3.3 3.3.1
Find solution to second order ODE with impulse as input.
Spring-damper-mass dropped from height h, find resulting EQM.
Find solution to second order ODE with 2 impulses as input, one delayed.
Find response of undamped system to half sin input force, using convolution.
As above, but the forcing function is triangle looking. Use convolution also.
Find Fourier series for sawtooth function.
Solving 2nd order ODE with impulse as input using Laplace transform.
Find shock response spectrum to half-sine input (hard).
Find \(H(s)\) , the transfer function for ODE.
Find the frequency response for the above \(H(s)\) , i.e. set \(s=jw\) and plot.
3.3.2 Problem
Calculate the solution to \(\ddot {x}+2\dot {x}+3x=\sin t+\delta \left ( t-\pi \right ) \) with IC \(x\left ( 0\right ) =0,\dot {x}\left ( 0\right ) =1\) and plot the solution.
Answer
\begin{align*} \ddot {x}+2\dot {x}+3x & =\sin t+\delta \left ( t-\pi \right ) \\ \ddot {x}+2\xi \omega _{n}\dot {x}+\omega _{n}^{2}x & =\sin t+\delta \left ( t-\pi \right ) \end{align*}
Hence \(\omega _{n}=\sqrt {3}\) \(2\xi \omega _{n}=2\) , hence \(\xi =\frac {1}{\sqrt {3}}\) \(=\allowbreak 0.577\,35\) , hence this is underdamped system .
Since \(x=x_{h}+x_{p}\) , then
\[ x_{h}=e^{-\xi \omega _{n}t}\left ( A\cos \omega _{d}t+B\sin \omega _{d}t\right ) \]
We have 2 particular solutions. The first \(x_{p_{1}}\) is due to \(\sin t\) and the second \(x_{p_{2}}\) is due to \(\delta \left ( t-\pi \right ) \) . When the
forcing function is \(\sin t\) , we guess
\[ x_{p_{1}}=c_{1}\cos t+c_{2}\sin t \]
and when the forcing function is \(\delta \left ( t-\pi \right ) \) the response is
\[ x_{p_{2}}=\frac {1}{\omega _{d}m}e^{-\xi \omega _{n}\left ( t-\pi \right ) }\sin \omega _{d}\left ( t-\pi \right ) \Phi \left ( t-\pi \right ) \]
From \(x_{p_{1}}\) we find \(\dot {x}_{p_{1}}\) and \(\ddot {x}_{p_{1}}\)
and plug these into \(\ddot {x}+2\dot {x}+3x=\sin t\) to find \(c_{1}\) and \(c_{2}\) , next we find \(A,B\) by using the IC, and then at the end we add the
solution \(x_{p_{2}}\) . Notice that \(x_{p_{2}}\) do not enter into the calculation of \(A,B\,\,\) since the impulse \(\delta \left ( t-\pi \right ) \) is not effective at \(t=0\) .
\begin{align*} \dot {x}_{p_{1}} & =-c_{1}\sin t+c_{2}\cos t\\ \ddot {x}_{p_{1}} & =-c_{1}\cos t-c_{2}\sin t \end{align*}
Hence
\begin{align*} \ddot {x}_{p_{1}}+2\dot {x}_{p_{1}}+3x_{p_{1}} & =\sin t\\ \left ( -c_{1}\cos t-c_{2}\sin t\right ) +2\left ( -c_{1}\sin t+c_{2}\cos t\right ) +3\left ( c_{1}\cos t+c_{2}\sin t\right ) & =\sin t\\ \sin t\left ( -c_{2}-2c_{1}+3c_{2}\right ) +\cos t\left ( -c_{1}+2c_{2}+3c_{1}\right ) & =\sin t \end{align*}
Hence \(\left ( -2c_{1}+2c_{2}\right ) =1\) and \(\left ( 2c_{2}+2c_{1}\right ) =0\) . This results in
\begin{align*} c_{1} & =-\frac {1}{4}\\ c_{2} & =\frac {1}{4}\end{align*}
Hence
\[ \fbox {$x_{p_{1}}=-\frac {1}{4}\cos t+\frac {1}{4}\sin t$}\]
Therefore
\[ x_{h}+x_{p_{1}}=e^{-\xi \omega _{n}t}\left ( A\cos \omega _{d}t+B\sin \omega _{d}t\right ) -\frac {1}{4}\cos t+\frac {1}{4}\sin t \]
Now we use IC’s to find \(A,B\) . At \(t=0\) we obtain
\[ \fbox {$A=\frac {1}{4}$}\]
And
\begin{align*} \dot {x}_{h}+\dot {x}_{p_{1}} & =-\xi \omega _{n}e^{-\xi \omega _{n}t}\left ( \frac {1}{4}\cos \omega _{d}t+B\sin \omega _{d}t\right ) \\ & +e^{-\xi \omega _{n}t}\left ( -\frac {1}{4}\omega _{d}\sin \omega _{d}t+\omega _{d}B\cos \omega _{d}t\right ) +\frac {1}{4}\sin t+\frac {1}{4}\cos t \end{align*}
At \(t=0\) we have
\begin{align*} 1 & =-\xi \omega _{n}\left ( \frac {1}{4}\right ) +\left ( \omega _{d}B\right ) +\frac {1}{4}\\ B & =\frac {\left ( 1+\frac {\xi \omega _{n}}{4}-\frac {1}{4}\right ) }{\omega _{d}}\end{align*}
But \(\omega _{d}=\omega _{n}\sqrt {1-\xi ^{2}}=\sqrt {3}\sqrt {1-\left ( \frac {1}{\sqrt {3}}\right ) ^{2}}=\sqrt {3}\sqrt {\frac {2}{3}}\) , Hence \(\omega _{d}=\sqrt {2}\)
\[ \fbox {$B=\frac {1}{\sqrt {2}}$}\]
Hence the final solution is
\begin{align*} x\left ( t\right ) & =x_{h}+x_{p_{1}}+x_{p_{2}}\\ & =e^{-\xi \omega _{n}t}\left ( \frac {1}{4}\cos \omega _{d}t+\frac {1}{\sqrt {2}}\sin \omega _{d}t\right ) -\frac {1}{4}\cos t+\frac {1}{4}\sin t+\frac {1}{\omega _{d}m}e^{-\xi \omega _{n}\left ( t-\pi \right ) }\sin \omega _{d}\left ( t-\pi \right ) \Phi \left ( t-\pi \right ) \end{align*}
Substitute values for the parameters above we obtain
\[ \fbox {$x\left ( t\right ) =e^{-t}\left ( \frac {1}{4}\cos \sqrt {2}t+\frac {1}{\sqrt {2}}\sin \sqrt {2}t\right ) -\frac {1}{4}\cos t+\frac {1}{4}\sin t+\frac {1}{\sqrt {2}}e^{-\left ( t-\pi \right ) }\sin \sqrt {2}\left ( t-\pi \right ) \Phi \left ( t-\pi \right ) $}\]
This is a plot of the solution superimposed on the forcing
functions
3.3.3
The magnitude of the impulse resulting when the mass hits the ground is given by the change of momentum that
occurs. Hence
\[ \hat {F}=Ft=m(v_{final}-v_{0}) \]
But assuming the mass is dropped from rest, hence \(v_{0}=0\) , and \(v_{final}=gt\,\ \) where \(t=\sqrt {2\frac {h}{g}}\) where \(h\) is the height that mass falls.
Hence
\begin{align*} \hat {F} & =mv_{final}\\ & =m\sqrt {2gh}\end{align*}
Hence the equation of motion is
\[ \fbox {$m\ddot {x}\left ( t\right ) +c\dot {x}\left ( t\right ) +kx\left ( t\right ) =m\sqrt {2gh}\delta \left ( t\right ) $}\]
Since underdamped, \(x\left ( t\right ) =h\left ( t\right ) =\frac {\hat {F}}{m\omega _{d}}e^{-\xi \omega _{n}t}\sin \omega _{d}t\) , hence the solution is
\begin{align*} x\left ( t\right ) & =\frac {m\sqrt {2gh}}{m\omega _{d}}e^{-\xi \omega _{n}t}\sin \omega _{d}t\\ & =\frac {\sqrt {2gh}}{\omega _{d}}e^{-\xi \omega _{n}t}\sin \omega _{d}t \end{align*}
Taking \(t=0\) as time of impact.
3.3.4 Problem
Compute response of the system \(3\ddot {x}\left ( t\right ) +6\dot {x}\left ( t\right ) +12x\left ( t\right ) =3\delta \left ( t\right ) -\delta \left ( t-1\right ) \) with IC \(x\left ( 0\right ) =0.01m\) and \(v\left ( 0\right ) =1m/s\) . Plot the response.
Answer
\begin{align*} 3\ddot {x}\left ( t\right ) +6\dot {x}\left ( t\right ) +12x\left ( t\right ) & =3\delta \left ( t\right ) -\delta \left ( t-1\right ) \\ \ddot {x}\left ( t\right ) +2\dot {x}\left ( t\right ) +4x\left ( t\right ) & =\delta \left ( t\right ) -\frac {1}{3}\delta \left ( t-1\right ) \\ \ddot {x}+2\xi \omega _{n}\dot {x}+\omega _{n}^{2}x & =\delta \left ( t\right ) -\frac {1}{3}\delta \left ( t-1\right ) \end{align*}
Where \(m=1,\omega _{n}^{2}=4\) , hence \(\omega _{n}=2\) and \(2\xi \omega _{n}=2\) , hence \(\xi =\frac {1}{2}\) . This is an underdamped system.
\(\omega _{d}=\omega _{n}\sqrt {1-\xi ^{2}}=2\sqrt {1-\left ( \frac {1}{2}\right ) ^{2}}=2\sqrt {\frac {3}{4}}\) , Hence \(\omega _{d}=\sqrt {3}\)
\[ x_{h}=e^{-\xi \omega _{n}t}\left ( A\cos \omega _{d}t+B\sin \omega _{d}t\right ) \]
The response due to the forcing function \(\delta \left ( t\right ) \) is given by
\[ x_{p_{1}}\left ( t\right ) =\frac {1}{\omega _{d}m}e^{-\xi \omega _{n}t}\sin \left ( \omega _{d}t\right ) \]
The response due to the other forcing function \(\delta \left ( t-1\right ) \) is
given by
\[ x_{p_{2}}\left ( t\right ) =-\frac {1}{3}\frac {1}{\omega _{d}m}e^{-\xi \omega _{n}\left ( t-1\right ) }\sin \omega _{d}\left ( t-1\right ) \Phi \left ( t-1\right ) \]
Now we determine \(A,B\,\) from IC’s
\begin{align*} x_{h}\left ( 0\right ) +x_{p_{1}}\left ( 0\right ) & =0.01\\ & =\left [ e^{-\xi \omega _{n}t}\left ( A\cos \omega _{d}t+B\sin \omega _{d}t\right ) +\frac {1}{\omega _{d}m}e^{-\xi \omega _{n}t}\sin \left ( \omega _{d}t\right ) \right ] _{t=0}\end{align*}
Hence \(A=0.01\) \(B\)
\begin{align*} \dot {x}_{h}\left ( t\right ) +\dot {x}_{p_{1}}\left ( t\right ) & =-\xi \omega _{n}e^{-\xi \omega _{n}t}\left ( A\cos \omega _{d}t+B\sin \omega _{d}t\right ) +e^{-\xi \omega _{n}t}\left ( -A\omega _{d}\sin \omega _{d}t+B\omega _{d}\cos \omega _{d}t\right ) \\ & +\frac {-\xi \omega _{n}}{\omega _{d}m}e^{-\xi \omega _{n}t}\sin \left ( \omega _{d}t\right ) +\frac {\omega _{d}}{\omega _{d}m}e^{-\xi \omega _{n}t}\cos \left ( \omega _{d}t\right ) \end{align*}
But \(\dot {x}_{h}\left ( 0\right ) +\dot {x}_{p_{1}}\left ( 0\right ) =1\) , hence from the above, and noting that \(m=1\)
\begin{align*} 1 & =-A\xi \omega _{n}+B\omega _{d}+1\\ B & =\frac {A\xi \omega _{n}}{\omega _{d}}\\ & =\frac {0.01\left ( \frac {1}{2}\right ) 2}{\sqrt {3}}\end{align*}
Hence
\[ B=\frac {1}{100\sqrt {3}} \]
Therefore
\[ x_{h}=e^{-\xi \omega _{n}t}\left ( \frac {1}{100}\cos \omega _{d}t+\frac {1}{100\sqrt {3}}\sin \omega _{d}t\right ) \]
Now we can combine the above solution to obtain the final solution
\begin{align*} x\left ( t\right ) & =x\left ( h\right ) +x_{p_{1}}\left ( t\right ) +x_{p_{2}}\left ( t\right ) \\ & =e^{-\xi \omega _{n}t}\left ( \frac {1}{100}\cos \omega _{d}t+\frac {1}{100\sqrt {3}}\sin \omega _{d}t\right ) \\ & +\frac {1}{\omega _{d}m}e^{-\xi \omega _{n}t}\sin \left ( \omega _{d}t\right ) \\ & -\frac {1}{3}\frac {1}{\omega _{d}m}e^{-\xi \omega _{n}\left ( t-1\right ) }\sin \omega _{d}\left ( t-1\right ) \Phi \left ( t-1\right ) \end{align*}
Substitute numerical values for the above parameters, we obtain
