Assuming zero initial conditions. The input to the system is made up of two inputs. We find the response to the first input, then add this response to the response due to the second input. The first input is\begin {align*} u_{1}\relax (t) & =F_{0}h\relax (t) -F_{0}h\left (t-T\right ) \\ & =F_{0}\left (h\relax (t) -h\left (t-T\right ) \right ) \end {align*}
Which is a rectangular pulse of width \(T\) starting at \(t=0.\) For example for \(T=10\) sec. and \(F_{0}=1\)
Assuming the response to unit step is \(g_{s}\relax (t) \) then the response to \(u_{1}\relax (t) \) is\[ g_{1}\relax (t) =F_{0}\left (\overbrace {g_{s}\relax (t) h\relax (t) }-\overbrace {g_{s}\left (t-T\right ) h\left (t-T\right ) }\right ) \]
From appending B, \(g_{s}\relax (t) =\frac {1}{m\omega _{n}^{2}}\left ( 1-\cos \left (\omega _{n}t\right ) \right ) \), hence the above becomes\begin {equation} g_{1}\relax (t) =F_{0}\left (\overbrace {\frac {1}{m\omega _{n}^{2}}\left (1-\cos \left (\omega _{n}t\right ) \right ) h\relax (t) }-\overbrace {\frac {1}{m\omega _{n}^{2}}\left (1-\cos \left (\omega _{n}\left ( t-T\right ) \right ) \right ) h\left (t-T\right ) }\right ) \label {eq:1.0} \end {equation}
Looking at the second input given by \(u_{2}\relax (t) =F_{0}e^{-\beta \left (t-T\right ) }h\left (t-T\right ) \)
From appendix B, the response to an exponential \(F_{0}e^{-\beta t}h\left ( t\right ) \) is \[ \frac {F_{0}}{m\left (\omega _{n}^{2}+\beta ^{2}\right ) }\left (e^{-\beta t}-\left (\cos \left (\omega _{n}t\right ) -\frac {\beta }{\omega _{n}}\sin \left (\omega _{n}t\right ) \right ) \right ) h\relax (t) \] Therefore the response to \(u_{2}\relax (t) \) is\begin {equation} g_{2}\relax (t) =\frac {F_{0}}{m\left (\omega _{n}^{2}+\beta ^{2}\right ) }\left (e^{-\beta \left (t-T\right ) }-\left (\cos \left (\omega _{n}\left ( t-T\right ) \right ) -\frac {\beta }{\omega _{n}}\sin \left (\omega _{n}\left ( t-T\right ) \right ) \right ) \right ) h\left (t-T\right ) \label {eq:1.1} \end {equation}
Adding Eqs ?? and ?? results in the final response\begin {align*} g\relax (t) & =g_{1}\relax (t) +g_{1}\relax (t) \\ & =F_{0}\left (\frac {1}{m\omega _{n}^{2}}\left (1-\cos \left (\omega _{n}t\right ) \right ) h\relax (t) -\frac {1}{m\omega _{n}^{2}}\left ( 1-\cos \left (\omega _{n}\left (t-T\right ) \right ) \right ) h\left ( t-T\right ) \right ) +\\ & \frac {F_{0}}{m\left (\omega _{n}^{2}+\beta ^{2}\right ) }\left ( e^{-\beta \left (t-T\right ) }-\left (\cos \left (\omega _{n}\left ( t-T\right ) \right ) -\frac {\beta }{\omega _{n}}\sin \left (\omega _{n}\left ( t-T\right ) \right ) \right ) \right ) h\left (t-T\right ) \end {align*}
For illustration, the following plot shows the response using some values. Using \(m=1\) kg\(,\omega _{n}=1\) rad/sec\(,T=10\) sec\(,\beta =1,F_{0}=1\) Volt.
