Question

(a) Suppose two identical pendulums are coupled by means of a spring with constant See Figure 7.R.12. Under the same assumptions made in the discussion preceding Example 3 in Section 7.6, it can be shown that when the displacement angles ${\mathbf{\theta }}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ and${\mathbf{\theta }}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ are small, the system of linear differential equations describing the motion is

$$\begin{gathered} {\mathit{\theta }}_{\mathbf{1}}^{\mathbf{*}}\mathbf{+}\frac{\mathbf{g}}{\mathbf{l}}{\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{-}\frac{\mathbf{k}}{\mathbf{m}}\left({\theta }_1-{\theta }_2\right) \\ {\mathit{\theta }}_{\mathbf{2}}^{\mathbf{*}}\mathbf{+}\frac{\mathbf{g}}{\mathbf{l}}{\mathit{\theta }}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{k}}{\mathbf{m}}\left({\theta }_1-{\theta }_2\right) \end{gathered}$$

Use the Laplace transform to solve the system when ${\mathbf{\theta }}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{\theta }}_{\mathbf{0}}\mathbf{,}{\mathbf{\theta }}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}{\mathbf{\theta }}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{\psi }}_{\mathbf{0}}\mathbf{,}{\mathbf{\theta }}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}$ where${\mathit{\theta }}_{\mathbf{0}}$ and ${\mathit{\psi }}_{\mathbf{0}}$ are constants. For convenience let ${\mathit{\omega }}^{\mathbf{2}}\mathbf{=}\mathit{g}\mathbf{/}\mathit{l}\mathbf{,}\mathit{K}\mathbf{=}\mathit{k}\mathbf{/}\mathit{m}$.

(b) Use the solution in part (a) to discuss the motion of the coupled pendulums in the special case when the initial conditions are ${\mathbf{\theta }}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{\theta }}_{\mathbf{0}}\mathbf{,}{\mathbf{\theta }}_{\mathbf{1}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}{\mathbf{\theta }}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{\theta }}_{\mathbf{0}}\mathbf{,}{\mathbf{\theta }}_{\mathbf{2}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}$When the initial conditions are ${\mathbf{\theta }}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{\theta }}_{\mathbf{0}}\mathbf{,}{\mathbf{\theta }}_{\mathbf{1}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}{\mathbf{\theta }}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}{\mathbf{\theta }}_{\mathbf{0}}\mathbf{,}{\mathbf{\theta }}_{\mathbf{2}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$.

Short answer

$$\begin{gathered} \left(a\right){\theta }_1\left(t\right)=\frac{{\theta }_0+{\psi }_0}{2}\cos \omega t+\frac{1}{2}\left({\theta }_0-{\psi }_0\right)\cos \sqrt{{\omega }^2+2K}t \\ {\theta }_2\left(t\right)=\frac{{\theta }_0+{\psi }_0}{2}\cos \omega t-\frac{{\theta }_0-{\psi }_0}{2}\cos \sqrt{{\omega }^2+2K}t \end{gathered}$$

(b)

$$\begin{gathered} {\theta }_1\left(t\right)={\theta }_0\cos \omega t \\ {\theta }_2\left(t\right)={\theta }_0\cos \omega t \\ {\theta }_1\left(t\right)={\theta }_0\cos \sqrt{{\omega }^2+2K}t \\ {\theta }_2\left(t\right)=-{\theta }_0\cos \sqrt{{\omega }^2+2K}t \end{gathered}$$

Step-by-step solution

To find the both side of differential equation

(a) We know that,

$\mathcal{L}\left\{{\theta }_1^{\text{'}\text{'}}\right\}=s^2\mathcal{L}\left\{{\theta }_1\right\}-s{\theta }_1\left(0\right)-{\theta }_1^{\text{'}}\left(0\right)----\left(1\right)$

$\mathcal{L}\left\{{\theta }_2^{\text{'}\text{'}}\right\}=s^2\mathcal{L}\left\{{\theta }_2\right\}-s{\theta }_2\left(0\right)-{\theta }_2^{\text{'}}\left(0\right)----\left(2\right)$

Take the Laplace transform on both sides of each given differential equation.

