Example 1 - Find the natural frequencies of the system shown in
the following figure with:
$${{m}_{1}}=m,{{m}_{2}}=2m,{{k}_{1}}=k,{{k}_{2}}=2k.$$
Determine the response of the system when k = 1000
N/m, m = 20 kg and when initial values
of displacements of the masses m_1 and m_2 are 1 and -1, respectively.
Solution: First step is to derive the differential
equation which describe the motion of the system.
$$\begin{align}
& {{m}_{1}}{{{\ddot{x}}}_{1}}+({{k}_{1}}+{{k}_{2}}){{x}_{1}}-{{k}_{2}}{{x}_{2}}=0, \\
& {{m}_{2}}{{{\ddot{x}}}_{2}}+{{k}_{2}}{{x}_{2}}-{{k}_{2}}{{x}_{1}}=0 \\
\end{align}$$
Solution of previous equation can be expressed in the
following form:
$$\begin{align}
& {{x}_{i}}(t)={{X}_{i}}\cos (\omega t+\phi ),\text{ i =1}\text{,2} \\
& {{x}_{1}}(t)={{X}_{1}}\cos (\omega t+\phi ) \\
& {{x}_{2}}(t)={{X}_{2}}\cos (\omega t+\phi ) \\
\end{align}$$
Inserting the solutions in previous differential
equation we get:
$$\begin{align}
& -{{m}_{1}}{{\omega }^{2}}{{X}_{1}}\cos (\omega t+\phi )+({{k}_{1}}+{{k}_{2}}){{X}_{1}}\cos (\omega t+\phi )-{{k}_{2}}{{X}_{2}}\cos (\omega t+\phi )={0}/{:}\;\cos (\omega t+\phi ), \\
& -{{m}_{2}}{{\omega }^{2}}{{X}_{2}}\cos (\omega t+\phi )+{{k}_{2}}{{X}_{2}}\cos (\omega t+\phi )-{{k}_{2}}{{X}_{1}}\cos (\omega t+\phi )={0}/{:}\;\cos (\omega t+\phi ) \\
& \left\{ -{{m}_{1}}{{\omega }^{2}}+({{k}_{1}}+{{k}_{2}}) \right\}{{X}_{1}}-{{k}_{2}}{{X}_{2}}=0, \\
& -{{k}_{2}}{{X}_{1}}+\left\{ -{{m}_{2}}{{\omega }^{2}}+{{k}_{2}} \right\}{{X}_{2}}=0 \\
& \det \left[ \begin{matrix}
-{{m}_{1}}{{\omega }^{2}}+({{k}_{1}}+{{k}_{2}}) & -{{k}_{2}} \\
-{{k}_{2}} & -{{m}_{2}}{{\omega }^{2}}+{{k}_{2}} \\
\end{matrix} \right]=0 \\
& {{m}_{1}}{{m}_{2}}{{\omega }^{4}}-{{k}_{2}}{{m}_{1}}{{\omega }^{2}}-({{k}_{1}}+{{k}_{2}}){{m}_{2}}{{\omega }^{2}}+{{k}_{2}}({{k}_{1}}+{{k}_{2}})-k_{2}^{2}=0 \\
& {{m}_{1}}{{m}_{2}}{{\omega }^{4}}-{{k}_{2}}{{m}_{1}}{{\omega }^{2}}-({{k}_{1}}+{{k}_{2}}){{m}_{2}}{{\omega }^{2}}+{{k}_{2}}({{k}_{1}}+{{k}_{2}})-k_{2}^{2}={0}/{:{{m}_{1}}{{m}_{2}}}\; \\