The differential equation is\begin {equation} \frac {1}{3}ML^{2}\theta ^{\prime \prime }\relax (t) +c\theta ^{\prime }\relax (t) +k\theta \relax (t) =F_{0}\left (h\relax (t) -h\left (t-T\right ) \right ) \label {eq:2.1} \end {equation}
The initial conditions are not given, and assumed to be zero, therefore \(\theta \relax (0) =0^{\circ }\) and \(\theta ^{\prime }\relax (0) =0\) rad/sec. The system is underdamped, hence \[ \omega _{d}=\omega _{n}\sqrt {1-\zeta ^{2}}\]
Let \(T_{d}\), be the damped period of oscillation given by\[ T_{d}=\frac {2\pi }{\omega _{d}}=\frac {2\pi }{\omega _{n}\sqrt {1-\zeta ^{2}}}\]
To obtain an expression for \(\omega _{n}\), Eq ?? is changed to a standard form \(\theta ^{\prime \prime }\relax (t) +2\zeta \omega _{n}\theta ^{\prime }\relax (t) +\omega _{n}^{2}\theta \relax (t) =\frac {F_{0}\left (h\relax (t) -h\left (t-T\right ) \right ) }{\frac {1}{3}ML^{2}}\)\begin {equation} \theta ^{\prime \prime }\relax (t) +\overset {2\zeta \omega _{n}}{\overbrace {\frac {3c}{ML^{2}}}}\theta ^{\prime }\relax (t) +\overset {\omega _{n}^{2}}{\overbrace {\frac {3k}{ML^{2}}}}\theta \left ( t\right ) =\frac {F_{0}\left (h\relax (t) -h\left (t-T\right ) \right ) }{\frac {1}{3}ML^{2}} \label {eq:2.2} \end {equation}
Therefore \[ \omega _{n}^{2}=\frac {3k}{ML^{2}}\]
Using \(k=20\) N/rad, \(L=0.02\) meter, \(M=0.003\) kg \[ \omega _{n}^{2}=\frac {3\left (20\right ) }{\left (0.003\right ) \left ( 0.02\right ) ^{2}}=5.0\times 10^{7}\]
or \[ \omega _{n}=\sqrt {5.0\times 10^{7}}=\fbox {$7071$ rad/sec}\]
and \[ T_{d}=\frac {2\pi }{7071.1\sqrt {1-0.02^{2}}}=\fbox {$0.8888$ ms}\] Therefore \[ T=2.5T_{d}=2.5\times 0.88857=\fbox {$2.221$ ms}\] To find \(F_{0}\) it is assumed the head was initially at rest. Therefore\begin {align*} F_{0} & =k\theta _{0}\\ & =20\left (\frac {\pi }{6}\right ) \text { }=\fbox {$10.472$ N-meter} \end {align*}
Eq ?? becomes\begin {align*} \theta ^{\prime \prime }\relax (t) +2\zeta \omega _{n}\theta ^{\prime }\left ( t\right ) +\omega _{n}^{2}\theta \relax (t) & =\frac {F_{0}\left ( h\relax (t) -h\left (t-2.5T_{d}\right ) \right ) }{\frac {1}{3}ML^{2}}\\ \theta ^{\prime \prime }\relax (t) +2\left (0.02\right ) \left ( 7071\right ) \theta ^{\prime }\relax (t) +\left (5\times 10^{7}\right ) \theta \relax (t) & =\frac {3\times 20\left (\frac {\pi }{6}\right ) \left (h\relax (t) -h\left (t-2.5T_{d}\right ) \right ) }{\left ( 0.003\right ) \left (0.02\right ) ^{2}}\\ \theta ^{\prime \prime }\relax (t) +283\theta ^{\prime }\relax (t) +5\times 10^{7}\theta \relax (t) & =2.618\times 10^{7}\left (h\left ( t\right ) -h\left (t-0.0022219\right ) \right ) \end {align*}
This is solved numerically for \(0<t<10T\) with the initial conditions \(\theta \relax (0) =0^{\circ }\) and \(\theta ^{\prime }\relax (0) =0\) rad/sec. Here is a plot of the solution and the input on a second plot.
A computational software was used to numerically solve the above differential equation for the solution \(\theta \relax (t) \).
From appendix B the response to underdamped second order system to a unit step \(u\relax (t) \) is\[ q_{s}\relax (t) =\frac {1}{M\omega _{n}^{2}}\left (1-e^{-\zeta \omega _{n}t}\left (\cos \left (\omega _{d}t\right ) +\frac {\zeta \omega _{n}}{\omega _{d}}\sin \left (\omega _{d}t\right ) \right ) \right ) h\left ( t\right ) \]
Hence the response \(U\relax (t) \) due to \(\frac {F_{0}}{\left (\frac {1}{3}mL^{2}\right ) }\left (h\relax (t) -h\left (t-T\right ) \right ) \) is given by\[ U\relax (t) =\frac {F_{0}}{\left (\frac {1}{3}L^{2}\right ) }\left ( \overbrace {q_{s}\relax (t) h\relax (t) }-\overbrace {q_{s}\left ( t-T\right ) h\left (t-T\right ) }\right ) \]
Notice the factor \(\frac {F_{0}}{\frac {1}{3}L^{2}}\). This was used since appendix B solution on based on equation of motion \(\theta ^{\prime \prime }\relax (t) +2\zeta \omega _{n}\theta ^{\prime }\relax (t) +\omega _{n}^{2}\theta \relax (t) =\frac {1}{m}\) while in this case, the equation of motion is \(\theta ^{\prime \prime }\relax (t) +2\zeta \omega _{n}\theta ^{\prime }\relax (t) +\omega _{n}^{2}\theta \left ( t\right ) =\frac {F_{0}}{\frac {1}{3}mL^{2}}\), hence a factor of \(\frac {F_{0}}{\frac {1}{3}L^{2}}\) is needed to scale the solution. Therefore the analytical solution is\begin {align*} U\relax (t) & =\frac {3F_{0}/L^{2}}{M\omega _{n}^{2}}\left ( 1-e^{-\zeta \omega _{n}t}\left (\cos \left (\omega _{d}t\right ) +\frac {\zeta \omega _{n}}{\omega _{d}}\sin \left (\omega _{d}t\right ) \right ) \right ) h\relax (t) \\ & -\frac {3F_{0}/L^{2}}{M\omega _{n}^{2}}\left (1-e^{-\zeta \omega _{n}\left ( t-2T\right ) }\left (\cos \left (\omega _{d}\left (t-T\right ) \right ) +\frac {\zeta \omega _{n}}{\omega _{d}}\sin \left (\omega _{d}\left (t-T\right ) \right ) \right ) \right ) h\left (t-T\right ) \end {align*}
To compare this solution with the numerical solution found in part(a), the two solutions are plotted side-by-side for the case \(T=2.5T_{d}\)
We see that solutions are in good approximate. Here is a plot of the difference. The error is in the order of \(10^{-7}\)
The analytical solutions for \(T=2.5T_{d}\) and \(T=3.0T_{d}\) are
We see when the step load duration is \(T=2.5T_{d}\), the disk head will vibrate with larger amplitudes than when the step duration was \(T=3T_{d}\).