\[ x\left ( t\right ) =\frac {e^{-t}}{100}\left ( \cos \sqrt {3}t+\frac {1}{\sqrt {3}}\sin \sqrt {3}t\right ) +\frac {1}{\sqrt {3}}e^{-t}\sin \left ( \sqrt {3}t\right ) -\frac {1}{3}\frac {1}{\sqrt {3}}e^{-\left ( t-1\right ) }\sin \left ( \sqrt {3}\left ( t-1\right ) \right ) \Phi \left ( t-1\right ) \]
This is a plot of the response
3.3.5
Let the response by \(x\left ( t\right ) \) . Hence \(x\left ( t\right ) =x_{h}\left ( t\right ) +x_{p}\left ( t\right ) \) , where \(x_{p}\left ( t\right ) \) is the particular solution, which is the response due the the above forcing
function. Using convolution
\[ x_{p}\left ( t\right ) ={\displaystyle \int \limits _{0}^{t}} f\left ( \tau \right ) h\left ( t-\tau \right ) d\tau \]
Where \(h\left ( t\right ) \) is the unit impulse response of a second order underdamped system which is
\[ h\left ( t\right ) =\frac {1}{m\omega _{d}}e^{-\xi \omega _{n}t}\sin \omega _{d}t \]
hence
\begin{align*} x_{p}\left ( t\right ) & =\frac {F_{0}}{m\omega _{d}}{\displaystyle \int \limits _{0}^{t}} \sin \left ( \tau \right ) e^{-\xi \omega _{n}\left ( t-\tau \right ) }\sin \left ( \omega _{d}\left ( t-\tau \right ) \right ) d\tau \\ & =\frac {F_{0}e^{-\xi \omega _{n}t}}{m\omega _{d}}{\displaystyle \int \limits _{0}^{t}} e^{\xi \omega _{n}\tau }\sin \left ( \tau \right ) \sin \left ( \omega _{d}\left ( t-\tau \right ) \right ) d\tau \end{align*}
Using \(\sin A\sin B=\frac {1}{2}\left [ \cos \left ( A-B\right ) -\cos \left ( A+B\right ) \right ] \,\ \) then
\[ \sin \left ( \tau \right ) \sin \left ( \omega _{d}\left ( t-\tau \right ) \right ) =\frac {1}{2}\left [ \cos \left ( \tau -\omega _{d}\left ( t-\tau \right ) \right ) -\cos \left ( \tau +\omega _{d}\left ( t-\tau \right ) \right ) \right ] \]
Then the integral becomes
\[ x_{p}\left ( t\right ) =\frac {F_{0}e^{-\xi \omega _{n}t}}{2m\omega _{d}}\left ( {\displaystyle \int \limits _{0}^{t}} e^{\xi \omega _{n}\tau }\cos \left ( \tau -\omega _{d}\left ( t-\tau \right ) \right ) d\tau -{\displaystyle \int \limits _{0}^{t}} e^{\xi \omega _{n}\tau }\cos \left ( \tau +\omega _{d}\left ( t-\tau \right ) \right ) d\tau \right ) \]
Consider the first integral \(I_{1}\) where
\[ I_{1}={\displaystyle \int \limits _{0}^{t}} e^{\xi \omega _{n}\tau }\cos \left ( \tau -\omega _{d}\left ( t-\tau \right ) \right ) d\tau \]
Integrate by parts, where \(\int udv=uv-\int vdu\) , Let \(dv=e^{\xi \omega _{n}\tau }\rightarrow v=\frac {e^{\xi \omega _{n}\tau }}{\xi \omega _{n}}\) and let \(u=\cos \left ( \tau -\omega _{d}\left ( t-\tau \right ) \right ) \rightarrow du=-\left ( 1+\omega _{d}\right ) \sin \left ( \tau -\omega _{d}\left ( t-\tau \right ) \right ) \) ,
hence
\begin{align} I_{1} & =\left [ \cos \left ( \tau -\omega _{d}\left ( t-\tau \right ) \right ) \frac {e^{\xi \omega _{n}\tau }}{\xi \omega _{n}}\right ] _{0}^{t}\nonumber \\ & -{\displaystyle \int \limits _{0}^{t}} \frac {e^{\xi \omega _{n}\tau }}{\xi \omega _{n}}\left [ -\left ( 1+\omega _{d}\right ) \sin \left ( \tau -\omega _{d}\left ( t-\tau \right ) \right ) \right ] d\tau \nonumber \\ & \nonumber \\ & =\left [ \cos \left ( t-\omega _{d}\left ( t-t\right ) \right ) \frac {e^{\xi \omega _{n}t}}{\xi \omega _{n}}-\cos \left ( 0-\omega _{d}\left ( t-0\right ) \right ) \frac {1}{\xi \omega _{n}}\right ] \nonumber \\ & +\frac {\left ( 1+\omega _{d}\right ) }{\xi \omega _{n}}{\displaystyle \int \limits _{0}^{t}} e^{\xi \omega _{n}\tau }\sin \left ( \tau -\omega _{d}\left ( t-\tau \right ) \right ) d\tau \nonumber \\ & \nonumber \\ & =\frac {1}{\xi \omega _{n}}\left [ \cos \left ( t\right ) e^{\xi \omega _{n}t}-\cos \left ( \omega _{d}t\right ) \right ] +\frac {\left ( 1+\omega _{d}\right ) }{\xi \omega _{n}}{\displaystyle \int \limits _{0}^{t}} e^{\xi \omega _{n}\tau }\sin \left ( \tau -\omega _{d}\left ( t-\tau \right ) \right ) d\tau \tag {1}\end{align}
Integrate by parts again the last integral above, where \(\int udv=uv-\int vdu\) , Let \(dv=e^{\xi \omega _{n}\tau }\rightarrow v=\frac {e^{\xi \omega _{n}\tau }}{\xi \omega _{n}}\) and let \(u=\sin \left ( \tau -\omega _{d}\left ( t-\tau \right ) \right ) \rightarrow du=\left ( 1+\omega _{d}\right ) \cos \left ( \tau -\omega _{d}\left ( t-\tau \right ) \right ) \) , hence
\begin{align}{\displaystyle \int \limits _{0}^{t}} e^{\xi \omega _{n}\tau }\sin \left ( \tau -\omega _{d}\left ( t-\tau \right ) \right ) d\tau & =\left [ \sin \left ( \tau -\omega _{d}\left ( t-\tau \right ) \right ) \frac {e^{\xi \omega _{n}\tau }}{\xi \omega _{n}}\right ] _{0}^{t}\nonumber \\ & -\int _{0}^{t}\frac {e^{\xi \omega _{n}\tau }}{\xi \omega _{n}}\left ( 1+\omega _{d}\right ) \cos \left ( \tau -\omega _{d}\left ( t-\tau \right ) \right ) d\tau \nonumber \\ & \nonumber \\ & =\frac {1}{\xi \omega _{n}}\left [ \sin \left ( t\right ) e^{\xi \omega _{n}t}+\sin \left ( \omega _{d}t\right ) \right ] -\nonumber \\ & \frac {\left ( 1+\omega _{d}\right ) }{\xi \omega _{n}}\int _{0}^{t}e^{\xi \omega _{n}\tau }\cos \left ( \tau -\omega _{d}\left ( t-\tau \right ) \right ) d\tau \tag {2}\end{align}
Substitute (2) into (1) we obtain
\begin{align*} I_{1} & =\frac {1}{\xi \omega _{n}}\left [ \cos \left ( t\right ) e^{\xi \omega _{n}t}-\cos \left ( \omega _{d}t\right ) \right ] +\\ & \frac {\left ( 1+\omega _{d}\right ) }{\xi \omega _{n}}\left ( \frac {1}{\xi \omega _{n}}\left [ \sin \left ( t\right ) e^{\xi \omega _{n}t}+\sin \left ( \omega _{d}t\right ) \right ] -\frac {\left ( 1+\omega _{d}\right ) }{\xi \omega _{n}}\int _{0}^{t}e^{\xi \omega _{n}\tau }\cos \left ( \tau -\omega _{d}\left ( t-\tau \right ) \right ) d\tau \right ) \\ & \\ & =\frac {1}{\xi \omega _{n}}\left [ \cos \left ( t\right ) e^{\xi \omega _{n}t}-\cos \left ( \omega _{d}t\right ) \right ] +\frac {\left ( 1+\omega _{d}\right ) }{\left ( \xi \omega _{n}\right ) ^{2}}\left [ \sin \left ( t\right ) e^{\xi \omega _{n}t}+\sin \left ( \omega _{d}t\right ) \right ] \\ & -\frac {\left ( 1+\omega _{d}\right ) ^{2}}{\left ( \xi \omega _{n}\right ) ^{2}}\int _{0}^{t}e^{\xi \omega _{n}\tau }\cos \left ( \tau -\omega _{d}\left ( t-\tau \right ) \right ) d\tau \\ & \\ & =\frac {1}{\xi \omega _{n}}\left [ \cos \left ( t\right ) e^{\xi \omega _{n}t}-\cos \left ( \omega _{d}t\right ) \right ] +\frac {\left ( 1+\omega _{d}\right ) }{\left ( \xi \omega _{n}\right ) ^{2}}\left [ \sin \left ( t\right ) e^{\xi \omega _{n}t}+\sin \left ( \omega _{d}t\right ) \right ] -\frac {\left ( 1+\omega _{d}\right ) ^{2}}{\left ( \xi \omega _{n}\right ) ^{2}}I_{1}\end{align*}
Hence
\begin{align*} I_{1}+\frac {\left ( 1+\omega _{d}\right ) ^{2}}{\left ( \xi \omega _{n}\right ) ^{2}}I_{1} & =\frac {1}{\xi \omega _{n}}\left [ \cos \left ( t\right ) e^{\xi \omega _{n}t}-\cos \left ( \omega _{d}t\right ) \right ] +\frac {\left ( 1+\omega _{d}\right ) }{\left ( \xi \omega _{n}\right ) ^{2}}\left [ \sin \left ( t\right ) e^{\xi \omega _{n}t}+\sin \left ( \omega _{d}t\right ) \right ] \\ I_{1}\left ( \frac {\left ( \xi \omega _{n}\right ) ^{2}+\left ( 1+\omega _{d}\right ) ^{2}}{\left ( \xi \omega _{n}\right ) ^{2}}\right ) & =\frac {1}{\xi \omega _{n}}\left [ \cos \left ( t\right ) e^{\xi \omega _{n}t}-\cos \left ( \omega _{d}t\right ) \right ] +\frac {\left ( 1+\omega _{d}\right ) }{\left ( \xi \omega _{n}\right ) ^{2}}\left [ \sin \left ( t\right ) e^{\xi \omega _{n}t}+\sin \left ( \omega _{d}t\right ) \right ] \\ I_{1} & =\left ( \frac {\left ( \xi \omega _{n}\right ) ^{2}}{\left ( \xi \omega _{n}\right ) ^{2}+\left ( 1+\omega _{d}\right ) ^{2}}\right ) \\ & \left ( \frac {1}{\xi \omega _{n}}\left [ \cos \left ( t\right ) e^{\xi \omega _{n}t}-\cos \left ( \omega _{d}t\right ) \right ] +\frac {\left ( 1+\omega _{d}\right ) }{\left ( \xi \omega _{n}\right ) ^{2}}\left [ \sin \left ( t\right ) e^{\xi \omega _{n}t}+\sin \left ( \omega _{d}t\right ) \right ] \right ) \\ & \\ & =\frac {\xi \omega _{n}\left [ \cos \left ( t\right ) e^{\xi \omega _{n}t}-\cos \left ( \omega _{d}t\right ) \right ] +\left ( 1+\omega _{d}\right ) \left [ \sin \left ( t\right ) e^{\xi \omega _{n}t}+\sin \left ( \omega _{d}t\right ) \right ] }{\left ( \xi \omega _{n}\right ) ^{2}+\left ( 1+\omega _{d}\right ) ^{2}}\end{align*}
Now consider the second integral \(I_{2}\) where
\[ I_{2}={\displaystyle \int \limits _{0}^{t}} e^{\xi \omega _{n}\tau }\cos \left ( \tau +\omega _{d}\left ( t-\tau \right ) \right ) d\tau \]
Integrate by parts, where \(\int udv=uv-\int vdu\) , Let \(dv=e^{\xi \omega _{n}\tau }\rightarrow v=\frac {e^{\xi \omega _{n}\tau }}{\xi \omega _{n}}\) and let \(u=\cos \left ( \tau +\omega _{d}\left ( t-\tau \right ) \right ) \rightarrow du=-\left ( 1-\omega _{d}\right ) \sin \left ( \tau +\omega _{d}\left ( t-\tau \right ) \right ) \) , hence
