\(k_{3}\) and \(k_{2}\) are in parallel, hence the effective stiffness is \[ k_{23}=k_{2}+k_{3}\] \(k_{23}\) and \(k_{1}\) are now in series, hence the effective stiffness is \[ \frac {1}{k_{123}}=\frac {1}{k_{1}}+\frac {1}{k_{23}}=\frac {k_{23}+k_{1}}{k_{1}k_{23}}=\frac {k_{2}+k_{3}+k_{1}}{k_{1}(k_{2}+k_{3})}=\frac {k_{2}+k_{3}+k_{1}}{k_{1}k_{2}+k_{1}k_{3}}\] Therefore \[ k_{123}=\frac {k_{1}k_{2}+k_{1}k_{3}}{k_{2}+k_{3}+k_{1}}\] \(k_{123}\) and \(k_{4}\) are now in parallel, hence the effective stiffness is \begin {align*} k_{1234} & =k_{4}+k_{123}\\ & =k_{4}+\frac {k_{1}k_{2}+k_{1}k_{3}}{k_{2}+k_{3}+k_{1}} \end {align*}
Hence the final effective stiffness is\[ k_{eq}=\frac {k_{4}(k_{2}+k_{3}+k_{1})+k_{1}k_{2}+k_{1}k_{3}}{k_{2}+k_{3}+k_{1}}\]
We start by drawing a free body diagram and taking displacement of mass from the static equilibrium position. Let the displacement of the mass be \(x\) and positive pointing upwards.
Let \(\triangle _{1}\) be the downward deflection at right end of the bottom beam. Let \(\triangle _{2}\) be the downward deflection at right end of top beam. The free body diagram is
Applying equilibrium of vertical forces \(\sum F_{v}=0\) for mass \(m\) and noting that inertial forces opposes motion, results in the equation of motion \begin {equation} mx^{\prime \prime }+k\left (x-\triangle _{1}\right ) =F \label {eq:2.1} \end {equation} To find an expression for \(\triangle _{1}\) in terms of \(x,\) we apply equilibrium of vertical forces at the right end of the lower beam1 \begin {equation} k\left (x-\triangle _{1}\right ) =k_{b}\triangle _{1}+k\left (\triangle _{1}-\triangle _{2}\right ) \label {eq:2.2} \end {equation} Similarly, applying equilibrium of vertical forces at the right end of the top beam \begin {equation} k\left (\triangle _{1}-\triangle _{2}\right ) =k_{b}\triangle _{2} \label {eq:2.3} \end {equation} Solving for \(\triangle _{1},\triangle _{2}\) from Eqs ??,?? (2 equations, 2 unknowns) gives \[ \triangle _{1}=\frac {k\left (k+k_{b}\right ) }{k^{2}+3kk_{b}+k_{b}^{2}}x \] Substituting the above value into Eq ?? results in the equation of motion\[ mx^{\prime \prime }+kx\left (1-\frac {k\left (k+k_{b}\right ) }{k^{2}+3kk_{b}+k_{b}^{2}}\right ) =F \]
Assuming periodic motion, the period is \(T=6\) ms, or \(6\times 10^{-3}\) sec. Hence \(\omega =\frac {\pi }{3}\) rad/ms Representing this as a cosine signal with phase gives\[ x(t)=A\cos (\omega t+\theta ) \] Then\begin {align} x(t) & =\operatorname {Re}[A+\cos (\omega t+\theta )]\nonumber \\ & =\operatorname {Re}[Ae^{i\theta }e^{i\omega t}]\nonumber \\ & =\operatorname {Re}[\bar {A}e^{i\omega t}]\label {eq:3.1} \end {align}
Where now \(\hat {A}=Ae^{i\theta }\).Using phasor diagram
Hence from the diagram we see that for \(x(t_{0})\) to be zero when \(t_{0}=1\) ms, we need to have \[ \omega t_{0}+\theta =\frac {\pi }{2}\] But \(\omega =\frac {\pi }{3}\) rad/ms, hence \[ \theta =\frac {\pi }{2}-\frac {\pi }{3}=\frac {\pi }{6}\] To find \(A\) we see that the maximum absolute value of \(x(t)\) is 20 mm hence \(A=20\) mm or \(20\times 10^{-3}\) meter. The equation of \(x(t)\) when substituting all numerical values becomes
\begin {equation} x(t)=20\cos \left (\frac {\pi }{3}t+\frac {\pi }{6}\right ) \label {eq:3.2} \end {equation} Where units used are radians, milliseconds and mm. This is a plot of the above function