To understand the reason for this, analysis was done on the undamped version of the solution for part \(b\)
From appendix B the response to undamped second order system to a unit step \(u\relax (t) \) is\[ q_{s}\relax (t) =\frac {1}{M\omega _{n}^{2}}\left (1-\cos \left ( \omega _{n}t\right ) \right ) h\relax (t) \]
Therefore the solution for \(0<t<T\) is \(\frac {3F_{0}/L^{2}}{M\omega _{n}^{2}}\left (1-\cos \left (\omega _{n}t\right ) \right ) \). This means at \(t=T\) which is when the step load is removed, \(\theta \relax (T) =\frac {3F_{0}/L^{2}}{M\omega _{n}^{2}}\left (1-\cos \left (\omega _{n}T\right ) \right ) \) and \(\theta ^{\prime }\relax (T) =-\frac {3F_{0}/L^{2}}{M\omega _{n}^{2}}\left (\omega _{n}\sin \left (\omega _{n}T\right ) \right ) \).
For \(t>T\), the load is not present any more and we have free vibration response but with the above initial conditions obtained at the end of the \(T\). The solution to free vibration of an undamped system for \(\tilde {t}=t-T\geq 0\) is given by \begin {align} \theta \left (\tilde {t}\right ) & =\frac {\theta ^{\prime }\relax (T) }{\omega _{n}}\sin \omega _{n}\tilde {t}+\theta \relax (T) \cos \omega _{n}\tilde {t}\nonumber \\ & =\frac {-\frac {3F_{0}/L^{2}}{M\omega _{n}^{2}}\left (\omega _{n}\sin \left ( \omega _{n}T\right ) \right ) }{\omega _{n}}\sin \omega _{n}\tilde {t}+\frac {3F_{0}/L^{2}}{M\omega _{n}^{2}}\left (1-\cos \left (\omega _{n}T\right ) \right ) \cos \omega _{n}\tilde {t}\nonumber \\ & =-\frac {3F_{0}/L^{2}}{M\omega _{n}^{2}}\sin \left (\omega _{n}T\right ) \sin \omega _{n}\tilde {t}+\left (\frac {3F_{0}/L^{2}}{M\omega _{n}^{2}}-\frac {3F_{0}/L^{2}}{M\omega _{n}^{2}}\cos \left (\omega _{n}T\right ) \right ) \cos \omega _{n}\tilde {t}\nonumber \\ & =-\frac {3F_{0}/L^{2}}{M\omega _{n}^{2}}\sin \left (\omega _{n}T\right ) \sin \omega _{n}\tilde {t}-\frac {3F_{0}/L^{2}}{M\omega _{n}^{2}}\cos \left ( \omega _{n}T\right ) \cos \omega _{n}\tilde {t}+\frac {3F_{0}/L^{2}}{M\omega _{n}^{2}}\cos \omega _{n}\tilde {t}\nonumber \\ & =-\frac {3F_{0}/L^{2}}{M\omega _{n}^{2}}\left (\sin \left (\omega _{n}T\right ) \sin \omega _{n}\tilde {t}+\cos \left (\omega _{n}T\right ) \cos \omega _{n}\tilde {t}\right ) +\frac {3F_{0}/L^{2}}{M\omega _{n}^{2}}\cos \omega _{n}\tilde {t} \label {eq:2.3} \end {align}
We have obtained a solution for the time after the step load was removed. We now investigate the result observed. We see that when \(T\) is close to an integer multiple of the period of the system, where we call the period of the system \(\tilde {T}\) to differentiate it from \(T\), then \[ \sin \left (\omega _{n}n\tilde {T}\right ) =\sin \left (\frac {2\pi }{\tilde {T}}n\tilde {T}\right ) =\sin \left (n2\pi \right ) =0 \] Also \[ \cos \left (\omega _{n}n\tilde {T}\right ) =\cos \left (\frac {2\pi }{\tilde {T}}n\tilde {T}\right ) =\cos \left (n2\pi \right ) =1 \] Hence the response given by equation ?? becomes\begin {align} \theta \left (\tilde {t}\right ) & =-\frac {3F_{0}/L^{2}}{M\omega _{n}^{2}}\cos \omega _{n}\tilde {t}+\frac {3F_{0}/L^{2}}{M\omega _{n}^{2}}\cos \omega _{n}\tilde {t}\nonumber \\ & =0\label {eq:2.4} \end {align}