\begin{align} I_{2} & =\left [ \cos \left ( \tau +\omega _{d}\left ( t-\tau \right ) \right ) \frac {e^{\xi \omega _{n}\tau }}{\xi \omega _{n}}\right ] _{0}^{t}-{\displaystyle \int \limits _{0}^{t}} \frac {e^{\xi \omega _{n}\tau }}{\xi \omega _{n}}\left [ -\left ( 1-\omega _{d}\right ) \sin \left ( \tau +\omega _{d}\left ( t-\tau \right ) \right ) \right ] d\tau \nonumber \\ & \nonumber \\ & =\left [ \cos \left ( t+\omega _{d}\left ( t-t\right ) \right ) \frac {e^{\xi \omega _{n}t}}{\xi \omega _{n}}-\cos \left ( 0+\omega _{d}\left ( t-0\right ) \right ) \frac {1}{\xi \omega _{n}}\right ] \nonumber \\ & +\frac {\left ( 1-\omega _{d}\right ) }{\xi \omega _{n}}{\displaystyle \int \limits _{0}^{t}} e^{\xi \omega _{n}\tau }\sin \left ( \tau +\omega _{d}\left ( t-\tau \right ) \right ) d\tau \nonumber \\ & \nonumber \\ & =\frac {1}{\xi \omega _{n}}\left [ \cos \left ( t\right ) e^{\xi \omega _{n}t}-\cos \left ( \omega _{d}t\right ) \right ] +\frac {\left ( 1-\omega _{d}\right ) }{\xi \omega _{n}}{\displaystyle \int \limits _{0}^{t}} e^{\xi \omega _{n}\tau }\sin \left ( \tau +\omega _{d}\left ( t-\tau \right ) \right ) d\tau \tag {3}\end{align}
Integrate by parts again the last integral above, where \(\int udv=uv-\int vdu\) , Let \(dv=e^{\xi \omega _{n}\tau }\rightarrow v=\frac {e^{\xi \omega _{n}\tau }}{\xi \omega _{n}}\) and let \(u=\sin \left ( \tau +\omega _{d}\left ( t-\tau \right ) \right ) \rightarrow du=\left ( 1-\omega _{d}\right ) \cos \left ( \tau +\omega _{d}\left ( t-\tau \right ) \right ) \) , hence
\begin{align}{\displaystyle \int \limits _{0}^{t}} e^{\xi \omega _{n}\tau }\sin \left ( \tau +\omega _{d}\left ( t-\tau \right ) \right ) d\tau & =\left [ \sin \left ( \tau +\omega _{d}\left ( t-\tau \right ) \right ) \frac {e^{\xi \omega _{n}\tau }}{\xi \omega _{n}}\right ] _{0}^{t}-\int _{0}^{t}\frac {e^{\xi \omega _{n}\tau }}{\xi \omega _{n}}\left ( 1-\omega _{d}\right ) \cos \left ( \tau +\omega _{d}\left ( t-\tau \right ) \right ) d\tau \nonumber \\ & =\frac {1}{\xi \omega _{n}}\left [ \sin \left ( t\right ) e^{\xi \omega _{n}t}-\sin \left ( \omega _{d}t\right ) \right ] -\frac {\left ( 1-\omega _{d}\right ) }{\xi \omega _{n}}\int _{0}^{t}e^{\xi \omega _{n}\tau }\cos \left ( \tau +\omega _{d}\left ( t-\tau \right ) \right ) d\tau \tag {4}\end{align}
Substitute (4) into (3) we obtain
\begin{align*} I_{2} & =\frac {1}{\xi \omega _{n}}\left [ \cos \left ( t\right ) e^{\xi \omega _{n}t}-\cos \left ( \omega _{d}t\right ) \right ] +\\ & \frac {\left ( 1-\omega _{d}\right ) }{\xi \omega _{n}}\left ( \frac {1}{\xi \omega _{n}}\left [ \sin \left ( t\right ) e^{\xi \omega _{n}t}-\sin \left ( \omega _{d}t\right ) \right ] -\frac {\left ( 1-\omega _{d}\right ) }{\xi \omega _{n}}\int _{0}^{t}e^{\xi \omega _{n}\tau }\cos \left ( \tau +\omega _{d}\left ( t-\tau \right ) \right ) d\tau \right ) \\ & \\ & =\frac {1}{\xi \omega _{n}}\left [ \cos \left ( t\right ) e^{\xi \omega _{n}t}-\cos \left ( \omega _{d}t\right ) \right ] +\frac {\left ( 1-\omega _{d}\right ) }{\left ( \xi \omega _{n}\right ) ^{2}}\left [ \sin \left ( t\right ) e^{\xi \omega _{n}t}-\sin \left ( \omega _{d}t\right ) \right ] \\ & -\frac {\left ( 1-\omega _{d}\right ) ^{2}}{\left ( \xi \omega _{n}\right ) ^{2}}\int _{0}^{t}e^{\xi \omega _{n}\tau }\cos \left ( \tau +\omega _{d}\left ( t-\tau \right ) \right ) d\tau \\ & \\ & =\frac {1}{\xi \omega _{n}}\left [ \cos \left ( t\right ) e^{\xi \omega _{n}t}-\cos \left ( \omega _{d}t\right ) \right ] +\frac {\left ( 1-\omega _{d}\right ) }{\left ( \xi \omega _{n}\right ) ^{2}}\left [ \sin \left ( t\right ) e^{\xi \omega _{n}t}+\sin \left ( \omega _{d}t\right ) \right ] -\frac {\left ( 1-\omega _{d}\right ) ^{2}}{\left ( \xi \omega _{n}\right ) ^{2}}I_{2}\end{align*}
Hence
\begin{align*} I_{2}+\frac {\left ( 1-\omega _{d}\right ) ^{2}}{\left ( \xi \omega _{n}\right ) ^{2}}I_{2} & =\frac {1}{\xi \omega _{n}}\left [ \cos \left ( t\right ) e^{\xi \omega _{n}t}-\cos \left ( \omega _{d}t\right ) \right ] +\frac {\left ( 1-\omega _{d}\right ) }{\left ( \xi \omega _{n}\right ) ^{2}}\left [ \sin \left ( t\right ) e^{\xi \omega _{n}t}+\sin \left ( \omega _{d}t\right ) \right ] \\ I_{2}\left ( \frac {\left ( \xi \omega _{n}\right ) ^{2}+\left ( 1-\omega _{d}\right ) ^{2}}{\left ( \xi \omega _{n}\right ) ^{2}}\right ) & =\frac {1}{\xi \omega _{n}}\left [ \cos \left ( t\right ) e^{\xi \omega _{n}t}-\cos \left ( \omega _{d}t\right ) \right ] +\frac {\left ( 1-\omega _{d}\right ) }{\left ( \xi \omega _{n}\right ) ^{2}}\left [ \sin \left ( t\right ) e^{\xi \omega _{n}t}+\sin \left ( \omega _{d}t\right ) \right ] \\ I_{2} & =\left ( \frac {\left ( \xi \omega _{n}\right ) ^{2}}{\left ( \xi \omega _{n}\right ) ^{2}+\left ( 1-\omega _{d}\right ) ^{2}}\right ) \\ & \left ( \frac {1}{\xi \omega _{n}}\left [ \cos \left ( t\right ) e^{\xi \omega _{n}t}-\cos \left ( \omega _{d}t\right ) \right ] +\frac {\left ( 1-\omega _{d}\right ) }{\left ( \xi \omega _{n}\right ) ^{2}}\left [ \sin \left ( t\right ) e^{\xi \omega _{n}t}+\sin \left ( \omega _{d}t\right ) \right ] \right ) \\ & \\ & =\frac {\xi \omega _{n}\left [ \cos \left ( t\right ) e^{\xi \omega _{n}t}-\cos \left ( \omega _{d}t\right ) \right ] +\left ( 1-\omega _{d}\right ) \left [ \sin \left ( t\right ) e^{\xi \omega _{n}t}+\sin \left ( \omega _{d}t\right ) \right ] }{\left ( \xi \omega _{n}\right ) ^{2}+\left ( 1-\omega _{d}\right ) ^{2}}\end{align*}
Using the above expressions for \(I_{1},I_{2}\) , we find (and multiplying the solution by \(\left ( \Phi \left ( t\right ) -\Phi \left ( t-\pi \right ) \right ) \) since the force is only active from \(t=0\) to \(t=\pi \) , we
obtain
\begin{align} x_{p}\left ( t\right ) & =\frac {F_{0}e^{-\xi \omega _{n}t}}{2m\omega _{d}}\left ( I_{1}-I_{2}\right ) \left ( \Phi \left ( t\right ) -\Phi \left ( t-\pi \right ) \right ) \nonumber \\ & \nonumber \\ & =\left ( \Phi \left ( t\right ) -\Phi \left ( t-\pi \right ) \right ) \ast \nonumber \\ & \frac {F_{0}e^{-\xi \omega _{n}t}}{2m\omega _{d}}\frac {\xi \omega _{n}\left [ \cos \left ( t\right ) e^{\xi \omega _{n}t}-\cos \left ( \omega _{d}t\right ) \right ] +\left ( 1+\omega _{d}\right ) \left [ \sin \left ( t\right ) e^{\xi \omega _{n}t}+\sin \left ( \omega _{d}t\right ) \right ] }{\left ( \xi \omega _{n}\right ) ^{2}+\left ( 1+\omega _{d}\right ) ^{2}}\nonumber \\ & -\frac {F_{0}e^{-\xi \omega _{n}t}}{2m\omega _{d}}\frac {\xi \omega _{n}\left [ \cos \left ( t\right ) e^{\xi \omega _{n}t}-\cos \left ( \omega _{d}t\right ) \right ] +\left ( 1-\omega _{d}\right ) \left [ \sin \left ( t\right ) e^{\xi \omega _{n}t}+\sin \left ( \omega _{d}t\right ) \right ] }{\left ( \xi \omega _{n}\right ) ^{2}+\left ( 1-\omega _{d}\right ) ^{2}} \tag {5}\end{align}
Hence \(x_{p}\left ( t\right ) =\left ( \Phi \left ( t\right ) -\Phi \left ( t-\pi \right ) \right ) \)
\(\left [ \frac {F_{0}}{2m\omega _{d}}e^{-\xi \omega _{n}t}\left ( \frac {\xi \omega _{n}\left [ \cos \left ( t\right ) e^{\xi \omega _{n}t}-\cos \left ( \omega _{d}t\right ) \right ] +\left ( 1+\omega _{d}\right ) \left [ \sin \left ( t\right ) e^{\xi \omega _{n}t}+\sin \left ( \omega _{d}t\right ) \right ] }{\left ( \xi \omega _{n}\right ) ^{2}+\left ( 1+\omega _{d}\right ) ^{2}}-\frac {\xi \omega _{n}\left [ \cos \left ( t\right ) e^{\xi \omega _{n}t}-\cos \left ( \omega _{d}t\right ) \right ] +\left ( 1-\omega _{d}\right ) \left [ \sin \left ( t\right ) e^{\xi \omega _{n}t}+\sin \left ( \omega _{d}t\right ) \right ] }{\left ( \xi \omega _{n}\right ) ^{2}+\left ( 1-\omega _{d}\right ) ^{2}}\right ) \right ] \)
But
\begin{align*} \left ( \xi \omega _{n}\right ) ^{2}+\left ( 1+\omega _{d}\right ) ^{2} & =\xi ^{2}\omega _{n}^{2}+1+\omega _{d}^{2}+2\omega _{d}\\ & =\xi ^{2}\omega _{n}^{2}+1+\omega _{n}^{2}\left ( 1-\xi ^{2}\right ) +2\omega _{d}\\ & =1+2\omega _{d}+\omega _{n}^{2}\end{align*}
and
\[ \left ( \xi \omega _{n}\right ) ^{2}+\left ( 1-\omega _{d}\right ) ^{2}=1-2\omega _{d}+\omega _{n}^{2}\]
Hence \(x_{p}\left ( t\right ) \) can now be written as
\begin{align*} x_{p}\left ( t\right ) & =\frac {F_{0}e^{-\xi \omega _{n}t}}{2m\omega _{d}}\frac {\xi \omega _{n}\cos \left ( t\right ) e^{\xi \omega _{n}t}-\xi \omega _{n}\cos \left ( \omega _{d}t\right ) +\left ( 1+\omega _{d}\right ) \sin \left ( t\right ) e^{\xi \omega _{n}t}+\left ( 1+\omega _{d}\right ) \sin \left ( \omega _{d}t\right ) }{1+2\omega _{d}+\omega _{n}^{2}}\\ & -\frac {F_{0}e^{-\xi \omega _{n}t}}{2m\omega _{d}}\frac {\xi \omega _{n}\cos \left ( t\right ) e^{\xi \omega _{n}t}-\xi \omega _{n}\cos \left ( \omega _{d}t\right ) +\left ( 1-\omega _{d}\right ) \sin \left ( t\right ) e^{\xi \omega _{n}t}+\left ( 1-\omega _{d}\right ) \sin \left ( \omega _{d}t\right ) }{1-2\omega _{d}+\omega _{n}^{2}}\end{align*}
And
\[ x_{h}\left ( t\right ) =e^{-\xi \omega _{n}t}\left ( A\cos \omega _{d}t+B\sin \omega _{d}t\right ) \]
Hence the overall solution is
\[ x\left ( t\right ) =e^{-\xi \omega _{n}t}\left ( A\cos \omega _{d}t+B\sin \omega _{d}t\right ) +x_{p}\left ( t\right ) \]
The above solution is a bit long due to integration by parts. I will not solve the
same problem using Laplace transformation method. The differential equation is
\[ \ddot {x}\left ( t\right ) +2\xi \omega _{n}\dot {x}\left ( t\right ) +\omega _{n}^{2}x\left ( t\right ) =f\left ( t\right ) \]
Take Laplace transform, we obtain
(assuming \(x\left ( 0\right ) =x_{0}\) and \(\dot {x}\left ( 0\right ) =v_{0}\) )
\begin{align} \left ( s^{2}X-sx\left ( 0\right ) -\dot {x}\left ( 0\right ) \right ) +2\xi \omega _{n}\left ( sX-x\left ( 0\right ) \right ) +\omega _{n}^{2}X & =F\left ( s\right ) \nonumber \\ \left ( s^{2}X-sx_{0}-v_{0}\right ) +2\xi \omega _{n}\left ( sX-x_{0}\right ) +\omega _{n}^{2}X & =F\left ( s\right ) \tag {7}\end{align}