& {{\omega }^{4}}-\left( \frac{{{k}_{1}}+{{k}_{2}}}{{{m}_{1}}}+\frac{{{k}_{2}}}{{{m}_{2}}} \right){{\omega }^{2}}+\frac{{{k}_{1}}{{k}_{2}}}{{{m}_{1}}{{m}_{2}}}=0 \\
\end{align}$$
Now it’s time to solve previously derived quadratic
equation:
$$\begin{align}
& {{\omega }^{4}}-\left( \frac{{{k}_{1}}+{{k}_{2}}}{{{m}_{1}}}+\frac{{{k}_{2}}}{{{m}_{2}}} \right){{\omega }^{2}}+\frac{{{k}_{1}}{{k}_{2}}}{{{m}_{1}}{{m}_{2}}}=0 \\
& \omega _{1,2}^{2}=\frac{\left( \frac{{{k}_{1}}+{{k}_{2}}}{{{m}_{1}}}+\frac{{{k}_{2}}}{{{m}_{2}}} \right)\pm \sqrt{{{\left( \frac{{{k}_{1}}+{{k}_{2}}}{{{m}_{1}}}+\frac{{{k}_{2}}}{{{m}_{2}}} \right)}^{2}}-4\frac{{{k}_{1}}{{k}_{2}}}{{{m}_{1}}{{m}_{2}}}}}{2}= \\
& \omega _{1,2}^{2}=\left( \frac{{{k}_{1}}+{{k}_{2}}}{2{{m}_{1}}}+\frac{{{k}_{2}}}{2{{m}_{2}}} \right)\pm \frac{1}{2}\sqrt{{{\left( \frac{{{k}_{1}}+{{k}_{2}}}{{{m}_{1}}}+\frac{{{k}_{2}}}{{{m}_{2}}} \right)}^{2}}-4\frac{{{k}_{1}}{{k}_{2}}}{{{m}_{1}}{{m}_{2}}}} \\
& \omega _{1,2}^{2}=\left( \frac{{{k}_{1}}+{{k}_{2}}}{2{{m}_{1}}}+\frac{{{k}_{2}}}{2{{m}_{2}}} \right)\pm \sqrt{\frac{1}{4}{{\left( \frac{{{k}_{1}}+{{k}_{2}}}{{{m}_{1}}}+\frac{{{k}_{2}}}{{{m}_{2}}} \right)}^{2}}-\frac{{{k}_{1}}{{k}_{2}}}{{{m}_{1}}{{m}_{2}}}} \\
\end{align}$$
The values of X1 and X2 remain to be determined.
These values depend on the natural frequencies ω1 and ω2. We shall denote these
values of X1 and X2 corresponding to ω1 as X1(1) and X2(1) and those
corresponding to ω2 as X1(2) and X2(2). All we need to do now is to define the
ratios
$$\begin{align}
& {{r}_{1}}=\frac{X_{2}^{(1)}}{X_{1}^{(1)}}=\frac{-{{m}_{1}}\omega _{1}^{2}+{{k}_{1}}+{{k}_{2}}}{{{k}_{2}}}=\frac{{{k}_{2}}}{-{{m}_{2}}\omega _{1}^{2}+{{k}_{2}}} \\
& {{r}_{2}}=\frac{X_{2}^{(2)}}{X_{1}^{(2)}}=\frac{-{{m}_{1}}\omega _{2}^{2}+{{k}_{1}}+{{k}_{2}}}{{{k}_{2}}}=\frac{{{k}_{2}}}{-{{m}_{2}}\omega _{2}^{2}+{{k}_{2}}} \\
\end{align}$$
General solution of differential equation can be
written in the following form:
$$\begin{align}
& {{x}_{1}}(t)=X_{1}^{(1)}\cos ({{\omega }_{1}}t+{{\phi }_{1}})+X_{1}^{(2)}\cos ({{\omega }_{2}}t+{{\phi }_{2}}) \\