But if \(T\) occurs at multiple of halves of the period of the system (for example, \(T=0.5\tilde {T},1.5\tilde {T},2.5\tilde {T}\), etc...) then now \[ \sin \left (\omega _{n}\left (n\frac {\tilde {T}}{2}\right ) \right ) \rightarrow \sin \left (\frac {2\pi }{\tilde {T}}\left (n\frac {\tilde {T}}{2}\right ) \right ) \rightarrow \sin \left (n\pi \right ) \rightarrow 0 \] However \[ \cos \left (\omega _{n}n\frac {\tilde {T}}{2}\right ) \rightarrow \cos \left ( \frac {2\pi }{\tilde {T}}n\frac {\tilde {T}}{2}\right ) \rightarrow \cos \left ( n\pi \right ) \rightarrow -1 \] We notice that the sign is now negative. This means equation ?? becomes\begin {align} \theta \left (\tilde {t}\right ) & =-\frac {3F_{0}/L^{2}}{M\omega _{n}^{2}}\cos \omega _{n}\tilde {t}-\frac {3F_{0}/L^{2}}{M\omega _{n}^{2}}\cos \omega _{n}\tilde {t}\nonumber \\ & =-\frac {6F_{0}/L^{2}}{M\omega _{n}^{2}}\cos \omega _{n}\tilde {t} \label {eq:2.5} \end {align}
Comparing Eqs ?? and ?? we see that when \(T\) is an integer multiple of the period of the system, then the response after \(T\) is minimal (zero for the case on undamped)
Whilewhen \(T\) occurs at multiple of halves of the period of the system,the response is large beyond the time \(T\).
The above analysis was done for undamped system, but the same idea carries to the underdamped case. This explains why the response dies out quickly when \(T=3T_{d}\) while it was large when \(T=2.5T_{d}\)
First lets look at the free vibration response (zero input response, called \(u_{zi}\)). The damping ratio \(\zeta =\frac {c}{c_{r}}=\frac {c}{2\sqrt {km}}=\frac {660}{2\sqrt {11000\times 1000}}=9.9499\times 10^{-2}\approx \)\(0.1\) and \(\omega _{n}=\sqrt {\frac {k}{m}}=\sqrt {\frac {10000}{1000}}\) hence \(\omega _{n}=3.162\) rad/sec, and \(\omega _{d}=\omega _{n}\sqrt {1-\zeta ^{2}}=3.1623\sqrt {1-0.1^{2}}\). Hence \(\omega _{d}=3.146\) rad/sec. The damped period of the system is \(T_{d}=\frac {2\pi }{\omega _{d}}=\frac {2\pi }{3.146}=\)\(1.997\) seconds and the natural period is \(T_{n}=\frac {2\pi }{\omega _{n}}=\frac {2\pi }{3.162}=\)\(1.987\) seconds.
Hence the system is underdamped and the solution is\[ u_{zi}=\operatorname {Re}\left (\hat {A}e^{\left (i\omega _{d}-\zeta \omega _{n}\right ) t}\right ) \]
Where \(\hat {A}=a+ib\) is the complex amplitude. At \(t=0\) we have \[ a=u_{zi}\relax (0) =-0.1 \] and \(u_{zi}^{\prime }\relax (0) \equiv u_{0}^{\prime }=\operatorname {Re}\left (\left (i\omega _{d}-\zeta \omega _{n}\right ) \left (a+ib\right ) \right ) =-b\omega _{d}-a\zeta \omega _{n}\) therefore \(b=\frac {-u_{0}^{\prime }-a\zeta \omega _{n}}{\omega _{d}}\). Since car was dropped from rest, then we take \(u_{0}^{\prime }=0\) which leads to \(b=-\frac {\left (-0.1\right ) \left ( 0.1\right ) 3.162}{3.146}=\)\(0.1\)
Hence, since \(a=u_{0}^{\prime }\relax (0) \equiv u_{0}\) and \[ b=\frac {-u_{0}^{\prime }-a\zeta \omega _{n}}{\omega _{d}}=0.1 \] then \begin {align} u_{zi}\relax (t) & =\operatorname {Re}\left (\left (a-ib\right ) e^{\left (i\omega _{d}-\zeta \omega _{n}\right ) t}\right ) \nonumber \\ & =\operatorname {Re}\left (e^{-\zeta \omega _{n}t}\left (u_{0}-i\left ( \frac {u_{0}^{\prime }+a\zeta \omega _{n}}{\omega _{d}}\right ) \right ) e^{i\omega _{d}t}\right ) \nonumber \\ & =e^{-\zeta \omega _{n}t}\left (u_{0}\relax (0) \cos \omega _{d}t+\left (\frac {u_{0}^{\prime }+a\zeta \omega _{n}}{\omega _{d}}\right ) \sin \omega _{d}t\right ) \label {eq:3.1} \end {align}