Now we find Laplace transform of \(f\left ( t\right ) \)
\begin{align*} F\left ( s\right ) & ={\displaystyle \int \limits _{0}^{\infty }} e^{-st}f\left ( t\right ) dt\\ & ={\displaystyle \int \limits _{0}^{\pi }} e^{-st}F_{0}\sin t\ dt\\ & =F_{0}\left [ {\displaystyle \int \limits _{0}^{\pi }} e^{-st}\sin t\ dt\right ] \end{align*}
Integration by parts gives
\begin{equation} F\left ( s\right ) =F_{0}\left [ \frac {1+e^{-\pi s}}{1+s^{2}}\right ] \tag {8}\end{equation}
\begin{align*} \left ( s^{2}X-sx_{0}-v_{0}\right ) +2\xi \omega _{n}\left ( sX-x_{0}\right ) +\omega _{n}^{2}X & =F_{0}\left [ \frac {1+e^{-\pi s}}{1+s^{2}}\right ] \\ X\left ( s^{2}+2\xi \omega _{n}s+\omega _{n}^{2}\right ) -sx_{0}-v_{0}-2\xi \omega _{n}x_{0} & =\frac {F_{0}\left ( 1+e^{-\pi s}\right ) }{1+s^{2}}\\ X\left ( s^{2}+2\xi \omega _{n}s+\omega _{n}^{2}\right ) & =\frac {F_{0}\left ( 1+e^{-\pi s}\right ) }{1+s^{2}}+sx_{0}+v_{0}+2\xi \omega _{n}x_{0}\\ & =\frac {F_{0}\left ( 1+e^{-\pi s}\right ) +\left ( 1+s^{2}\right ) sx_{0}+v_{0}\left ( 1+s^{2}\right ) +2\xi \omega _{n}x_{0}\left ( 1+s^{2}\right ) }{1+s^{2}}\end{align*}
Hence
\begin{align*} X & =\frac {F_{0}\left ( 1+e^{-\pi s}\right ) +\left ( 1+s^{2}\right ) sx_{0}+v_{0}\left ( 1+s^{2}\right ) +2\xi \omega _{n}x_{0}\left ( 1+s^{2}\right ) }{\left ( 1+s^{2}\right ) \left ( s^{2}+2\xi \omega _{n}s+\omega _{n}^{2}\right ) }\\ & =\frac {F_{0}+v_{0}+\frac {F_{0}}{e^{\pi s}}+sx_{0}+s^{2}v_{0}+s^{3}x_{0}+2\xi \omega _{n}\allowbreak x_{0}+2s^{2}\xi \omega _{n}x_{0}}{\left ( 1+s^{2}\right ) \left ( s^{2}+2\xi \omega _{n}s+\omega _{n}^{2}\right ) }\end{align*}
Now we can use inverse Laplace transform on the above. It is easier to do partial fraction decomposition and use
tables. I used CAS to do this and this is the result. I plot the solution \(x\left ( t\right ) \) . I used the following values to be able to
obtain a plot \(\xi =0.5,\omega _{n}=2,F_{0}=10,x_{0}=1,v_{0}=0\)
3.3.6
The acceleration \(\ddot {x}\) of the mass is measured w.r.t. to the inertial frame, but the spring length is measured relative to
the ground which is moving with displacement \(y\left ( t\right ) \) , hence the equation of motion of the mass \(m\) is given by
\[ m\ddot {x}\left ( t\right ) +k\left ( x\left ( t\right ) -y\left ( y\right ) \right ) =0 \]
Therefore
\begin{equation} m\ddot {x}\left ( t\right ) +k\ x\left ( t\right ) =k\ y\left ( t\right ) \tag {1}\end{equation}
\(y\left ( t\right ) \) is given as
\[ y\left ( t\right ) =\left \{ \begin {array} [c]{ccc}2.5t & & 0\leq t\leq 0.2\\ 0.75-1.25t & & 0.2<t\leq 0.6\\ 0 & & 0.6<t \end {array} \right . \]
The solution to (1) is given by \(x\left ( t\right ) =x_{h}\left ( t\right ) +x_{p}\left ( t\right ) \) where \(x_{p}\left ( t\right ) \) can be found using convolution, and \(x_{h}\left ( t\right ) \) is as usual
given by
\[ x_{h}=A\cos \omega _{n}+B\sin \omega _{n}\]
Let us first find \(x_{p}\left ( t\right ) \) . Note that the impulse response \(h\left ( t\right ) \) to undamped system is given by
\[ h\left ( t\right ) =\frac {1}{m\omega _{n}}\sin \omega _{n}t \]
Hence for
\(0\leq t\leq 0.2\) ,
\begin{align} x_{p\left ( 0\cdots 0.2\right ) }\left ( t\right ) & =\int _{0}^{t}f\left ( \tau \right ) \left ( kh\left ( t-\tau \right ) \right ) d\tau \nonumber \\ & =\int _{0}^{t}2.5\tau \left ( \frac {k}{m\omega _{n}}\sin \omega _{n}\left ( t-\tau \right ) \right ) d\tau \nonumber \\ & =\frac {2.5k}{m\omega _{n}}\int _{0}^{t}\tau \sin \omega _{n}\left ( t-\tau \right ) \ d\tau \nonumber \\ & =2.5\omega _{n}\int _{0}^{t}\tau \sin \omega _{n}\left ( t-\tau \right ) \ d\tau \tag {2}\end{align}
Integration by parts, \(\int udv=uv-\int vdu\) where \(u=\tau \) , \(dv=\sin \omega _{n}\left ( t-\tau \right ) \) , hence \(v=\frac {-\cos \left ( \omega _{n}\left ( t-\tau \right ) \right ) }{-\omega _{n}}\) , therefore (2) becomes
\begin{align*} x_{p\left ( 0\cdots 0.2\right ) }\left ( t\right ) & =2.5\omega _{n}\left ( \left [ \tau \frac {\cos \omega _{n}\left ( t-\tau \right ) }{\omega _{n}}\right ] _{0}^{t}-\int _{0}^{t}\frac {\cos \left ( \omega _{n}\left ( t-\tau \right ) \right ) }{\omega _{n}}\ d\tau \right ) \\ & =2.5\omega _{n}\left ( \frac {t}{\omega _{n}}+\frac {1}{\omega _{n}^{2}}\left [ \sin \left ( \omega _{n}\left ( t-\tau \right ) \right ) \right ] _{0}^{t}\ \right ) \\ & =2.5\omega _{n}\left ( \frac {t}{\omega _{n}}+\frac {1}{\omega _{n}^{2}}\left [ \sin \omega _{n}\left ( t-t\right ) -\sin \omega _{n}\left ( t\right ) \right ] \ \right ) \\ & =2.5\left ( t-\frac {\sin \omega _{n}t}{\omega _{n}}\right ) \end{align*}
For \(0.2<t\leq 0.6\)
\begin{align} x_{p\left ( 0.2\cdots 0.6\right ) }\left ( t\right ) & =\omega _{n}\int _{0}^{0.2}2.5\tau \sin \omega _{n}\left ( t-\tau \right ) \ d\tau +\int _{0.2}^{t}f\left ( \tau \right ) \left ( kh\left ( t-\tau \right ) \right ) d\tau \nonumber \\ & =2.5\omega _{n}\int _{0}^{0.2}\tau \sin \omega _{n}\left ( t-\tau \right ) \ d\tau +\int _{0.2}^{t}\left ( 0.75-1.25\tau \right ) \left ( \frac {k}{m\omega _{n}}\sin \omega _{n}\left ( t-\tau \right ) \right ) d\tau \nonumber \\ & =2.5\omega _{n}\int _{0}^{0.2}\tau \sin \omega _{n}\left ( t-\tau \right ) \ d\tau +\nonumber \\ & \int _{0.2}^{t}0.75\frac {k}{m\omega _{n}}\sin \omega _{n}\left ( t-\tau \right ) d\tau \nonumber \\ & -\int _{0.2}^{t}1.25\tau \frac {k}{m\omega _{n}}\sin \omega _{n}\left ( t-\tau \right ) d\tau \tag {3}\end{align}
For the first integral in (3), we obtain
\begin{align*} I_{1} & =2.5\omega _{n}\int _{0}^{0.2}\tau \sin \omega _{n}\left ( t-\tau \right ) \ d\tau \\ & =2.5\omega _{n}\left ( \left [ \tau \frac {\cos \omega _{n}\left ( t-\tau \right ) }{\omega _{n}}\right ] _{0}^{0.2}-\int _{0}^{0.2}\frac {\cos \left ( \omega _{n}\left ( t-\tau \right ) \right ) }{\omega _{n}}\ d\tau \right ) \\ & =2.5\omega _{n}\left ( 0.2\frac {\cos \omega _{n}\left ( t-0.2\right ) }{\omega _{n}}+\frac {1}{\omega _{n}^{2}}\left [ \sin \left ( \omega _{n}\left ( t-\tau \right ) \right ) \right ] _{0}^{0.2}\ \right ) \\ & =2.5\omega _{n}\left ( 0.2\frac {\cos \omega _{n}\left ( t-0.2\right ) }{\omega _{n}}+\frac {1}{\omega _{n}^{2}}\left ( \sin \omega _{n}\left ( t-0.2\right ) -\sin \omega _{n}t\right ) \right ) \\ & =0.5\cos \omega _{n}\left ( t-0.2\right ) +\frac {2.5}{\omega _{n}}\sin \omega _{n}\left ( t-0.2\right ) -\frac {2.5}{\omega _{n}}\sin \omega _{n}t \end{align*}
For the second integral in (3) we obtain
\begin{align*} I_{2} & =0.75\omega _{n}\int _{0.2}^{t}\sin \omega _{n}\left ( t-\tau \right ) d\tau \\ & =0.75\left [ \cos \omega _{n}\left ( t-\tau \right ) \right ] _{0.2}^{t}\\ & =0.75\left ( 1-\cos \omega _{n}\left ( t-0.2\right ) \right ) \end{align*}
For the third integral in (3) we obtain
\begin{align*} I_{3} & =\int _{0.2}^{t}1.25\tau \frac {k}{m\omega _{n}}\sin \omega _{n}\left ( t-\tau \right ) d\tau \\ & =1.25\omega _{n}\int _{0.2}^{t}\tau \sin \omega _{n}\left ( t-\tau \right ) d\tau \end{align*}
Integration by parts gives
\begin{align*} I_{3} & =1.25\omega _{n}\left ( \left [ \tau \frac {\cos \omega _{n}\left ( t-\tau \right ) }{\omega _{n}}\right ] _{0.2}^{t}-\int _{0.2}^{t}\frac {\cos \left ( \omega _{n}\left ( t-\tau \right ) \right ) }{\omega _{n}}\ d\tau \right ) \\ & =1.25\omega _{n}\left ( \frac {t}{\omega _{n}}-0.2\frac {\cos \omega _{n}\left ( t-0.2\right ) }{\omega _{n}}+\frac {1}{\omega _{n}^{2}}\left [ \sin \omega _{n}\left ( t-\tau \right ) \right ] _{0.2}^{t}\ \right ) \\ & =1.25\omega _{n}\left ( \frac {t}{\omega _{n}}-0.2\frac {\cos \omega _{n}\left ( t-0.2\right ) }{\omega _{n}}+\frac {1}{\omega _{n}^{2}}\left [ -\sin \omega _{n}\left ( t-0.2\right ) \right ] \ \right ) \\ & =1.25\left ( t-0.2\cos \omega _{n}\left ( t-0.2\right ) -\frac {1}{\omega _{n}}\sin \omega _{n}\left ( t-0.2\right ) \ \right ) \end{align*}
Hence
\begin{align*} x_{p\left ( 0.2\cdots 0.6\right ) }\left ( t\right ) & =I_{1}+I_{2}-I_{3}\\ & =0.5\cos \omega _{n}\left ( t-0.2\right ) +\frac {2.5}{\omega _{n}}\sin \omega _{n}\left ( t-0.2\right ) -\frac {2.5}{\omega _{n}}\sin \omega _{n}t+\\ & 0.75\left ( 1-\cos \omega _{n}\left ( t-0.2\right ) \right ) \\ & -1.25\left ( t-0.2\cos \omega _{n}\left ( t-0.2\right ) -\frac {1}{\omega _{n}}\sin \omega _{n}\left ( t-0.2\right ) \ \right ) \\ & \\ & =0.5\cos \omega _{n}\left ( t-0.2\right ) +\frac {2.5}{\omega _{n}}\sin \omega _{n}\left ( t-0.2\right ) -\frac {2.5}{\omega _{n}}\sin \omega _{n}t+\\ & 0.75-0.75\cos \omega _{n}\left ( t-0.2\right ) \\ & -1.25t+0.25\cos \omega _{n}\left ( t-0.2\right ) +\frac {1.25}{\omega _{n}}\sin \omega _{n}\left ( t-0.2\right ) \\ & \\ & =0.75-1.25t+\frac {3.75}{\omega _{n}}\sin \omega _{n}\left ( t-0.2\right ) -\frac {2.5}{\omega _{n}}\sin \omega _{n}t \end{align*}
For \(t>0.6\)
\begin{align} x_{p\left ( 0.6\cdots t\right ) }\left ( t\right ) & =2.5\omega _{n}\int _{0}^{0.2}\tau \sin \omega _{n}\left ( t-\tau \right ) \ d\tau +\int _{0.2}^{0.6}\left ( 0.75-1.25\tau \right ) \left ( \frac {k}{m\omega _{n}}\sin \omega _{n}\left ( t-\tau \right ) \right ) d\tau \nonumber \\ & =2.5\omega _{n}\int _{0}^{0.2}\tau \sin \omega _{n}\left ( t-\tau \right ) \ d\tau +\nonumber \\ & \int _{0.2}^{0.6}0.75\frac {k}{m\omega _{n}}\sin \omega _{n}\left ( t-\tau \right ) d\tau \nonumber \\ & -\int _{0.2}^{0.6}1.25\tau \frac {k}{m\omega _{n}}\sin \omega _{n}\left ( t-\tau \right ) d\tau \tag {4}\end{align}
