ) }{9+\omega _{n}^{2}}\right ) _{0}^{t}\nonumber \\ & =\frac{e^{-3t}}{9+\omega _{n}^{2}}\left ( e^{3\tau }\left ( 3\sin \left ( \omega _{n}\tau \right ) -\omega _{n}\cos \left ( \omega _{n}\tau \right ) \right ) \right ) _{0}^{t}\nonumber \\ & =\frac{e^{-3t}}{9+\omega _{n}^{2}}\left ( e^{3t}\left ( 3\sin \left ( \omega _{n}t\right ) -\omega _{n}\cos \left ( \omega _{n}t\right ) \right ) -\left ( -\omega _{n}\right ) \right ) \nonumber \\ & =\frac{e^{-3t}}{9+\omega _{n}^{2}}\left ( e^{3t}\left ( 3\sin \left ( \omega _{n}t\right ) -\omega _{n}\cos \left ( \omega _{n}t\right ) \right ) +\omega _{n}\right ) \nonumber \\ & =\frac{1}{9+\omega _{n}^{2}}\left ( 3\sin \left ( \omega _{n}t\right ) -\omega _{n}\cos \left ( \omega _{n}t\right ) +\omega _{n}e^{-3t}\right ) \tag{4} \end{align}$$

Substituting (2,4) into (1) gives the final result

$$\begin{equation} x_{conv}\left ( t\right ) =\left ( 5\times 10^{4}\right ) \frac{A}{m\omega _{n}}\left ( \frac{1}{\omega _{n}}\left ( 1-\cos \left ( \omega _{n}t\right ) \right ) -\frac{3\sin \left ( \omega _{n}t\right ) -\omega _{n}\cos \left ( \omega _{n}t\right ) +\omega _{n}e^{-3t}}{9+\omega _{n}^{2}}\right ) \tag{5} \end{equation}$$ Because initial conditions are zero then $$\begin{align*} x\left ( t\right ) & =x_{h}\left ( t\right ) +x_{cov\left ( t\right ) }\\ & =x_{cov\left ( t\right ) } \end{align*}$$

Comparing (5) above to equation (4) found using option (1) shows they are the same as expected.