For the numerical values gives, we now can plot this solution \[ u_{zi}\relax (t) =e^{-0.1\left (3.162\right ) t}\left (-0.1\cos 3.146t+0.1\sin 3.146t\right ) \]
The phase is given by \(\tan ^{-1}\left (\frac {b}{a}\right ) =\) \(\tan ^{-1}\left (\frac {0.1}{-0.1}\right ) =2.356\) \(rad=135^{0}\), In complex plane, \(u_{h}\relax (t) \) is
Now we add the zero initial conditions response, also called zero state response \(u_{zs}\) for an input which is an impulse using appendix B.\[ u_{zs}\relax (t) =e^{-\zeta \omega _{n}t}\left (\frac {F_{0}}{m\omega _{d}}\sin \omega _{d}t\right ) h\relax (t) \]
Hence \(u_{zs}\) for an impulse that occurs at time \(T\) is\begin {equation} u_{zs}\relax (t) =e^{-\zeta \omega _{n}\left (t-T\right ) }\left ( \frac {F_{0}}{m\omega _{d}}\sin \omega _{d}\left (t-T\right ) \right ) h\left ( t-T\right ) \label {eq:3.2} \end {equation}
Hence the solution is found by combining Eq. ?? and Eq ??\begin {align*} u\relax (t) & =u_{zi}+u_{zs}\\ & =e^{-\zeta \omega _{n}t}\left (u_{0}\relax (0) \cos \omega _{d}t+\left (\frac {u_{0}^{\prime }+a\zeta \omega _{n}}{\omega _{d}}\right ) \sin \omega _{d}t\right ) h\relax (t) +e^{-\zeta \omega _{n}\left ( t-T\right ) }\left (\frac {F_{0}}{m\omega _{d}}\sin \omega _{d}\left ( t-T\right ) \right ) h\left (t-T\right ) \end {align*}
We need now to solve for \(T\) and \(F_{0}\) in order to meet the requirements that \(u\relax (t) \) should become zero between for \(2<t<5\). To do this in the complex plane, we draw the zero state response as a vector \begin {align*} u_{zs} & =e^{-\zeta \omega _{n}\left (t-T\right ) }\left (\frac {F_{0}}{m\omega _{d}}\sin \omega _{d}\left (t-T\right ) \right ) h\left (t-T\right ) \\ & =\operatorname {Re}\left (\frac {F_{0}e^{-\zeta \omega _{n}\left (t-T\right ) }}{m\omega _{d}}\frac {1}{i}e^{i\omega _{d}\left (t-T\right ) }\right ) h\left ( t-T\right ) \\ & =\operatorname {Re}\left (\frac {F_{0}e^{-\zeta \omega _{n}\left (t-T\right ) }}{m\omega _{d}}e^{i\left (\omega _{d}\left (t-T\right ) -\frac {\pi }{2}\right ) }\right ) h\left (t-T\right ) \end {align*}
Hence \(u_{zs}\) vector has magnitude \(\frac {F_{0}e^{-\zeta \omega _{n}\left (t-T\right ) }}{m\omega _{d}}\) and phase \(\omega _{d}t-\omega _{d}T-\frac {\pi }{2}\) Now to solve the problem of finding \(T\) and \(F_{0}\): To make the response become \(zero\) we need the magnitude of \(u_{zs}\) to be equal but opposite in sign to the magnitude of \(u_{zi}\) so that the projection on the x-axis cancel out (the projection on the x-axis of the vector is the real part which is the solution). Therefore, for the projection of \(u_{zs}\) to be the same as the projection of \(u_{zi}\) but of different sign, the following diagram shows all the possible \(T\) values that allows this. We will pick the first \(T\) value which is larger than \(2\) seconds to use.
From the above diagram, we need \(\omega _{d}T+\frac {\pi }{2}=2\pi -\frac {\pi }{4}\), hence \(T=\frac {2\pi -\frac {\pi }{4}-\frac {\pi }{4}}{\omega _{d}}=\frac {\frac {3}{2}\pi }{\pi }=1.5\) seconds. Hence this value of \(T\) is not acceptable. We now look for the next possible \(T\).