For the first integral in (4), we obtain
\[ I_{1}=0.5\cos \omega _{n}\left ( t-0.2\right ) +\frac {2.5}{\omega _{n}}\sin \omega _{n}\left ( t-0.2\right ) -\frac {2.5}{\omega _{n}}\sin \omega _{n}t \]
For the second integral in (4) we obtain
\begin{align*} I_{2} & =0.75\omega _{n}\int _{0.2}^{0.6}\sin \omega _{n}\left ( t-\tau \right ) d\tau \\ & =0.75\left [ \cos \omega _{n}\left ( t-\tau \right ) \right ] _{0.2}^{0.6}\\ & =0.75\left ( \cos \omega _{n}\left ( t-0.6\right ) -\cos \omega _{n}\left ( t-0.2\right ) \right ) \\ & =0.75\cos \omega _{n}\left ( t-0.6\right ) -0.75\cos \omega _{n}\left ( t-0.2\right ) \end{align*}
For the third integral in (4) we obtain
\begin{align*} I_{3} & =\int _{0.2}^{0.6}1.25\tau \frac {k}{m\omega _{n}}\sin \omega _{n}\left ( t-\tau \right ) d\tau \\ & =1.25\omega _{n}\int _{0.2}^{0.6}\tau \sin \omega _{n}\left ( t-\tau \right ) d\tau \end{align*}
Integration by parts gives
\begin{align*} I_{3} & =1.25\omega _{n}\left ( \left [ \tau \frac {\cos \omega _{n}\left ( t-\tau \right ) }{\omega _{n}}\right ] _{0.2}^{0.6}-\int _{0.2}^{0.6}\frac {\cos \left ( \omega _{n}\left ( t-\tau \right ) \right ) }{\omega _{n}}\ d\tau \right ) \\ & =1.25\omega _{n}\left ( 0.6\frac {\cos \omega _{n}\left ( t-0.6\right ) }{\omega _{n}}-0.2\frac {\cos \omega _{n}\left ( t-0.2\right ) }{\omega _{n}}-\frac {1}{\omega _{n}^{2}}\left ( \sin \omega _{n}\left ( t-0.6\right ) -\sin \omega _{n}\left ( t-0.2\right ) \right ) \ \right ) \\ & =0.75\cos \omega _{n}\left ( t-0.6\right ) -0.25\cos \omega _{n}\left ( t-0.2\right ) -\frac {1.25}{\omega _{n}}\sin \omega _{n}\left ( t-0.6\right ) +\frac {1.25}{\omega _{n}}\sin \omega _{n}\left ( t-0.2\right ) \
\end{align*}
Hence
\begin{align*} x_{p\left ( 0.6\cdots t\right ) }\left ( t\right ) & =I_{1}+I_{2}-I_{3}\\ & =0.5\cos \omega _{n}\left ( t-0.2\right ) +\frac {2.5}{\omega _{n}}\sin \omega _{n}\left ( t-0.2\right ) -\frac {2.5}{\omega _{n}}\sin \omega _{n}t\\ & +0.75\cos \omega _{n}\left ( t-0.6\right ) -0.75\cos \omega _{n}\left ( t-0.2\right ) \\ & -0.75\cos \omega _{n}\left ( t-0.6\right ) +0.25\cos \omega _{n}\left ( t-0.2\right ) +\frac {1.25}{\omega _{n}}\sin \omega _{n}\left ( t-0.6\right ) -\frac {1.25}{\omega _{n}}\sin \omega _{n}\left ( t-0.2\right ) \\ & \\ & =\frac {3.75}{\omega _{n}}\sin \omega _{n}\left ( t-0.2\right ) -\frac {2.5}{\omega _{n}}\sin \omega _{n}t-\frac {1.25}{\omega _{n}}\sin \omega _{n}\left ( t-0.6\right ) \end{align*}
Hence, the overall response is, assuming zero initial conditions, is given by
\[ x\left ( t\right ) =\left \{ \begin {array} [c]{ccc}2.5\left ( t-\frac {\sin \omega _{n}t}{\omega _{n}}\right ) & & 0\leq t\leq 0.2\\ 0.75-1.25t+\frac {3.75}{\omega _{n}}\sin \omega _{n}\left ( t-0.2\right ) -\frac {2.5}{\omega _{n}}\sin \omega _{n}t & & 0.2<t\leq 0.6\\ \frac {3.75}{\omega _{n}}\sin \omega _{n}\left ( t-0.2\right ) -\frac {2.5}{\omega _{n}}\sin \omega _{n}t-\frac {1.25}{\omega _{n}}\sin \omega _{n}\left ( t-0.6\right ) & & t>0.6 \end {array} \right . \]
Noting that \(\omega _{n}=\sqrt {\frac {k}{m}}=\sqrt {\frac {1500}{5000}}=0.54772,\) the above becomes
\[ x\left ( t\right ) =\left \{ \begin {array} [c]{ccc}2.5t-4.\,\allowbreak 564\,4\sin \omega _{n}t & & 0\leq t\leq 0.2\\ 0.75-1.25t+6.\,\allowbreak 846\,6\sin \omega _{n}\left ( t-0.2\right ) -4.\,\allowbreak 564\,4\sin \omega _{n}t & & 0.2<t\leq 0.6\\ 6.\,\allowbreak 846\,6\sin \omega _{n}\left ( t-0.2\right ) -4.\,\allowbreak 564\,4\sin \omega _{n}t-2.\,\allowbreak 282\,2\sin \omega _{n}\left ( t-0.6\right ) & & t>0.6 \end {array} \right . \]
This is a
plot of the solution superimposed on top of the forcing function
3.3.7
Let \(f\left ( t\right ) \) be the function shown above. Let \(\tilde {f}\left ( t\right ) \) be its approximation using Fourier series. Hence
\[ \tilde {f}\left ( t\right ) =\frac {a_{0}}{2}+{\displaystyle \sum \limits _{n=1}^{\infty }} a_{n}\cos \left ( \frac {2\pi }{T}nt\right ) +b_{n}\sin \left ( \frac {2\pi }{T}nt\right ) \]
Where \(T\) is the period of \(f\left ( t\right ) \)
and
\begin{align*} a_{0} & =\frac {2}{T}{\displaystyle \int \limits _{0}^{T}} f\left ( t\right ) dt\\ a_{n} & =\frac {2}{T}{\displaystyle \int \limits _{0}^{T}} f\left ( t\right ) \cos \left ( \frac {2\pi }{T}nt\right ) dt\ \ \ \ \ n=1,2,\cdots \\ b_{n} & =\frac {2}{T}{\displaystyle \int \limits _{0}^{T}} f\left ( t\right ) \sin \left ( \frac {2\pi }{T}nt\right ) dt\ \ \ \ \ n=1,2,\cdots \end{align*}
For \(f\left ( t\right ) \) we see that \(T=2\pi \) and \(f\left ( t\right ) =\frac {t}{T}\) for \(0\leq t\leq T\) , hence
\begin{align*} a_{0} & =\frac {2}{T}{\displaystyle \int \limits _{0}^{T}} \frac {t}{T}dt\\ & =\frac {2}{2\pi }{\displaystyle \int \limits _{0}^{T}} \frac {t}{2\pi }dt\\ & =\frac {1}{2\pi ^{2}}\left [ \frac {t^{2}}{2}\right ] _{0}^{2\pi }\\ & =\frac {1}{4\pi ^{2}}\left [ 4\pi ^{2}\right ] \\ & =1 \end{align*}
And
\begin{align*} a_{n} & =\frac {2}{T}{\displaystyle \int \limits _{0}^{2\pi }} \frac {t}{T}\cos \left ( nt\right ) dt\ \ \ \ \ n=1,2,\cdots \\ & =\frac {2}{2\pi }{\displaystyle \int \limits _{0}^{2\pi }} \frac {1}{2\pi }t\cos \left ( nt\right ) dt\ \ \ \ \ \\ & =\frac {1}{2\pi ^{2}}{\displaystyle \int \limits _{0}^{2\pi }} t\cos \left ( nt\right ) dt\ \\ & =\frac {1}{2\pi ^{2}}\left ( \left [ t\frac {\sin nt}{n}\right ] _{0}^{2\pi }-\frac {1}{n}{\displaystyle \int \limits _{0}^{2\pi }} \sin nt\ dt\right ) \\ & =\ \frac {1}{2\pi ^{2}}\left ( 0+\frac {1}{n}\left [ \frac {\cos nt}{n}\right ] _{0}^{2\pi }\ \right ) \\ & =\frac {1}{2\pi ^{2}}\left ( \frac {1}{n^{2}}\left [ \cos 2n\pi -1\right ] \right ) \\ & =\frac {1}{2n^{2}\pi ^{2}}\left ( \cos 2n\pi -1\right ) \\ & =0 \end{align*}
And
\begin{align*} b_{n} & =\frac {2}{T}{\displaystyle \int \limits _{0}^{2\pi }} \frac {t}{T}\sin \left ( nt\right ) dt\ \ \ \ \ n=1,2,\cdots \\ & =\frac {2}{2\pi }{\displaystyle \int \limits _{0}^{2\pi }} \frac {t}{2\pi }\sin \left ( nt\right ) dt\\ & =\frac {1}{2\pi ^{2}}\left ( \left [ -\frac {t\cos nt}{n}\right ] _{0}^{2\pi }+{\displaystyle \int \limits _{0}^{2\pi }} \frac {\cos nt}{n}dt\right ) \\ & =\frac {1}{2\pi ^{2}}\left ( \left [ \frac {-2\pi \cos 2\pi n}{n}\right ] -\frac {1}{n}\left [ \frac {\sin nt}{n}\right ] _{0}^{2\pi }\right ) \\ & =\frac {1}{2\pi ^{2}}\left ( \frac {-2\pi \cos 2\pi n}{n}\right ) \\ & =\frac {-\cos 2\pi n}{n\pi }\\ & =\frac {-1}{n\pi }\end{align*}
Hence
\begin{align*} \tilde {f}\left ( t\right ) & =\frac {a_{0}}{2}+{\displaystyle \sum \limits _{n=1}^{\infty }} a_{n}\cos \left ( \frac {2\pi }{T}nt\right ) +b_{n}\sin \left ( \frac {2\pi }{T}nt\right ) \\ & =\frac {1}{2}+{\displaystyle \sum \limits _{n=1}^{\infty }} \frac {-1}{n\pi }\sin \left ( nt\right ) \end{align*}
These are few terms in the series
\[ \tilde {f}\left ( t\right ) =\frac {1}{2}-\frac {1}{\pi }\sin t-\frac {1}{2\pi }\sin 2t-\frac {1}{3\pi }\sin 3t-\cdots \]
This is a plot of the above for increasing number of \(n\)
3.3.8 Problem
Solve the following system using Laplace transform \(100\ddot {x}\left ( t\right ) +2000x\left ( t\right ) =50\delta \left ( t\right ) \) where the units are in Newtons and the initial conditions are
both zero.
Answer
Divide the equation by \(50\) we obtain
\[ 2\ddot {x}\left ( t\right ) +40x\left ( t\right ) =\delta \left ( t\right ) \]
Let \(m=2,k=40,\) hence the equation becomes
\[ m\ddot {x}\left ( t\right ) +kx\left ( t\right ) =\delta \left ( t\right ) \]
Applying Laplace transform
\[ m\left ( s^{2}X\left ( s\right ) -sx_{0}-v_{0}\right ) +kX\left ( s\right ) =1 \]
But due to zero
initial conditions, the above simplifies to
\begin{align*} ms^{2}X\left ( s\right ) +kX\left ( s\right ) & =1\\ X\left ( s\right ) \left [ ms^{2}+k\right ] & =1\\ X\left ( s\right ) & =\frac {1}{ms^{2}+k}\end{align*}
From tables, the inverse Laplace transform of \(\frac {\alpha }{s^{2}+\alpha ^{2}}\) is \(\sin \alpha t\) , but
\[ \frac {1}{ms^{2}+k}=\frac {\frac {1}{m}}{s^{2}+\frac {k}{m}}=\frac {1}{m}\frac {1}{\sqrt {\frac {k}{m}}}\left ( \frac {\sqrt {\frac {k}{m}}}{s^{2}+\frac {k}{m}}\right ) \]
Hence, letting \(\alpha =\sqrt {\frac {k}{m}}\) \(\frac {1}{ms^{2}+k}\) is the
same as the inverse laplace transform of\(\frac {1}{m}\frac {1}{\alpha }\left ( \frac {\alpha }{s^{2}+\alpha ^{2}}\right ) \) which is \(\frac {1}{m}\frac {1}{\alpha }\sin \alpha t\)
But \(\alpha =\omega _{n}\) , hence
\[ x\left ( t\right ) =\frac {1}{m\omega _{n}}\sin \omega _{n}t \]
or
\begin{align*} x\left ( t\right ) & =\frac {1}{2\sqrt {\frac {40}{2}}}\sin \sqrt {\frac {40}{2}}t\\ & =0.111\,8\sin \left ( 4.\,\allowbreak 472\,1t\right ) \end{align*}
3.3.9 Problem
Calculate the response spectrum of an undamped system to the forcing function
\(F\left ( t\right ) =\left \{ \begin {array} [c]{ccc}F_{0}\sin \frac {\pi t}{t_{1}} & & 0\leq t\leq t_{1}\\ 0 & & t>t_{1}\end {array} \right . \) assuming zero initial conditions.