parms = f -> 1/(6 10^-3);
Plot[0.02 Cos[2 Pi f t + (Pi/6)] /. parms, {t,0,0.005},
AxesLabel -> {t,x[t]}, ImageSize -> 300]
At \(t=0\), from ?? \(x(0)=\operatorname {Re}[\hat {A}]=A\cos (\theta )=20\cos (\frac {\pi }{6})\) hence \[ x(0)=17.321\text { {mm}}\] From ?? \(x^{\prime }(t)=\operatorname {Re}[\omega \hat {A}e^{i\omega t}]\) hence \(x^{\prime }(0)=\operatorname {Re}[\omega \hat {A}]=\omega A\cos (\theta )=20\frac {\pi }{3}\cos (\frac {\pi }{6})\) giving \[ x^{\prime }(0)=18.138\text { m/sec}\]
This can be solved using calculus2 \begin {align*} x^{\prime }\relax (t) & =-2\pi fA\sin \left (2\pi ft+\theta \right ) \\ 0 & =-2\pi fA\sin \left (2\pi ft+\frac {\pi }{6}\right ) \\ & =-\left (\frac {2\pi }{6\times 10^{-3}}\right ) \left (20\times 10^{-3}\right ) \sin \left (\frac {2\pi }{6\times 10^{-3}}t+\frac {\pi }{6}\right ) \\ 0 & =\sin \left (\frac {2\pi }{6\times 10^{-3}}t+\frac {\pi }{6}\right ) \end {align*}
We solve for \(t\) and find \(t\)=2.5 ms. But this can be solved more easily by looking at the phasor diagram
The minimum \(x(t)\) (in negative sense and not in absolute value sense) occurs when \(\omega t_{min}+\theta =\pi \), hence \(t_{min}=\frac {\pi -\theta }{\omega }\), therefore \[ t_{min}=2.5 \]
This is solved in a similar way by treating the speed as the rotating vector in complex plan. Since \(x^{\prime }(t)=\operatorname {Re}\left (A\omega e^{i\left (\omega t+\theta +\frac {\pi }{2}\right ) }\right ) \) then in complex plan as follows
The difference is that the velocity vector has phase of \(\theta +\frac {\pi }{2}\) instead of \(\theta \) as was the case with the position vector, and the amplitude is \(A\omega \) instead of \(A\). Hence the first time the speed vector will have the maximum value is when \[ \theta +\frac {\pi }{2}+\omega t=2\pi \] Hence \begin {align*} t & =\frac {2\pi -\frac {\pi }{2}-\theta }{\omega }\\ & =\frac {2\pi -\frac {\pi }{2}-\frac {\pi }{6}}{\frac {\pi }{3}} \end {align*}
Hence \(t=4\) ms and the amplitude is given by \(A\omega =20\frac {\pi }{3}\) hence \(A\omega =20.944\) meter/sec
Now treating the acceleration as the rotating vector in complex plan\begin {align*} x(t) & =\operatorname {Re}\left (Ae^{i\left (\theta +\omega t\right ) }\right ) \\ x^{\prime }(t) & =\operatorname {Re}\left (iA\omega e^{i\left (\theta +\omega t\right ) }\right ) \\ x^{\prime \prime }(t) & =\operatorname {Re}\left (-A\omega ^{2}e^{i\left ( \theta +\omega t\right ) }\right ) \end {align*}
But \(-1=e^{i\pi }\) This adds a \(\pi \) to the phase resulting in\[ x^{\prime \prime }(t)=\operatorname {Re}\left (A\omega ^{2}e^{i\left ( \theta +\omega t+\pi \right ) }\right ) \] Representing \(x^{\prime \prime }(t)\) in complex plan gives
The first time the \(x''(t)\) vector will have the maximum value is when \[ \theta + \pi + \omega t = 2 \pi \] Hence \begin {align*} t & = \frac {2 \pi - \pi - \theta }{\omega } \\ & = \frac {\pi - \frac {\pi }{6} }{\frac {\pi }{6}} \end {align*}
Hence \(t=2.5\ \text {ms}\) and the amplitude is \begin {align*} A\omega ^{2} & = 20\ \text {mm} \left (\frac {\pi }{3}\ \text {rad/msec} \right ) ^{2}\\ & = 21.933 \times 10^{3}\ \text {meter/sec}^{2} \end {align*}
The function of the signal is converted to complex exponential. A \(\sin \) or \(\cos \) can be used to represent the signal as long as we are consistent. Assuming the signal is \(x(t)=A\cos (\omega t+\theta )\), plotting the general representation of the position vector in complex plan gives
