Examples of Two Degrees of Freedom Systems 1-5

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: 

$$\begin{align} & {{x}_{1}}\left( 0 \right)=1,{{{\dot{x}}}_{1}}\left( 0 \right)=0,{{x}_{2}}\left( 0 \right)=-1,{{{\dot{x}}}_{2}}\left( 0 \right)=0, \\ & X_{1}^{(1)}=\frac{1}{{{r}_{2}}-{{r}_{1}}}{{\left[ {{\left\{ {{r}_{2}}{{x}_{1}}(0)-{{x}_{2}}(0) \right\}}^{2}}+\frac{{{\left\{ -{{r}_{2}}{{{\dot{x}}}_{1}}(0)-{{{\dot{x}}}_{2}}(0) \right\}}^{2}}}{\omega _{1}^{2}} \right]}^{1/2}}=-0.36602 \\ & X_{1}^{(2)}=\frac{1}{{{r}_{2}}-{{r}_{1}}}{{\left[ {{\left\{ -{{r}_{1}}{{x}_{1}}(0)-{{x}_{2}}(0) \right\}}^{2}}+\frac{{{\left\{ {{r}_{1}}{{{\dot{x}}}_{1}}(0)-{{{\dot{x}}}_{2}}(0) \right\}}^{2}}}{\omega _{2}^{2}} \right]}^{1/2}}=-1.36603 \\ & {{\phi }_{1}}={{\tan }^{-1}}\left\{ \frac{-{{r}_{2}}{{{\dot{x}}}_{1}}(0)+{{{\dot{x}}}_{2}}(0)}{{{\omega }_{1}}\left[ {{r}_{2}}{{x}_{1}}\left( 0 \right)-{{x}_{2}}(0) \right]} \right\}=0 \\ & {{\phi }_{2}}={{\tan }^{-1}}\left\{ \frac{{{r}_{1}}{{{\dot{x}}}_{1}}(0)+{{{\dot{x}}}_{2}}(0)}{{{\omega }_{2}}\left[ {{r}_{1}}{{x}_{1}}\left( 0 \right)-{{x}_{2}}(0) \right]} \right\}=0 \\ \end{align}$$
$$\begin{align} & {{x}_{1}}(t)=-0.36602cos3.6603t-1.36603\cos 13.6603t \\ & {{x}_{2}}(t)=-0.5cos3.6603t+0.5\cos 13.6603t \\ \end{align}$$