Answer
Solution sketch: Find the response \(x\left ( t\right ) \) of the system to the above input. Then find \(t\) where this response is maximum,
call this \(x_{\max }\) , then plot \(\left ( x_{\max }\frac {k}{F_{0}}\right ) \) vs. \(\frac {t\omega _{n}}{2\pi }\)
The system EQM is
\[ x^{\prime \prime }\left ( t\right ) +\omega _{n}^{2}x\left ( t\right ) =\frac {F\left ( t\right ) }{m}\]
For \(0<t\leq t_{1}\) ,
\begin{align*} x_{1}\left ( t\right ) & =x_{h}\left ( t\right ) +x_{p}\left ( t\right ) \\ & =A\cos \omega _{n}t+B\sin \omega _{n}t+x_{p}\left ( t\right ) \end{align*}
Guess \(x_{p}\left ( t\right ) =c_{1}\cos \omega t+c_{2}\sin \omega t\) \(x_{p}^{\prime }\left ( t\right ) =-\omega c_{1}\sin \omega t+\omega c_{2}\cos \omega t\) and \(x_{p}^{\prime \prime }\left ( t\right ) =-\omega ^{2}c_{1}\cos \omega t-\omega ^{2}c_{2}\sin \omega t\) , hence substitute these into the EQM and compare, we obtain
\[ \left ( -\omega ^{2}c_{1}\cos \omega t-\omega ^{2}c_{2}\sin \omega t\right ) +\omega _{n}^{2}\left ( c_{1}\cos \omega t+c_{2}\sin \omega t\right ) =\frac {F_{0}}{m}\sin \frac {\pi t}{t_{1}}\]
The input is half sin where \(\omega t=\frac {\pi t}{t_{1}}\) ,
hence \(\omega =\frac {\pi }{t_{1}}\) \(,\) hence the above becomes
\[ \left ( -\omega ^{2}c_{1}+\omega _{n}^{2}c_{1}\right ) \cos \omega t+\left ( -\omega ^{2}c_{2}+\omega _{n}^{2}c_{2}\right ) \sin \omega t=\frac {F_{0}}{m}\sin \omega t \]
Hence \(c_{1}=0\) \(c_{2}\left ( -\omega ^{2}+\omega _{n}^{2}\right ) =\frac {F_{0}}{m}\) or \(c_{2}=\frac {\frac {F_{0}}{m}}{\omega _{n}^{2}-\omega ^{2}}\) \(,\) Then the solution becomes
\[ x_{1}\left ( t\right ) =A\cos \omega _{n}t+B\sin \omega _{n}t+\frac {\frac {F_{0}}{m}}{\omega _{n}^{2}-\omega ^{2}}\sin \omega t \]
And since \(x\left ( 0\right ) =0\) then \(A=0\) and take derivative we
obtain
\[ x_{1}^{\prime }\left ( t\right ) =\omega _{n}B\cos \omega _{n}t+\omega \frac {\frac {F_{0}}{m}}{\omega _{n}^{2}-\omega ^{2}}\cos \omega t \]
And since \(x^{\prime }\left ( 0\right ) =0\) then the above results in
\begin{align*} 0 & =\omega _{n}B+\omega \frac {\frac {F_{0}}{m}}{\omega _{n}^{2}-\omega ^{2}}\\ B & =\frac {\omega }{\omega _{n}}\frac {\frac {F_{0}}{m}}{\omega ^{2}-\omega _{n}^{2}}\end{align*}
Hence the solution becomes
\begin{align*} x_{1}\left ( t\right ) & =\frac {\omega }{\omega _{n}}\frac {\frac {F_{0}}{m}}{\omega ^{2}-\omega _{n}^{2}}\sin \omega _{n}t+\frac {\frac {F_{0}}{m}}{\omega _{n}^{2}-\omega ^{2}}\sin \omega t\\ & =\frac {\frac {F_{0}}{m}}{\omega _{n}^{2}-\omega ^{2}}\left ( \sin \omega t-\frac {\omega }{\omega _{n}}\sin \omega _{n}t\right ) \\ & =\frac {\frac {F_{0}}{m}}{\omega _{n}^{2}\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left ( \sin \omega t-\frac {\omega }{\omega _{n}}\sin \omega _{n}t\right ) \\ & =\frac {\frac {F_{0}}{m}}{\frac {k}{m}\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left ( \sin \omega t-\frac {\omega }{\omega _{n}}\sin \omega _{n}t\right ) \end{align*}
Hence
\begin{equation} x_{1}\left ( t\right ) =\frac {\frac {F_{0}}{k}}{\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left ( \sin \omega t-\frac {\omega }{\omega _{n}}\sin \omega _{n}t\right ) \ \ \ \ \ \ \ \ \ 0<t\leq t_{1} \tag {1}\end{equation}
\[ x_{1}^{\prime }\left ( t\right ) =\frac {F_{0}}{k}\frac {1}{\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left ( \omega \cos \omega t-\omega \cos \omega _{n}t\right ) =0 \]
For \(\omega \neq \omega _{n}\) , we need to
solve
\[ \cos \omega t-\cos \omega _{n}t=0 \]
Using \(\cos A-\cos B=-2\sin \left ( \frac {A+B}{2}\right ) \sin \left ( \frac {A-B}{2}\right ) \) , then the above becomes
\begin{align*} -2\sin \left ( \frac {\left ( \omega +\omega _{n}\right ) t}{2}\right ) \sin \left ( \frac {\left ( \omega -\omega _{n}\right ) t}{2}\right ) & =0\\ \sin \left ( \frac {\left ( \omega +\omega _{n}\right ) t}{2}\right ) \sin \left ( \frac {\left ( \omega -\omega _{n}\right ) t}{2}\right ) & =0 \end{align*}
Hence, either \(\frac {\left ( \omega +\omega _{n}\right ) t_{p}}{2}=n\pi \) or \(\frac {\left ( \omega -\omega _{n}\right ) t_{p}}{2}=n\pi \) for \(n=\pm 1,\pm 2,\cdots \) or the time \(t_{p}\) which makes the maximum \(x\left ( t\right ) \) is one of the following
\[ t_{p}=\left \{ \begin {array} [c]{c}\frac {2n\pi }{\omega +\omega _{n}}\\ \frac {2n\pi }{\omega -\omega _{n}}\end {array} \right . \ \ \ \ \ n=\pm 1,\pm 2,\cdots \]
We now need to
find which one of the above 2 solution gives a larger maximum. Using the first solution \(t_{p}=\frac {2n\pi }{\omega +\omega _{n}}\) , then (1)
becomes
\begin{align*} x_{\max }\left ( t_{p_{1}}\right ) & =\frac {F_{0}}{k\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left ( \sin \omega \left ( \frac {2n\pi }{\omega +\omega _{n}}\right ) -\frac {\omega }{\omega _{n}}\sin \omega _{n}\left ( \frac {2n\pi }{\omega +\omega _{n}}\right ) \right ) \\ & =\frac {F_{0}}{k\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left ( \sin \left ( \frac {2n\pi \frac {\omega }{\omega _{n}}}{1+\frac {\omega }{\omega _{n}}}\right ) -\frac {\omega }{\omega _{n}}\sin \left ( \frac {2n\pi }{1+\frac {\omega }{\omega _{n}}}\right ) \right ) \end{align*}
And at \(t_{p}=\frac {2n\pi }{\omega -\omega _{n}},\) then (1) becomes
\begin{align*} x_{\max }\left ( t_{p_{2}}\right ) & =\frac {F_{0}}{k\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left ( \sin \omega \left ( \frac {2n\pi }{\omega -\omega _{n}}\right ) -\frac {\omega }{\omega _{n}}\sin \omega _{n}\left ( \frac {2n\pi }{\omega -\omega _{n}}\right ) \right ) \\ & =\frac {F_{0}}{k\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left ( \sin \left ( \frac {2n\pi \frac {\omega }{\omega _{n}}}{\frac {\omega }{\omega _{n}}-1}\right ) -\frac {\omega }{\omega _{n}}\sin \left ( \frac {2n\pi }{\frac {\omega }{\omega _{n}}-1}\right ) \right ) \end{align*}
Need now to find which of the above is larger. Let us take the difference and see if the result is positive or negative
(is there an easier way?)
\begin{align*} x\left ( t_{p_{1}}\right ) -x\left ( t_{p_{2}}\right ) & =\frac {F_{0}/k}{\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left ( \sin \left ( \frac {2n\pi \frac {\omega }{\omega _{n}}}{1+\frac {\omega }{\omega _{n}}}\right ) -\sin \left ( \frac {2n\pi \frac {\omega }{\omega _{n}}}{\frac {\omega }{\omega _{n}}-1}\right ) -\frac {\omega }{\omega _{n}}\sin \left ( \frac {2n\pi }{1+\frac {\omega }{\omega _{n}}}\right ) +\frac {\omega }{\omega _{n}}\sin \left ( \frac {2n\pi }{\frac {\omega }{\omega _{n}}-1}\right ) \right ) \\ & =\frac {F_{0}/k}{\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left ( \sin \left ( \frac {2n\pi \frac {\omega }{\omega _{n}}}{1+\frac {\omega }{\omega _{n}}}\right ) -\sin \left ( \frac {2n\pi \frac {\omega }{\omega _{n}}}{\frac {\omega }{\omega _{n}}-1}\right ) -\frac {\omega }{\omega _{n}}\sin \left ( \frac {2n\pi }{1+\frac {\omega }{\omega _{n}}}\right ) +\frac {\omega }{\omega _{n}}\sin \left ( \frac {2n\pi }{\frac {\omega }{\omega _{n}}-1}\right ) \right ) \end{align*}
Not sure how to continue. Now let us look at \(t>t_{1}.\) The solution here is
\[ x_{2}\left ( t\right ) =A\cos \omega _{n}t+B\sin \omega _{n}t \]
But with IC given by \(x_{1}\left ( t_{1}\right ) \) and \(x_{1}^{\prime }\left ( t_{1}\right ) ,\) hence from (1)
\[ x_{1}\left ( t_{1}\right ) =\frac {\frac {F_{0}}{k}}{\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left ( \sin \omega t_{1}-\frac {\omega }{\omega _{n}}\sin \omega _{n}t_{1}\right ) \]
and
\[ x_{1}^{\prime }\left ( t_{1}\right ) =\frac {F_{0}}{k}\frac {1}{\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left ( \omega \cos \omega t_{1}-\omega \cos \omega _{n}t_{1}\right ) \]
Hence
\begin{equation} x_{2}\left ( t_{1}\right ) =A\cos \omega _{n}t_{1}+B\sin \omega _{n}t_{1}=\frac {F_{0}/k}{\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left ( \sin \omega t_{1}-\frac {\omega }{\omega _{n}}\sin \omega _{n}t_{1}\right ) \tag {3}\end{equation}
\begin{equation} x_{2}^{\prime }\left ( t_{1}\right ) =-\omega _{n}A\sin \omega _{n}t+B\omega _{n}\cos \omega _{n}t=\frac {F_{0}/k}{\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left ( \omega \cos \omega t_{1}-\omega \cos \omega _{n}t_{1}\right ) \tag {4}\end{equation}
\(A\) and \(B.\) Combining (3) and (4) we obtain
\[\begin {bmatrix} \cos \omega _{n}t_{1} & \sin \omega _{n}t_{1}\\ -\omega _{n}\sin \omega _{n}t & \omega _{n}\cos \omega _{n}t \end {bmatrix}\begin {bmatrix} A\\ B \end {bmatrix} =\begin {bmatrix} \frac {F_{0}/k}{\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left ( \sin \omega t_{1}-\frac {\omega }{\omega _{n}}\sin \omega _{n}t_{1}\right ) \\ \frac {F_{0}/k}{\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left ( \omega \cos \omega t_{1}-\omega \cos \omega _{n}t_{1}\right ) \end {bmatrix} \]
This is in the form \(Ax=b\) , solve for \(x,\)
we obtain
\[\begin {bmatrix} A\\ B \end {bmatrix} =\frac {1}{\omega _{n}}\begin {bmatrix} \omega _{n}\cos \omega _{n}t_{1} & -\sin \omega _{n}t_{1}\\ \omega _{n}\sin \omega _{n}t & \cos \omega _{n}t \end {bmatrix}\begin {bmatrix} \left ( \sin \omega t_{1}-\frac {\omega }{\omega _{n}}\sin \omega _{n}t_{1}\right ) \\ \left ( \omega \cos \omega t_{1}-\omega \cos \omega _{n}t_{1}\right ) \end {bmatrix} \left ( \frac {F_{0}/k}{\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\right ) \]
Hence
\begin{align*} A & =\frac {1}{\omega _{n}}\frac {F_{0}/k}{\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left [ \omega _{n}\cos \omega _{n}t_{1}\left ( \sin \omega t_{1}-\frac {\omega }{\omega _{n}}\sin \omega _{n}t_{1}\right ) -\sin \omega _{n}t_{1}\left ( \omega \cos \omega t_{1}-\omega \cos \omega _{n}t_{1}\right ) \right ] \\ & =\frac {1}{\omega _{n}}\frac {F_{0}/k}{\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left [ \omega _{n}\cos \omega _{n}t_{1}\sin \omega t_{1}-\omega \cos \omega _{n}t_{1}\sin \omega _{n}t_{1}-\omega \sin \omega _{n}t_{1}\cos \omega t_{1}+\omega \sin \omega _{n}t_{1}\cos \omega _{n}t_{1}\right ] \\ & =\frac {1}{\omega _{n}}\frac {F_{0}/k}{\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left [ \omega _{n}\cos \omega _{n}t_{1}\sin \omega t_{1}-\omega \sin \omega _{n}t_{1}\cos \omega t_{1}\right ] \end{align*}
And
\begin{align*} B & =\frac {1}{\omega _{n}}\frac {F_{0}/k}{\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left [ \omega _{n}\sin \omega _{n}t_{1}\left ( \sin \omega t_{1}-\frac {\omega }{\omega _{n}}\sin \omega _{n}t_{1}\right ) +\cos \omega _{n}t_{1}\left ( \omega \cos \omega t_{1}-\omega \cos \omega _{n}t_{1}\right ) \right ] \\ & =\frac {1}{\omega _{n}}\frac {F_{0}/k}{\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left [ \omega _{n}\sin \omega _{n}t_{1}\sin \omega t_{1}-\omega \sin \omega _{n}t_{1}\sin \omega _{n}t_{1}+\omega \cos \omega _{n}t_{1}\cos \omega t_{1}-\omega \cos \omega _{n}t_{1}\cos \omega _{n}t_{1}\right ] \\ & =\frac {1}{\omega _{n}}\frac {F_{0}/k}{\left [ 1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}\right ] }\left [ \omega _{n}\sin \omega _{n}t_{1}\sin \omega t_{1}-\omega \sin \omega _{n}t_{1}\sin \omega _{n}t_{1}+\omega \cos \omega _{n}t_{1}\cos \omega t_{1}-\omega \cos \omega _{n}t_{1}\cos \omega _{n}t_{1}\right ] \end{align*}
Ask about the above, why can’t I get the answer shown in notes?