& {{x}_{2}}(t)={{r}_{1}}X_{2}^{(1)}\cos ({{\omega }_{1}}t+{{\phi }_{1}})+{{r}_{2}}X_{1}^{(2)}\cos ({{\omega }_{2}}t+{{\phi }_{2}}) \\
\end{align}$$
The next step is to find variables inside the general
solution according to these formulas:
$$\begin{align}
& X_{1}^{(1)}=\frac{1}{{{r}_{2}}-{{r}_{1}}}{{\left[ {{\left\{ {{r}_{2}}{{x}_{1}}(0)-{{x}_{2}}(0) \right\}}^{2}}+\frac{{{\left\{ -{{r}_{2}}{{{\dot{x}}}_{1}}(0)-{{{\dot{x}}}_{2}}(0) \right\}}^{2}}}{\omega _{1}^{2}} \right]}^{1/2}} \\
& X_{1}^{(2)}=\frac{1}{{{r}_{2}}-{{r}_{1}}}{{\left[ {{\left\{ -{{r}_{1}}{{x}_{1}}(0)-{{x}_{2}}(0) \right\}}^{2}}+\frac{{{\left\{ {{r}_{1}}{{{\dot{x}}}_{1}}(0)-{{{\dot{x}}}_{2}}(0) \right\}}^{2}}}{\omega _{2}^{2}} \right]}^{1/2}} \\
& {{\phi }_{1}}={{\tan }^{-1}}\left\{ \frac{-{{r}_{2}}{{{\dot{x}}}_{1}}(0)+{{{\dot{x}}}_{2}}(0)}{{{\omega }_{1}}\left[ {{r}_{2}}{{x}_{1}}\left( 0 \right)-{{x}_{2}}(0) \right]} \right\} \\
& {{\phi }_{2}}={{\tan }^{-1}}\left\{ \frac{{{r}_{1}}{{{\dot{x}}}_{1}}(0)+{{{\dot{x}}}_{2}}(0)}{{{\omega }_{2}}\left[ {{r}_{1}}{{x}_{1}}\left( 0 \right)-{{x}_{2}}(0) \right]} \right\} \\
\end{align}$$
For m1 = m, m2 = m,
k1 = k and k2 = 2k we get:
$$\begin{align}
& \omega _{1,2}^{2}=\left( \frac{{{k}_{1}}+{{k}_{2}}}{2{{m}_{1}}}+\frac{{{k}_{2}}}{2{{m}_{2}}} \right)\pm \sqrt{\frac{1}{4}{{\left( \frac{{{k}_{1}}+{{k}_{2}}}{{{m}_{1}}}+\frac{{{k}_{2}}}{{{m}_{2}}} \right)}^{2}}-\frac{{{k}_{1}}{{k}_{2}}}{{{m}_{1}}{{m}_{2}}}} \\
& \omega _{1,2}^{2}=\left( \frac{k+2k}{2m}+\frac{2k}{4m} \right)\pm \sqrt{\frac{1}{4}{{\left( \frac{k+2k}{m}+\frac{2k}{2m} \right)}^{2}}-\frac{2{{k}^{2}}}{2{{m}^{2}}}} \\
& \omega _{1,2}^{2}=\left( \frac{6k+2k}{4m} \right)\pm \sqrt{\frac{1}{4}{{\left( \frac{6k+2k}{2m} \right)}^{2}}-\frac{2{{k}^{2}}}{2{{m}^{2}}}} \\
& \omega _{1,2}^{2}=\left( \frac{2k}{m} \right)\pm \sqrt{\frac{1}{4}{{\left( \frac{4k}{m} \right)}^{2}}-\frac{2{{k}^{2}}}{2{{m}^{2}}}}=\left( \frac{2k}{m} \right)\pm \sqrt{\frac{4{{k}^{2}}}{{{m}^{2}}}-\frac{2{{k}^{2}}}{2{{m}^{2}}}}=\left( \frac{2k}{m} \right)\pm \sqrt{\frac{8{{k}^{2}}-2{{k}^{2}}}{2{{m}^{2}}}} \\