From the above diagram we see it will be \(\omega _{d}T+\frac {\pi }{2}=2\pi +\frac {\pi }{4}\) hence \(T=\frac {2\pi +\frac {\pi }{4}-\frac {\pi }{2}}{\pi }=1.75\) seconds. Hence this is still too early to apply the impulse. We look at the next possible case. We see that now we must rotate the vector all the way it was in the first diagram above to get the projection on the x-axis canceling the projection of the free vibration vector. Hence now the relation to solve for is\[ \omega _{d}T+\frac {\pi }{2}=4\pi -\frac {\pi }{4}\]
Where in the above we added full \(2\pi \) to the first case we considered above. This gives \[ T=\frac {4\pi -\frac {\pi }{4}-\frac {\pi }{2}}{\pi }=3.25\text { sec}\] .We have found \(T\) which brings the system to halt after at least 2 seconds has elapsed. Now we find \(F_{0}\) This is done by equating the amplitudes of the vectors as follows\[ \frac {F_{0}e^{-\zeta \omega _{n}\left (t-T\right ) }}{m\omega _{d}}=e^{-\zeta \omega _{n}t}\left \vert \hat {A}\right \vert \]
Now for\(t=T=3.25\) second, plug-in numerical values\begin {align*} \frac {F_{0}}{1000\left (3.146\right ) } & =e^{-\left (0.1\right ) \left ( 3.162\right ) \left (3.25\right ) }\sqrt {0.1^{2}+0.1^{2}}\\ \frac {F_{0}}{3146.0} & =5.0607\times 10^{-2}\\ F_{0} & =159.21 \end {align*}
To verify, here is a plot of the response when the impulse hit with
Let \(A\) be the area of the cross section and \(\rho \) the mass density and \(L\) the length, then actual mass is \begin {align*} m_{actual} & =LA\rho \\ & =50\left (18\times 0.0254\right ) \left (4\times 0.3048\right ) \left ( 7800\right ) \\ & =217393\text { kg} \end {align*}
Hence we will use \[ m=\frac {217393}{3}=72464\text { kg}\]
The actual stiffness for a simply supported mean with loading at the center is \(\frac {48EI}{L^{3}}\) where \(I\) is the area moment of inertia. Hence\[ I=\frac {wh^{3}}{12}=\frac {\left (4\times 0.3048\right ) \left (18\times 0.0254\right ) ^{3}}{12}=0.00971\text { m}^{4}\] Therefore the stiffness of the beam is\begin {align*} k & =\frac {48EI}{L^{3}}\\ & =\frac {48\left (210\times 10^{9}\right ) \left (0.00971\right ) }{50^{3}}=783014\text { N/m} \end {align*}
The natural frequency is\begin {align*} \omega _{n} & =\sqrt {\frac {k}{m}}=\sqrt {\frac {783014}{72464}}=3.287\text {rad/sec}\\ f_{n} & =0.523\text {\ Hz} \end {align*}
Therefore, assuming the loading is given by \(F_{0}\cos \left (\bar {\omega }t\right ) \) where \(\bar {\omega }\) is the forcing frequency. The dynamic response at any time is given by\[ \left \vert \hat {X}\right \vert =\frac {F_{0}/k}{\sqrt {\left (1-r^{2}\right ) ^{2}+\left (2\zeta r\right ) ^{2}}}\]
Where \(r=\frac {\bar {\omega }}{\omega _{n}}.\) We start by drawing \(\left \vert \hat {X}\right \vert \) vs. \(\bar {\omega }\) for the load of \(1000\ \)N by changing \(\bar {\omega }\) from \(0\) to \(8\pi \), Hence for a single student the displacement vs. forcing frequency is
Hence we see that for one student, the maximum displacement is around \(6\) cm when the student is jumping at resonance frequency.
To answer the question of how many students are needed to cause \(\left \vert X\right \vert \) to be \(50\) cm then that will depend on what forcing frequency is used. Now we will find the minimum number of students needed.
The minimum number will be when they all jump at the resonance frequency which is found from solving for \(\bar {\omega }_{resonance}\) in\begin {align*} \frac {\bar {\omega }_{resonance}}{\omega _{n}} & =\sqrt {1-2\zeta ^{2}}\\ \bar {\omega }_{resonance} & =\omega _{n}\sqrt {1-2\zeta ^{2}}\\ & =3.287\sqrt {1-2\left (0.01\right ) ^{2}}\\ & =3.28667\text { rad/sec} \end {align*}
Therefore, at this forcing frequency, we now solve for \(F_{0}\) to determine the number of students\begin {align*} \left \vert \hat {X}\right \vert & =\frac {F_{0}/k}{\sqrt {\left (1-\left ( \frac {\bar {\omega }_{resonance}}{\omega _{n}}\right ) ^{2}\right ) ^{2}+\left ( 2\zeta \frac {\bar {\omega }_{resonance}}{\omega _{n}}\right ) ^{2}}}\\ F_{0} & =k\left \vert \hat {X}\right \vert \sqrt {\left (1-\left (\frac {\bar {\omega }_{resonance}}{\omega _{n}}\right ) ^{2}\right ) ^{2}+\left ( 2\zeta \frac {\bar {\omega }_{resonance}}{\omega _{n}}\right ) ^{2}}\\ & =\left (783014\right ) \left (0.5\right ) \sqrt {\left (1-\left ( \frac {3.28667}{3.287}\right ) ^{2}\right ) ^{2}+\left ( 2\left (0.01\right ) \frac {3.28667}{3.287}\right ) ^{2}}\\ & =7829.75\text { N} \end {align*}
Therefore we need at least \(8\) students all jumping at \(3.287\) rad/sec to cause a displacement of at least \(50\) cm.
To make the structure avoid resonance, we need to make sure the ratio \(\frac {\varpi }{\omega _{n}}\) stays away from one. This is the ratio of the forcing frequency to the natural frequency. One way is to make \(\omega _{n}\) much larger than any expected \(\varpi \) that can occur is typical use of this structure.
But to make \(\omega _{n}=\sqrt {\frac {k}{m}}\) large, means either making \(m\) small or making \(k\) large. It is hard to reduce the mass of the structure. Therefore, making the structure more stiff will be a better solution.
The bridge can be made more stiff in many ways, such as by adding additional truss structure to it (assuming this will add minimal weight). For this example, suppose we double the stiffness. Hence \(\omega _{n}=\sqrt {\frac {2k}{m}}=\sqrt {\frac {2\left (783014\right ) }{72464\text { }}}= 4.649\)rad/sec.