3.3.10 To find the response \(x\left ( t\right ) \) use convolution. Since this is an undamped system, then the impulse response is
\[ h\left ( t\right ) =\frac {1}{m\omega _{n}}\sin \omega _{n}t \]
Hence, for
\(0\leq t\leq t_{1}\)
\begin{align*} x\left ( t\right ) & =\int _{0}^{t}F\left ( \tau \right ) h\left ( t-\tau \right ) d\tau \ \ \ \\ & =\ \int _{0}^{t}\left ( F_{0}\sin \frac {\pi \tau }{t_{1}}\right ) \frac {1}{m\omega _{n}}\sin \omega _{n}\left ( t-\tau \right ) d\tau \\ & =\frac {F_{0}}{m\omega _{n}}\int _{0}^{t}\sin \frac {\pi \tau }{t_{1}}\sin \omega _{n}\left ( t-\tau \right ) d\tau \end{align*}
Using \(\sin A\sin B=\frac {1}{2}\left ( \cos \left ( A-B\right ) -\cos \left ( A+B\right ) \right ) \) , then
\[ \sin \overset {A}{\overbrace {\frac {\pi \tau }{t_{1}}}}\sin \overset {B}{\overbrace {\omega _{n}\left ( t-\tau \right ) }}=\frac {1}{2}\cos \left ( \frac {\pi \tau }{t_{1}}-\omega _{n}\left ( t-\tau \right ) \right ) -\frac {1}{2}\cos \left ( \frac {\pi \tau }{t_{1}}+\omega _{n}\left ( t-\tau \right ) \right ) \]
Hence the convolution integral becomes
\begin{align*} x\left ( t\right ) & =\frac {F_{0}}{m\omega _{n}}\int _{0}^{t}\frac {1}{2}\cos \left ( \frac {\pi \tau }{t_{1}}-\omega _{n}\left ( t-\tau \right ) \right ) -\frac {1}{2}\cos \left ( \frac {\pi \tau }{t_{1}}+\omega _{n}\left ( t-\tau \right ) \right ) d\tau \\ & =\frac {F_{0}}{2m\omega _{n}}\left [ \int _{0}^{t}\cos \left ( \frac {\pi \tau }{t_{1}}-\omega _{n}\left ( t-\tau \right ) \right ) d\tau -\int _{0}^{t}\cos \left ( \frac {\pi \tau }{t_{1}}+\omega _{n}\left ( t-\tau \right ) \right ) \right ] \\ & =\frac {F_{0}}{2m\omega _{n}}\left \{ \left [ \frac {\sin \left ( \frac {\pi \tau }{t_{1}}-\omega _{n}\left ( t-\tau \right ) \right ) }{\frac {\pi }{t_{1}}+\omega _{n}}\right ] _{0}^{t}-\left [ \frac {\sin \left ( \frac {\pi \tau }{t_{1}}+\omega _{n}\left ( t-\tau \right ) \right ) }{\frac {\pi }{t_{1}}-\omega _{n}}\right ] _{0}^{t}\right \} \\ & =\frac {F_{0}}{2m\omega _{n}}\left \{ \left ( \frac {\sin \left ( \frac {\pi t}{t_{1}}\right ) }{\frac {\pi }{t_{1}}+\omega _{n}}-\frac {\sin \left ( -\omega _{n}t\right ) }{\frac {\pi }{t_{1}}+\omega _{n}}\right ) -\left ( \frac {\sin \left ( \frac {\pi t}{t_{1}}\right ) }{\frac {\pi }{t_{1}}-\omega _{n}}-\frac {\sin \left ( \omega _{n}t\right ) }{\frac {\pi }{t_{1}}-\omega _{n}}\right ) \right \} \\ & =\frac {F_{0}}{2m\omega _{n}}\left \{ \frac {\sin \frac {\pi t}{t_{1}}}{\frac {\pi }{t_{1}}+\omega _{n}}+\frac {\sin \omega _{n}t}{\frac {\pi }{t_{1}}+\omega _{n}}-\frac {\sin \frac {\pi t}{t_{1}}}{\frac {\pi }{t_{1}}-\omega _{n}}+\frac {\sin \omega _{n}t}{\frac {\pi }{t_{1}}-\omega _{n}}\right \} \end{align*}
And for \(t>t_{1}\)
\begin{align*} x\left ( t\right ) & =\int _{0}^{t_{1}}F\left ( \tau \right ) h\left ( t-\tau \right ) d\tau \ \ +\int _{t_{1}}^{t}0\times h\left ( t-\tau \right ) d\tau \\ & =\ \int _{0}^{t_{1}}\left ( F_{0}\sin \frac {\pi \tau }{t_{1}}\right ) \frac {1}{m\omega _{n}}\sin \omega _{n}\left ( t-\tau \right ) d\tau \\ & =\frac {F_{0}}{m\omega _{n}}\int _{0}^{t_{1}}\sin \frac {\pi \tau }{t_{1}}\sin \omega _{n}\left ( t-\tau \right ) d\tau \end{align*}
As was done earlier, perform integration by parts, we obtain
\begin{align*} x\left ( t\right ) & =\frac {F_{0}}{2m\omega _{n}}\left \{ \left [ \frac {\sin \left ( \frac {\pi \tau }{t_{1}}-\omega _{n}\left ( t-\tau \right ) \right ) }{\frac {\pi }{t_{1}}+\omega _{n}}\right ] _{0}^{t_{1}}-\left [ \frac {\sin \left ( \frac {\pi \tau }{t_{1}}+\omega _{n}\left ( t-\tau \right ) \right ) }{\frac {\pi }{t_{1}}-\omega _{n}}\right ] _{0}^{t_{1}}\right \} \\ & =\frac {F_{0}}{2m\omega _{n}}\left \{ \frac {\sin \left ( \pi -\omega _{n}\left ( t-t_{1}\right ) \right ) }{\frac {\pi }{t_{1}}+\omega _{n}}-\frac {\sin \left ( -\omega _{n}t\right ) }{\frac {\pi }{t_{1}}+\omega _{n}}-\frac {\sin \left ( \pi +\omega _{n}\left ( t-t_{1}\right ) \right ) }{\frac {\pi }{t_{1}}-\omega _{n}}+\frac {\sin \left ( \omega _{n}t\right ) }{\frac {\pi }{t_{1}}-\omega _{n}}\right \} \\ & =\frac {F_{0}}{2m\omega _{n}}\left \{ \frac {\sin \left ( \pi -\omega _{n}\left ( t-t_{1}\right ) \right ) }{\frac {\pi }{t_{1}}+\omega _{n}}+\frac {\sin \left ( \omega _{n}t\right ) }{\frac {\pi }{t_{1}}+\omega _{n}}-\frac {\sin \left ( \pi +\omega _{n}\left ( t-t_{1}\right ) \right ) }{\frac {\pi }{t_{1}}-\omega _{n}}+\frac {\sin \left ( \omega _{n}t\right ) }{\frac {\pi }{t_{1}}-\omega _{n}}\right \} \end{align*}
But \(\sin \left ( \pi -\alpha \right ) =\sin \alpha \) and \(\sin \left ( \pi +\alpha \right ) =-\sin \alpha \) , hence the above becomes
\[ x\left ( t\right ) =\frac {F_{0}}{2m\omega _{n}}\left \{ \frac {\sin \omega _{n}\left ( t-t_{1}\right ) }{\frac {\pi }{t_{1}}+\omega _{n}}+\frac {\sin \left ( \omega _{n}t\right ) }{\frac {\pi }{t_{1}}+\omega _{n}}+\frac {\sin \omega _{n}\left ( t-t_{1}\right ) }{\frac {\pi }{t_{1}}-\omega _{n}}+\frac {\sin \left ( \omega _{n}t\right ) }{\frac {\pi }{t_{1}}-\omega _{n}}\right \} \]
Therefore, the final solution is
\begin{equation} x\left ( t\right ) =\left \{ \begin {array} [c]{ccc}\frac {F_{0}}{2m\omega _{n}}\left \{ \frac {\sin \frac {\pi t}{t_{1}}}{\frac {\pi }{t_{1}}+\omega _{n}}+\frac {\sin \omega _{n}t}{\frac {\pi }{t_{1}}+\omega _{n}}-\frac {\sin \frac {\pi t}{t_{1}}}{\frac {\pi }{t_{1}}-\omega _{n}}+\frac {\sin \omega _{n}t}{\frac {\pi }{t_{1}}-\omega _{n}}\right \} & & 0\leq t\leq t_{1}\\ \frac {F_{0}}{2m\omega _{n}}\left \{ \frac {\sin \omega _{n}\left ( t-t_{1}\right ) }{\frac {\pi }{t_{1}}+\omega _{n}}+\frac {\sin \left ( \omega _{n}t\right ) }{\frac {\pi }{t_{1}}+\omega _{n}}+\frac {\sin \omega _{n}\left ( t-t_{1}\right ) }{\frac {\pi }{t_{1}}-\omega _{n}}+\frac {\sin \left ( \omega _{n}t\right ) }{\frac {\pi }{t_{1}}-\omega _{n}}\right \} & & t>t_{1}\end {array} \right . \tag {1}\end{equation}
\begin{equation} x\left ( t\right ) =\left \{ \begin {array} [c]{ccc}\frac {F_{0}}{2m\omega _{n}}\left \{ \sin \frac {\pi t}{t_{1}}\left ( \frac {1}{\frac {\pi }{t_{1}}+\omega _{n}}-\frac {1}{\frac {\pi }{t_{1}}-\omega _{n}}\right ) +\sin \omega _{n}t\left ( \frac {1}{\frac {\pi }{t_{1}}+\omega _{n}}+\frac {1}{\frac {\pi }{t_{1}}-\omega _{n}}\right ) \right \} & & 0\leq t\leq t_{1}\\ \frac {F_{0}}{2m\omega _{n}}\left \{ \sin \omega _{n}\left ( t-t_{1}\right ) \left ( \frac {1}{\frac {\pi }{t_{1}}+\omega _{n}}+\frac {1}{\frac {\pi }{t_{1}}-\omega _{n}}\right ) +\sin \left ( \omega _{n}t\right ) \left ( \frac {1}{\frac {\pi }{t_{1}}+\omega _{n}}+\frac {1}{\frac {\pi }{t_{1}}-\omega _{n}}\right ) \right \} & & t>t_{1}\end {array} \right . \tag {1}\end{equation}
\begin{equation} x\left ( t\right ) =\left \{ \begin {array} [c]{ccc}\frac {F_{0}}{2m\omega _{n}}\left \{ \sin \frac {\pi t}{t_{1}}\left ( \frac {\left ( \frac {\pi }{t_{1}}-\omega _{n}\right ) -\left ( \frac {\pi }{t_{1}}+\omega _{n}\right ) }{\left ( \frac {\pi }{t_{1}}+\omega _{n}\right ) \left ( \frac {\pi }{t_{1}}-\omega _{n}\right ) }\right ) +\sin \omega _{n}t\left ( \frac {\left ( \frac {\pi }{t_{1}}-\omega _{n}\right ) +\left ( \frac {\pi }{t_{1}}+\omega _{n}\right ) }{\left ( \frac {\pi }{t_{1}}+\omega _{n}\right ) \left ( \frac {\pi }{t_{1}}-\omega _{n}\right ) }\right ) \right \} & & 0\leq t\leq t_{1}\\ \frac {F_{0}}{2m\omega _{n}}\left \{ \sin \omega _{n}\left ( t-t_{1}\right ) \left ( \frac {\left ( \frac {\pi }{t_{1}}-\omega _{n}\right ) +\left ( \frac {\pi }{t_{1}}+\omega _{n}\right ) }{\left ( \frac {\pi }{t_{1}}+\omega _{n}\right ) \left ( \frac {\pi }{t_{1}}-\omega _{n}\right ) }\right ) +\sin \left ( \omega _{n}t\right ) \left ( \frac {\left ( \frac {\pi }{t_{1}}-\omega _{n}\right ) +\left ( \frac {\pi }{t_{1}}+\omega _{n}\right ) }{\left ( \frac {\pi }{t_{1}}+\omega _{n}\right ) \left ( \frac {\pi }{t_{1}}-\omega _{n}\right ) }\right ) \right \} & & t>t_{1}\end {array} \right . \tag {1}\end{equation}
\begin{equation} x\left ( t\right ) =\left \{ \begin {array} [c]{ccc}\frac {F_{0}}{2m\omega _{n}}\left \{ \sin \frac {\pi t}{t_{1}}\left ( \frac {-2\omega _{n}}{\left ( \frac {\pi }{t_{1}}\right ) ^{2}-\omega _{n}^{2}}\right ) +\sin \omega _{n}t\left ( \frac {2\frac {\pi }{t_{1}}}{\left ( \frac {\pi }{t_{1}}\right ) ^{2}-\omega _{n}^{2}}\right ) \right \} & & 0\leq t\leq t_{1}\\ \frac {F_{0}}{2m\omega _{n}}\left \{ \sin \omega _{n}\left ( t-t_{1}\right ) \left ( \frac {2\frac {\pi }{t_{1}}}{\left ( \frac {\pi }{t_{1}}\right ) ^{2}-\omega _{n}^{2}}\right ) +\sin \left ( \omega _{n}t\right ) \left ( \frac {2\frac {\pi }{t_{1}}}{\left ( \frac {\pi }{t_{1}}\right ) ^{2}-\omega _{n}^{2}}\right ) \right \} & & t>t_{1}\end {array} \right . \tag {1}\end{equation}
\begin{equation} x\left ( t\right ) =\left \{ \begin {array} [c]{ccc}\left ( \frac {1}{\left ( \frac {\pi }{t_{1}}\right ) ^{2}-\omega _{n}^{2}}\right ) \frac {F_{0}}{2m\omega _{n}}\left \{ -2\omega _{n}\sin \frac {\pi t}{t_{1}}+2\frac {\pi }{t_{1}}\sin \omega _{n}t\right \} & & 0\leq t\leq t_{1}\\ \left ( \frac {\omega }{\left ( \frac {\pi }{t_{1}}\right ) ^{2}-\omega _{n}^{2}}\right ) \frac {F_{0}}{m\omega _{n}}\left \{ \sin \omega _{n}\left ( t-t_{1}\right ) -\sin \left ( \omega _{n}t\right ) \right \} & & t>t_{1}\end {array} \right . \tag {1}\end{equation}
\begin{equation} x\left ( t\right ) =\left \{ \begin {array} [c]{ccc}\left ( \frac {1}{\left ( \frac {\pi }{t_{1}}\right ) ^{2}-\omega _{n}^{2}}\right ) \frac {F_{0}}{m\omega _{n}}\left \{ -\omega _{n}\sin \frac {\pi t}{t_{1}}+\frac {\pi }{t_{1}}\sin \omega _{n}t\right \} & & 0\leq t\leq t_{1}\\ \left ( \frac {\omega }{\left ( \frac {\pi }{t_{1}}\right ) ^{2}-\omega _{n}^{2}}\right ) \frac {F_{0}}{m\omega _{n}}\left \{ \sin \omega _{n}\left ( t-t_{1}\right ) -\sin \omega _{n}t\right \} & & t>t_{1}\end {array} \right . \tag {1}\end{equation}
\(x_{\max }\) is, we need to find \(x_{\max }\) . Take the derivative, we obtain
\(\dot {x}\left ( t\right ) =\left \{ \begin {array} [c]{ccc}\left ( \frac {1}{\left ( \frac {\pi }{t_{1}}\right ) ^{2}-\omega _{n}^{2}}\right ) \frac {F_{0}}{m\omega _{n}}\left \{ -\omega _{n}\frac {\pi }{t_{1}}\cos \frac {\pi t}{t_{1}}+\frac {\pi }{t_{1}}\omega _{n}\cos \omega _{n}t\right \} & & 0\leq t\leq t_{1}\\ \left ( \frac {\omega }{\left ( \frac {\pi }{t_{1}}\right ) ^{2}-\omega _{n}^{2}}\right ) \frac {F_{0}}{m\omega _{n}}\left \{ \omega _{n}\cos \omega _{n}\left ( t-t_{1}\right ) -\omega _{n}\cos \omega _{n}t\right \} & & t>t_{1}\end {array} \right . \)
Now let \(\dot {x}\left ( t\right ) =0\) for \(t>t_{1}\) to find \(t_{peak}\) .