& \omega _{1,2}^{2}=\left( \frac{2k}{m} \right)\pm \sqrt{\frac{8{{k}^{2}}-2{{k}^{2}}}{2{{m}^{2}}}}=\left( \frac{2k}{m} \right)\pm \sqrt{\frac{3{{k}^{2}}}{{{m}^{2}}}} \\
& \omega _{1}^{2}=(2-\sqrt{3})\frac{k}{m};\omega _{2}^{2}=(2+\sqrt{3})\frac{k}{m} \\
\end{align}$$
For k = 1000 N/m m = 20kg:
$$\begin{align}
& \omega _{1}^{2}=(2-\sqrt{3})\frac{k}{m}=(2-\sqrt{3})\frac{1000}{20}=(2-\sqrt{3})50=13.397 \\
& {{\omega }_{1}}=3.6603\text{ rad/sec} \\
& \omega _{2}^{2}=(2+\sqrt{3})\frac{k}{m}=(2+\sqrt{3})50=186.602 \\
& {{\omega }_{2}}=13.6603\text{ rad/sec} \\
\end{align}$$
$$\begin{align}
& {{r}_{1}}=\frac{X_{2}^{(1)}}{X_{1}^{(1)}}=\frac{{{k}_{2}}}{-{{m}_{2}}\omega _{1}^{2}+{{k}_{2}}}=\frac{2k}{-2m\omega _{1}^{2}+2k}=\frac{k}{-m\omega _{1}^{2}+k}=1.36604 \\
& {{r}_{2}}=\frac{X_{2}^{(2)}}{X_{1}^{(2)}}=\frac{{{k}_{2}}}{-{{m}_{2}}\omega _{2}^{2}+{{k}_{2}}}=\frac{k}{-m\omega _{2}^{2}+k}=-0.36602 \\
\end{align}$$
With:
$$\begin{align}
& {{x}_{1}}\left( 0 \right)=1,{{{\dot{x}}}_{1}}\left( 0 \right)=0,{{x}_{2}}\left( 0 \right)=-1,{{{\dot{x}}}_{2}}\left( 0 \right)=0, \\
& X_{1}^{(1)}=\frac{1}{{{r}_{2}}-{{r}_{1}}}{{\left[ {{\left\{ {{r}_{2}}{{x}_{1}}(0)-{{x}_{2}}(0) \right\}}^{2}}+\frac{{{\left\{ -{{r}_{2}}{{{\dot{x}}}_{1}}(0)-{{{\dot{x}}}_{2}}(0) \right\}}^{2}}}{\omega _{1}^{2}} \right]}^{1/2}}=-0.36602 \\
& X_{1}^{(2)}=\frac{1}{{{r}_{2}}-{{r}_{1}}}{{\left[ {{\left\{ -{{r}_{1}}{{x}_{1}}(0)-{{x}_{2}}(0) \right\}}^{2}}+\frac{{{\left\{ {{r}_{1}}{{{\dot{x}}}_{1}}(0)-{{{\dot{x}}}_{2}}(0) \right\}}^{2}}}{\omega _{2}^{2}} \right]}^{1/2}}=-1.36603 \\
& {{\phi }_{1}}={{\tan }^{-1}}\left\{ \frac{-{{r}_{2}}{{{\dot{x}}}_{1}}(0)+{{{\dot{x}}}_{2}}(0)}{{{\omega }_{1}}\left[ {{r}_{2}}{{x}_{1}}\left( 0 \right)-{{x}_{2}}(0) \right]} \right\}=0 \\
& {{\phi }_{2}}={{\tan }^{-1}}\left\{ \frac{{{r}_{1}}{{{\dot{x}}}_{1}}(0)+{{{\dot{x}}}_{2}}(0)}{{{\omega }_{2}}\left[ {{r}_{1}}{{x}_{1}}\left( 0 \right)-{{x}_{2}}(0) \right]} \right\}=0 \\
\end{align}$$
$$\begin{align}
& {{x}_{1}}(t)=-0.36602cos3.6603t-1.36603\cos 13.6603t \\
& {{x}_{2}}(t)=-0.5cos3.6603t+0.5\cos 13.6603t \\
\end{align}$$
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