Therefore now \(\bar {\omega }_{resonance}=\omega _{n}\sqrt {1-2\zeta ^{2}}=4.649\sqrt {1-2\left (0.01\right ) ^{2}}= 4.65\) rad/sec. Now the same number of students \(\relax (8) \) as before, jumping at same frequency of \(3.28667\) will cause displacement of\begin {align*} \left \vert \hat {X}\right \vert & =\frac {8F_{0}/k}{\sqrt {\left (1-\left ( \frac {\bar {\omega }}{\omega _{n}}\right ) ^{2}\right ) ^{2}+\left (2\zeta \frac {\bar {\omega }}{\omega _{n}}\right ) ^{2}}}\\ & =\frac {8000/783014}{\sqrt {\left (1-\left (\frac {3.28667}{4.649}\right ) ^{2}\right ) ^{2}+\left (2\left (0.01\right ) \frac {3.28667}{4.649}\right ) ^{2}}}\\ & =0.02\text { meter} \end {align*}
Therefore by making the bridge twice as stiff, now the same \(8\) students at \(\bar {\omega }=3.287\) will cause only \(2\) cm displacement instead of \(50\) cm.
A radar display is to be tested by mounting it on spring-dashpot suspension and subjecting it to harmonic force \(Q=F\cos \left (\bar {\omega }t\right ) \). The mounted mass is 8 kg and \(\zeta =0.25\). A free vibration shows that damped natural frequency \(f_{d}=5\)hz\(.\)It is observed that when the force is applied at very low frequency the displacement amplitude is 2 mm. The test is to be performed at \(5.2\) Hz. What will be the steady state response?
We are given are the following \begin {align*} m & =8\text { kg}\\ \zeta & =0.25\\ \omega _{d} & =\omega _{n}\sqrt {1-\zeta ^{2}}=2\pi \relax (5) \text { rad/sec}\\ F_{0}/k & =0.002\text { meter}\\ \bar {\omega } & =2\pi \left (5.2\right ) \text { rad/sec} \end {align*}
Hence \(\omega _{n}=\frac {\omega _{d}}{\sqrt {1-\zeta ^{2}}}=\frac {2\pi \left ( 5\right ) }{\sqrt {1-0.25^{2}}}=32.446\) rad/sec. The steady state response is given by\[ u_{ss}=\operatorname {Re}\left (\hat {X}e^{i\bar {\omega }t}\right ) \]
where \(\hat {X}=\left \vert \hat {X}\right \vert e^{i\theta }\). Hence\begin {align*} \left \vert \hat {X}\right \vert & =\frac {F_{0}/k}{\sqrt {\left (1-\left ( \frac {\bar {\omega }}{\omega _{n}}\right ) ^{2}\right ) ^{2}+\left (2\zeta \frac {\bar {\omega }}{\omega _{n}}\right ) ^{2}}}\\ & =\frac {0.002}{\sqrt {\left (1-\left (\frac {2\pi \left (5.2\right ) }{32.446}\right ) ^{2}\right ) ^{2}+\left (2\left (0.25\right ) \frac {2\pi \left (5.2\right ) }{32.446}\right ) ^{2}}}\\ & =0.00397 \end {align*}
and\begin {align*} \theta & =\tan ^{-1}\left (\frac {2\zeta r}{1-r^{2}}\right ) \\ & =\tan ^{-1}\left (\frac {2\left (0.25\right ) }{0}\right ) \\ & \tan ^{-1}\left (\infty \right ) \end {align*}
Since \(0\leq \theta \leq \pi \) then the phase is\[ \theta =\frac {\pi }{2}\]
Hence\begin {align*} u & =\operatorname {Re}\left (\hat {X}e^{i\bar {\omega }t}\right ) \\ & =\operatorname {Re}\left (0.00397e^{i\frac {\pi }{2}}e^{i\bar {\omega }t}\right ) \\ & =0.00397\cos \left (\bar {\omega }t+\frac {\pi }{2}\right ) \\ & =-0.00397\sin \left (\bar {\omega }t\right ) \end {align*}
A one degree of freedom system whose mass is 10 kg and whose natural frequency is \(1\) khz is subjected to a harmonic excitation \(1.2\sin \bar {\omega }t\) kN. The steady state amplitude when \(\bar {\omega }=1\) khz is observed to be \(2.4\) mm. Determine the steady state response at \(\bar {\omega }=0.95\) khz and \(\bar {\omega }=1.05\) khz.