The complex representation of the position vector is \[ x(t)=\operatorname {Re}\left [ Ae^{i(\omega t+\theta )}\right ] \] We are given that \(\omega =\frac {2\pi }{T}=\frac {2\pi }{16}\), and since \(x(t_{0})\) has first zero at \(t_{0}=5.5\) ms this means from looking at the above diagram that \[ \theta +\omega t_{0}=\frac {\pi }{2}\] Hence \(\theta =\frac {\pi }{2}-(\omega t_{0})=\frac {\pi }{2}-(\frac {\pi }{8}\frac {55}{10})\) which gives \[ \theta =\frac {-3\pi }{16}\text {{radians}}\] Hence the signal is \begin {align*} x(t) & =\operatorname {Re}\left [ Ae^{i(\omega t+\theta )}\right ] \\ & =\operatorname {Re}\left [ 1.2e^{i\left (\frac {\pi }{8}t-\frac {3\pi }{16}\right ) }\right ] \\ & =\operatorname {Re}\left [ 1.2e^{-i\frac {3\pi }{16}}e^{i\frac {\pi }{8}t}\right ] \\ & =\operatorname {Re}\left [ \hat {A}e^{i\frac {\pi }{8}t}\right ] \end {align*}
Where \(\hat {A}=1.2e^{-i\frac {3\pi }{16}}\) is the complex amplitude in polar coordinates. In rectangular coordinates it becomes \begin {align*} \hat {A} & =1.2e^{-i\frac {3\pi }{16}}\\ & =1.2\left (\cos \left (\frac {3\pi }{16}\right ) -i\sin \left (\frac {3\pi }{16}\right ) \right ) \\ & =1.2\left (0.831-i0.5556\right ) \\ & =\fbox { $0.9977-i0.6667$ } \end {align*}
Hence \begin {align*} x(t) &= \operatorname {Re}\left [ \left (0.998-i0.668\right ) \left (\cos \frac {\pi }{8}t+i\sin \frac {\pi }{8}t\right ) \right ] \\ &= \operatorname {Re}\Bigl [ \left (0.998\cos \frac {\pi }{8}t+0.668\sin \frac {\pi }{8}t\right )+ \\ & i\left (0.998\sin \frac {\pi }{8}t-0.668\cos \frac {\pi }{8}t\right ) \Bigr ] \end {align*}
Here is a plot of the signal for 20 ms
w = Pi/8;
f = 1.2 Cos[w t - 3 Pi/16];
Plot[f, {t, 0, 20}, AxesLabel -> {t, x[t]},
ImageSize -> 300,
GridLines -> Automatic,
GridLinesStyle->{{Dashed,Gray},{Dashed,Gray}},
PlotStyle -> Red]
From above it was found that\[ x(t)=\operatorname {Re}\left [ Ae^{i\left (\omega t+\theta \right ) }\right ] \] Hence \begin {align*} x^{\prime }(t) & =\operatorname {Re}\left [ i\omega Ae^{i(\omega t+\theta )}\right ] \\ & =\operatorname {Re}\left [ e^{i\frac {\pi }{2}}\omega Ae^{i\theta }e^{i\omega t}\right ] \\ & =\operatorname {Re}\left [ \omega Ae^{i(\frac {\pi }{2}+\theta )}e^{i\omega t}\right ] \\ & =\operatorname {Re}\left [ \hat {A}e^{i\omega t}\right ] \end {align*}
Where \(\hat {A}=\omega Ae^{i(\frac {\pi }{2}+\theta )}\) Replacing numerical values gives \(\hat {A}=\frac {\pi }{8}\left (1.2\right ) e^{i(\frac {\pi }{2}-\frac {3}{16}\pi )}=0.471e^{i0.983}\) and \begin {align*} x^{\prime }(t) & =\operatorname {Re}\left [ 0.471e^{i0.983}e^{i\omega t}\right ] \\ & =\operatorname {Re}\left [ 0.471e^{i0.983}e^{i\frac {\pi }{8}t}\right ] \\ & =\operatorname {Re}\left [ 0.471e^{i0.983}e^{i0.3923t}\right ] \end {align*}
In rectangular coordinates, the above becomes
To find the maximum rate of the signal\[ x^{\prime }\relax (t) =\operatorname {Re}\left [ \hat {A}e^{i\omega t}\right ] \]
Then the maximum \(x^{\prime }\relax (t) \) is \(\left \vert \hat {A}\right \vert \) which is\begin {align*} \left \vert \hat {A}\right \vert & =\left \vert 0.261+0.392i\right \vert \\ & =\sqrt {0.261^{2}+0.392^{2}}\\ & =0.471 \end {align*}
Hence maximum \(x^{\prime }\relax (t) \) is \(0.471\) v/ms or \(471\) volt/sec.