\begin{align} \left ( \frac {\omega }{\left ( \frac {\pi }{t_{1}}\right ) ^{2}-\omega _{n}^{2}}\right ) \frac {F_{0}}{m\omega _{n}}\left \{ \omega _{n}\cos \omega _{n}\left ( t_{p}-t_{1}\right ) -\omega _{n}\cos \omega _{n}t_{p}\right \} & =0\nonumber \\ \cos \omega _{n}\left ( t_{p}-t_{1}\right ) -\cos \omega _{n}t_{p} & =0 \tag {2}\end{align}
But
\[ \cos \omega _{n}\left ( t_{p}-t_{1}\right ) =\cos \omega _{n}t_{p}\cos \omega _{n}t_{1}+\sin \omega _{n}t_{p}\sin \omega _{n}t_{1}\]
Substitute the above into (2) we obtain
\[ \left ( \cos \omega _{n}t_{p}\cos \omega _{n}t_{1}+\sin \omega _{n}t_{p}\sin \omega _{n}t_{1}\right ) -\cos \omega _{n}t_{p}=0 \]
Divide by \(\cos \omega _{n}t_{p}\)
\begin{align*} \cos \omega _{n}t_{1}+\tan \omega _{n}t_{p}\sin \omega _{n}t_{1}-1 & =0\\ \tan \omega _{n}t_{p} & =\frac {\left ( 1-\cos \omega _{n}t_{1}\right ) }{\sin \omega _{n}t_{1}}\\ \omega _{n}t_{p} & =\tan ^{-1}\left ( \frac {1-\cos \omega _{n}t_{1}}{\sin \omega _{n}t_{1}}\right ) \end{align*}
Hence, the hypotenuse is \(\sqrt {\left ( 1-\cos \omega _{n}t_{1}\right ) ^{2}+\sin ^{2}\omega _{n}t_{1}}=\sqrt {2\left ( 1-\cos \omega _{n}t_{1}\right ) }\) and so \(\sin \omega _{n}t_{p}=-\sqrt {\frac {1}{2}\left ( 1-\cos \omega _{n}t_{1}\right ) }\) and \(\cos \omega _{n}t_{p}=\frac {-\sin \omega _{n}t_{1}}{\sqrt {2\left ( 1-\cos \omega _{n}t_{1}\right ) }}\) and using these into (1) we find \(x_{\max }\) when \(t>t_{1}\) as
\[ x_{\max }=\left ( \frac {\omega }{\left ( \frac {\pi }{t_{1}}\right ) ^{2}-\omega _{n}^{2}}\right ) \frac {F_{0}}{m\omega _{n}}\left \{ \sin \omega _{n}\left ( t_{p}-t_{1}\right ) -\sin \omega _{n}t_{p}\right \} \]
But \(\sin \omega _{n}\left ( t_{p}-t_{1}\right ) =\sin \omega _{n}t_{p}\cos \omega _{n}t_{1}-\cos \omega _{n}t_{p}\sin \omega _{n}t_{1}\) , hence
\begin{align*} x_{\max } & =\left ( \frac {\omega }{\omega ^{2}-\omega _{n}^{2}}\right ) \frac {F_{0}}{m\omega _{n}}\left \{ \sin \omega _{n}t_{p}\cos \omega _{n}t_{1}-\cos \omega _{n}t_{p}\sin \omega _{n}t_{1}-\sin \omega _{n}t_{p}\right \} \\ & =\left ( \frac {\omega }{\omega ^{2}-\omega _{n}^{2}}\right ) \frac {F_{0}}{m\omega _{n}}\left \{ -\sqrt {\frac {1}{2}\left ( 1-\cos \omega _{n}t_{1}\right ) }\cos \omega _{n}t_{1}+\frac {\sin \omega _{n}t_{1}}{\sqrt {2\left ( 1-\cos \omega _{n}t_{1}\right ) }}\sin \omega _{n}t_{1}+\sqrt {\frac {1}{2}\left ( 1-\cos \omega _{n}t_{1}\right ) }\right \} \\ & =\left ( \frac {\omega }{\omega ^{2}-\omega _{n}^{2}}\right ) \frac {F_{0}}{m\omega _{n}}\left \{ \frac {-\left ( 1-\cos \omega _{n}t_{1}\right ) \cos \omega _{n}t_{1}+\sin ^{2}\omega _{n}t_{1}+\left ( 1-\cos \omega _{n}t_{1}\right ) }{\sqrt {2\left ( 1-\cos \omega _{n}t_{1}\right ) }}\right \} \\ & =\left ( \frac {\omega }{\omega ^{2}-\omega _{n}^{2}}\right ) \frac {F_{0}}{m\omega _{n}}\left \{ \frac {-\cos \omega _{n}t_{1}+\cos ^{2}\omega _{n}t_{1}+\sin ^{2}\omega _{n}t_{1}+1-\cos \omega _{n}t_{1}}{\sqrt {2\left ( 1-\cos \omega _{n}t_{1}\right ) }}\right \} \\ & =\left ( \frac {\omega }{\omega ^{2}-\omega _{n}^{2}}\right ) \frac {F_{0}}{m\omega _{n}}\left \{ \frac {-\cos \omega _{n}t_{1}+1+1-\cos \omega _{n}t_{1}}{\sqrt {2\left ( 1-\cos \omega _{n}t_{1}\right ) }}\right \} \\ & =\left ( \frac {\omega }{\omega ^{2}-\omega _{n}^{2}}\right ) \frac {2F_{0}}{m\omega _{n}}\left \{ \frac {1-\cos \omega _{n}t_{1}}{\sqrt {2\left ( 1-\cos \omega _{n}t_{1}\right ) }}\right \} \\ & =\left ( \frac {\omega }{\omega ^{2}-\omega _{n}^{2}}\right ) \frac {F_{0}}{2m\omega _{n}}\left \{ \sqrt {1-\cos \omega _{n}t_{1}}\right \} \end{align*}
Hence
\begin{align*} x_{\max } & =\left ( \frac {\omega }{\omega ^{2}-\omega _{n}^{2}}\right ) \frac {F_{0}}{2m\omega _{n}}\sqrt {1-\cos \omega _{n}t_{1}}\\ & =\left ( \frac {\omega }{\omega ^{2}-\omega _{n}^{2}}\right ) \frac {\omega _{n}F_{0}}{2m\omega _{n}^{2}}\sqrt {1-\cos \omega _{n}t_{1}}\\ & =\left ( \frac {\frac {\omega }{\omega _{n}}}{\left ( \left ( \frac {\omega }{\omega _{n}}\right ) ^{2}-1\right ) }\right ) \frac {F_{0}}{2k}\sqrt {1-\cos \omega _{n}t_{1}}\end{align*}
Hence
\[ \fbox {$x_{\max }\frac {k}{F_{0}}=\left ( \frac {r}{r^{2}-1}\right ) \frac {1}{2}\sqrt {1-\cos \omega _{n}t_{1}}$}\]
Where \(r=\frac {\omega }{\omega _{n}}\)
A plot of \(x_{\max }\frac {k}{F_{0}}\) vs. \(r\) gives the response spectrum
3.3.11 Problem Calculate the compliance transfer function for a system described by \(ax^{\prime \prime \prime \prime }+bx^{\prime \prime \prime }+cx^{\prime \prime }+dx^{\prime }+ex=f\left ( t\right ) \) where \(f\left ( t\right ) \) is the input and \(x\left ( t\right ) \) is the
displacement.
Answer
Take Laplace transform (assuming zero IC) we obtain
\[ as^{4}X\left ( s\right ) +bs^{3}X\left ( s\right ) +cs^{2}X\left ( s\right ) +dsX\left ( s\right ) +eX\left ( s\right ) =F\left ( s\right ) \]
Hence
\[ X\left ( s\right ) \left [ as^{4}+bs^{3}+cs^{2}+ds+e\right ] =F\left ( s\right ) \]
Hence
\[ \fbox {$H\left ( s\right ) =\frac {X\left ( s\right ) }{F\left ( s\right ) }=\frac {1}{as^{4}+bs^{3}+cs^{2}+ds+e}$}\]
3.3.12 Problem
Calculate the frequency response function for the system of problem 3.49 for \(a=1,b=4,c=11,d=16,e=8\)
Answer
\begin{align*} H\left ( s\right ) & =\frac {1}{as^{4}+bs^{3}+cs^{2}+ds+e}\\ & =\frac {1}{s^{4}+4s^{3}+11s^{2}+16s+11}\end{align*}
Let \(s=j\omega \)
\begin{align*} H\left ( j\omega \right ) & =\frac {1}{\left ( j\omega \right ) ^{4}+4\left ( j\omega \right ) ^{3}+11\left ( j\omega \right ) ^{2}+16\left ( j\omega \right ) +11}\\ & =\frac {1}{\omega ^{4}-4j\omega ^{3}-11\omega ^{2}+16j\omega +11}\\ & =\frac {1}{\left ( \omega ^{4}-11\omega ^{2}+11\right ) +j\left ( 16\omega -4\omega ^{3}\right ) }\end{align*}
Hence
\begin{align*} \left \vert H\left ( j\omega \right ) \right \vert & =\frac {1}{\sqrt {\left ( \omega ^{4}-11\omega ^{2}+11\right ) ^{2}+\left ( 16\omega -4\omega ^{3}\right ) ^{2}}}\\ & =\allowbreak \frac {1}{\sqrt {\omega ^{8}-6\omega ^{6}+15\omega ^{4}+14\omega ^{2}+121}}\end{align*}
and
\[ Phase(H\left ( j\omega \right ) )=-\tan ^{-1}\left ( \frac {16\omega -4\omega ^{3}}{\omega ^{4}-11\omega ^{2}+11}\right ) \]
This is a plot of the magnitude and phase
EDU >> num=1;
EDU >> den=[1 4 11 16 11]
EDU >> sys=tf(num,den)
Transfer function:
1
-------------------------------- s ^4 + 4 s^3 + 11 s^2 + 16 s + 11
EDU >> w = logspace(-1,1);
EDU >> freqs(num,den,w)
3.3.13 PDF