We are given\begin {align*} m & =10\text { kg}\\ \omega _{n} & =2\pi \left (1000\right ) \text { rad/sec}\\ F_{0} & =1200\text { N}\\ \left \vert X\right \vert & =2.4\times 10^{-3}\text { meter when }\bar {\omega }=\omega _{n} \end {align*}
Since \(\omega _{n}^{2}=\frac {k}{m}\), hence \(k=\omega _{n}^{2}m=\left ( 2\pi \left (1000\right ) \right ) ^{2}\left (10\right ) \), therefore\[ k=3.949\times 10^{8}\text { N/m}\]
Now when \(\bar {\omega }=\omega _{n}\) we have\begin {align*} \left \vert \hat {X}\right \vert & =\frac {F_{0}/k}{\sqrt {\left (1-\left ( \frac {\bar {\omega }}{\omega _{n}}\right ) ^{2}\right ) ^{2}+\left (2\zeta \frac {\bar {\omega }}{\omega _{n}}\right ) ^{2}}}\\ 2.4\times 10^{-3} & =\frac {1200/\left (3.949\times 10^{8}\right ) }{\sqrt {\left (2\zeta \right ) ^{2}}}\\ & =\frac {3.039\times 10^{-6}}{2\zeta } \end {align*}
Hence \begin {align*} \zeta & =\left (\frac {3.039\times 10^{-6}}{2\times 2.4\times 10^{-3}}\right ) \\ & =0.000633 \end {align*}
when \(\bar {\omega }=2\pi \left (950\right ) \) now \(r=\frac {\bar {\omega }}{\omega _{n}}<1\) hence dynamic magnification factor is positive. Therefore loading and displacement will be in phase with each others. (i.e. displacement is in same direction as force). Since the force is \(\sin \) then the response will be \(\sin \) with same frequency but different phase and amplitude. Hence let \[ u_{ss}=X\sin \left (\bar {\omega }t-\theta \right ) \]
Where\begin {align*} X & =\frac {F_{0}/k}{\sqrt {\left (1-\left (\frac {\bar {\omega }}{\omega _{n}}\right ) ^{2}\right ) ^{2}+\left (2\zeta \frac {\bar {\omega }}{\omega _{n}}\right ) ^{2}}}\\ & =\frac {1200/\left (3.949\times 10^{8}\right ) }{\sqrt {\left ( 1-\left (\frac {2\pi \left (950\right ) }{2\pi \left (1000\right ) }\right ) ^{2}\right ) ^{2}+\left (2\left (0.000633\right ) \frac {2\pi \left ( 950\right ) }{2\pi \left (1000\right ) }\right ) ^{2}}}\\ & =3.116\times 10^{-5}\text { meter} \end {align*}
and \begin {align*} \theta & =\tan ^{-1}\left (\frac {2\zeta r}{1-r^{2}}\right ) \\ & =\tan ^{-1}\left (\frac {2\left (0.000633\right ) \frac {2\pi \left ( 950\right ) }{2\pi \left (1000\right ) }}{1-\left (\frac {2\pi \left ( 950\right ) }{2\pi \left (1000\right ) }\right ) ^{2}}\right ) \\ & =\tan ^{-1}\left (1.234\times 10^{-2}\right ) \\ & =0.01235\text { radians}\\ & =0.71^{\circ } \end {align*}
Hence steady state response is\[ u_{ss}=3.116\times 10^{-5}\sin \left (\bar {\omega }t-0.71^{\circ }\right ) \]
Hence we see that the displacement is lagging the load by \(0.71^{\circ }\). On complex plane it looks as follows
When \(\bar {\omega }=2\pi \left (1050\right ) \) now \(r=\frac {\bar {\omega }}{\omega _{n}}>1\) hence dynamic magnification factor is negative. Therefore loading and displacement will be out of phase with loading. (i.e .displacement is in opposite direction to force). Doing the same calculations are done as above\[ u_{ss}=X\sin \left (\bar {\omega }t-\theta \right ) \]
where \(X\)\begin {align*} X & =\frac {F_{0}/k}{\sqrt {\left (1-\left (\frac {\bar {\omega }}{\omega _{n}}\right ) ^{2}\right ) ^{2}+\left (2\zeta \frac {\bar {\omega }}{\omega _{n}}\right ) ^{2}}}\\ & =\frac {1200/\left (3.949\times 10^{8}\right ) }{\sqrt {\left ( 1-\left (\frac {2\pi \left (1050\right ) }{2\pi \left (1000\right ) }\right ) ^{2}\right ) ^{2}+\left (2\left (0.000633\right ) \frac {2\pi \left ( 1050\right ) }{2\pi \left (1000\right ) }\right ) ^{2}}}\\ & =2.964\times 10^{-5}\text { meter} \end {align*}
and \begin {align*} \theta & =\tan ^{-1}\left (\frac {2\zeta r}{1-r^{2}}\right ) \\ & =\tan ^{-1}\left (\frac {2\left (0.000633\right ) \frac {2\pi \left ( 1050\right ) }{2\pi \left (1000\right ) }}{1-\left (\frac {2\pi \left ( 1050\right ) }{2\pi \left (1000\right ) }\right ) ^{2}}\right ) \\ & =\tan ^{-1}\left (\frac {0.0013293}{-0.1025}\right ) \\ & =3.12862\text { radians}\\ & =179.257^{\circ } \end {align*}
Hence steady state response is\[ u_{ss}=2.964\times 10^{-5}\sin \left (\bar {\omega }t-179.257^{\circ }\right ) \]
On complex plane it looks as follows
Here is a plot by hand for the above 2 cases. First, the period that the loading is using \(T=\frac {2\pi }{\varpi }=\frac {1}{950}=1.0526\times 10^{-3}\)sec \[ T=1.053\ \text {ms}\]