Maximum velocity in simple harmonic motion occurs when \(x\relax (t) =0\). This occurs at \(t=5.5\) ms and at \(8\) ms henceforth. Hence maximum speed occurs at \[ t=5.5+n\relax (8) \] for \(n=0,1,2,\cdots \) this results in\[ t=5.5,13,5,21.5,\cdots \text {ms}\] Here is a plot of \(x^{\prime }\relax (t) \) in units of volt/ms
f = 0.261 Cos[0.392 t] - 0.392 Sin[0.392 t];
Plot[f, {t, 0, 30},
AxesLabel -> {Row[{t, "(ms)"}], x'[t]},
ImageSize -> 300, GridLines -> Automatic,
GridLinesStyle -> {{Dashed, Gray}, {Dashed, Gray}},
PlotStyle -> Red]
This is a plot of the signal
f = 0.01 Sin[50 t] - 0.02 Cos[50 t - 0.3 Pi];
Plot[f, {t, 0, 0.2},
AxesLabel -> {Row[{t , " (sec)"}], x[t]},
ImageSize -> 300,
GridLines -> Automatic,
GridLinesStyle->{{Dashed,Gray},{Dashed,Gray}},
PlotStyle -> Red]
\begin {align*} q & =0.01\sin \left (50t\right ) -0.02\cos \left (50t-0.3\pi \right ) \\ & =\operatorname {Re}\left [ \frac {0.01}{i}e^{i50t}-0.02e^{i\left ( 50t-0.3\pi \right ) }\right ] \\ & =\operatorname {Re}\left [ 0.01e^{-i\frac {\pi }{2}}e^{i50t}-0.02e^{i50t}e^{-i0.3\pi }\right ] \\ & =\operatorname {Re}\left [ \left (0.01e^{-i\frac {\pi }{2}}-0.02e^{-i0.3\pi }\right ) e^{i50t}\right ] \\ & =\operatorname {Re}\left [ \hat {A}e^{i50t}\right ] \end {align*}
Hence the complex amplitude is \[ \hat {A}=0.01e^{-i\frac {\pi }{2}}-0.02e^{-i0.3\pi } \]
From above, we see that \[ \omega =50\text { rad/sec} \] Hence \(f=\frac {50}{2\pi }\)Hz, or the period \(T=\frac {2\pi }{50}=0.126\sec \), therefore the time period separating the zeros is \(\frac {0.126}{2}=0.063\sec \) or \(63\) ms
The complex phase \(\hat {A}\) can be found by adding the vector \(0.01e^{-i\frac {\pi }{2}}\) and \(-0.02e^{-i\frac {3\pi }{10}}\) by completing the parallelogram as shown in this diagram. \(\hat {A}=-0.02\cos 0.7\pi +i\left (-0.01+0.02\sin 0.7\pi \right ) \), hence the angle \(\alpha \) that \(\hat {A}\) makes with the horizontal is\begin {align*} \tan ^{-1}\left (\frac {-0.01+0.02\sin 0.7\pi }{-0.02\cos 0.7\pi }\right ) &= \arctan \left (0.526\right ) \\ &= 0.484\ \text {radian} \\ &= 27.73\ \text {degree} \end {align*}
and the amplitude is \[ \sqrt {\left (-0.01+0.02\sin 0.7\pi \right )^{2}+\left (0.02\cos 0.7\pi \right )^{2}}=0.0133 \text {V} \] To find the earliest time \(q\) will be zero, we need to find the time the complex position vector will take to rotate and reach the imaginary axis.
Hence we need to solve \begin {align*} \pi -\alpha +\omega t_{0} & =\frac {3}{2}\pi \\ t_{0} & =\frac {\frac {3}{2}\pi -\pi +0.48402}{50}\\ & =0.0411\text { s} \end {align*}
Therefore\[ t=41.1\text { ms}\]
The largest value of \(q\) is the absolute value of its complex amplitude. We found this above as \[ \left \vert \hat {A}\right \vert =0.0133\ \text {Volt}\] To find when this occur first time, the time the position vector will align with the real axis in the positive direction is found. Hence solving for \(t_{0}\) from \begin {align*} \pi -\alpha +\omega t_{0} &= 2\pi \\ t_{0} &= \frac {2\pi -\pi +0.484}{50} \end {align*}
Gives \(t = 72.5\ \text {ms}\). Another way would be to take derivative of \(qt)\) and set that to zero and solve for first \(t\) which satisfy the equation.
\begin {align*} x_{1} & =8\sin \left (10t-\frac {5}{6}\pi \right ) \\ x_{2} & =12\cos \left (10t+\phi \right ) \end {align*}
Let \(\omega =10\), hence\begin {align} x & =x_{1}+x_{2}\nonumber \\ & =\operatorname {Re}\left [ \frac {8}{i}e^{i\left (\omega t-\frac {5}{6}\pi \right ) }\right ] +\operatorname {Re}\left [ 12e^{i\left (\omega t+\phi \right ) }\right ] \nonumber \\ & =\operatorname {Re}\left [ \frac {8}{i}e^{i\left (\omega t-\frac {5}{6}\pi \right ) }+12e^{i\left (\omega t+\phi \right ) }\right ] \nonumber \\ & =\operatorname {Re}\left [ 8e^{-i\frac {\pi }{2}}e^{i\left (\omega t-\frac {5}{6}\pi \right ) }+12e^{i\left (\omega t+\phi \right ) }\right ] \nonumber \\ & =\operatorname {Re}\left [ 8e^{-i\frac {\pi }{2}}e^{i\omega t}e^{-i\frac {5}{6}\pi }+12e^{i\omega t}e^{i\phi }\right ] \nonumber \\ & =\operatorname {Re}\left [ \left (8e^{-i\left (\frac {4}{3}\pi \right ) }+12e^{i\phi }\right ) e^{i\omega t}\right ] \nonumber \\ & =\operatorname {Re}\left [ \hat {A}e^{i\omega t}\right ] \label {eq:6.1} \end {align}
Where \begin {align*} \hat {A} & =8e^{-i\left (\frac {4}{3}\pi \right ) }+12e^{i\phi }\\ & =\left (-4+6.928i\right ) +12\left (\cos \phi +i\sin \phi \right ) \\ & =\left (-4+12\cos \phi \right ) +i\left (6.928+\sin \phi \right ) \end {align*}
Hence Eq ?? becomes\[ x=\operatorname {Re}\left [ \left \{ \left (-4+12\cos \phi \right ) +i\left ( 6.928+\sin \phi \right ) \right \} e^{i\omega t}\right ] \] To convert to \(\sin \) we multiply and divide by \(i\) hence\begin {align} x & =\operatorname {Re}\left [ \left \{ \left (-4+12\cos \phi \right ) +i\left (6.\allowbreak 928+\sin \phi \right ) \right \} i\frac {e^{i\omega t}}{i}\right ] \nonumber \\ & =\operatorname {Re}\left [ \left \{ -\left (6.\allowbreak 928+\sin \phi \right ) +i\left (-4+12\cos \phi \right ) \right \} \frac {e^{i\omega t}}{i}\right ] \label {eq:6.2} \end {align}
The complex number \(-\left (6.\allowbreak 928+\sin \phi \right ) +i\left ( -4+12\cos \phi \right ) \) can be written in polar form as \(ke^{i\beta }\) where \(K=\sqrt {\left (6.928+\sin \phi \right ) ^{2}+\left (-4+12\cos \phi \right ) ^{2}}\) and \(\beta =\tan ^{-1}\left (\frac {-4+12\cos \phi }{-\left ( 6.928+\sin \phi \right ) }\right ) \), hence Eq ?? becomes\begin {align*} x & =\operatorname {Re}\left [ ke^{i\beta }\frac {e^{i\omega t}}{i}\right ] \\ & =\operatorname {Re}\left [ k\frac {e^{i\left (\omega t+\beta \right ) }}{i}\right ] \\ & =k\sin \left (\omega t+\beta \right ) \end {align*}
or in full form \begin {multline*} x= \sqrt { \left (6.928+ \sin \phi \right )^{2}+ \left (-4+12\cos \phi \right )^{2}} \\ \sin \left (\omega t+\tan ^{-1}\left (\frac {-4+12\cos \phi }{-\left (6.928+\sin \phi \right ) }\right ) \right ) \end {multline*} For pure sine function we need \(\frac {-4+12\cos \phi }{-\left (6.928+\sin \phi \right ) }=0\) or \(12\cos \phi =4\) or \(\cos \phi =\frac {1}{3}\), hence\begin {align*} \phi & =1.23096\text { radian}\\ & =70.529^{\circ } \end {align*}
The amplitude can also be found from the complex amplitude above when \(\phi =1.23096\) as follows\begin {align*} \left \vert 8e^{-i\left (\frac {4}{3}\pi \right ) }+12e^{i1.23096}\right \vert & =\left \vert -6.592\times 10^{-6}+18.242i\right \vert \\ & =\sqrt {\left (-6.592\times 10^{-6}\right ) ^{2}+\left (18.242\right ) ^{2}}\\ & =18.242 \end {align*}