$-K\mathcal{L}\left\{{\theta }_1-{\theta }_2\right\}=\mathcal{L}\left\{{\theta }_1^{\text{'}\text{'}}\right\}+{\omega }^2\mathcal{L}\left\{{\theta }_1\right\}$

$K\mathcal{L}\left\{{\theta }_1-{\theta }_2\right\}=\mathcal{L}\left\{{\theta }_2^{\text{'}\text{'}}\right\}+{\omega }^2\mathcal{L}\left\{{\theta }_2\right\}$

$-K\mathcal{L}\left\{{\theta }_1-{\theta }_2\right\}=s^2\mathcal{L}\left\{{\theta }_1\right\}-s{\theta }_1\left(0\right)-{\theta }_1^{\text{'}}\left(0\right)+{\omega }^2\mathcal{L}\left\{{\theta }_1\right\}$

$K\mathcal{L}\left\{{\theta }_1-{\theta }_2\right\}=s^2\mathcal{L}\left\{{\theta }_2\right\}-s{\theta }_2\left(0\right)-{\theta }_2^{\text{'}}\left(0\right)+{\omega }^2\mathcal{L}\left\{{\theta }_2\right\}$

$-K\mathcal{L}\left\{{\theta }_1-{\theta }_2\right\}=s^2\mathcal{L}\left\{{\theta }_1\right\}-s{\theta }_0+{\omega }^2\mathcal{L}\left\{{\theta }_1\right\}$ $\left({\theta }_1\left(0\right)={\theta }_0,{\theta }_1^{\text{'}}\left(0\right)=0\right)$

$K\mathcal{L}\left\{{\theta }_1-{\theta }_2\right\}=s^2\mathcal{L}\left\{{\theta }_2\right\}-s{\psi }_0+{\omega }^2\mathcal{L}\left\{{\theta }_2\right\}$ $\left({\theta }_2\left(0\right)={\psi }_0,{\theta }_2^{\text{'}}\left(0\right)=0\right)$

Denote $\mathcal{L}\left\{{\theta }_1\right\}={\Theta }_1\left(s\right)$and $\mathcal{L}\left\{{\theta }_2\right\}={\Theta }_2\left(s\right)$in the above system.

$$\begin{gathered} -K{\Theta }_1\left(s\right)+K{\Theta }_2\left(s\right)=s^2{\Theta }_1\left(s\right)-s{\theta }_0+{\omega }^2{\Theta }_1\left(s\right) \\ K{\Theta }_1\left(s\right)-K{\Theta }_2\left(s\right)=s^2{\Theta }_2\left(s\right)-s{\psi }_0+{\omega }^2{\Theta }_2\left(s\right) \end{gathered}$$

System simplifies to the form.

$$\begin{gathered} -K+s^2+{\omega }^2{\Theta }_1\left(s\right)+K{\Theta }_2\left(s\right)=-s{\theta }_0----\left(3\right) \\ K{\Theta }_1\left(s\right)-K+s^2+{\omega }^2{\Theta }_2\left(s\right)=-s{\psi }_0----\left(4\right) \end{gathered}$$

Multiply equation (3) by $K+s^2+{\omega }^2$and equation (4) by K

$$\begin{gathered} -K+s^2+{{\omega }^2}^2{\Theta }_1\left(s\right)+KK+s^2+{\omega }^2{\Theta }_2\left(s\right)=-s{\theta }_{0}K+s^2+\omega ----\left(5\right) \\ {K}^2{\Theta }_1\left(s\right)-KK+s^2+{\omega }^2{\Theta }_2\left(s\right)=-s{\psi }_{0}K----\left(6\right) \end{gathered}$$

Now add (5) and (6) equations to get

$$\begin{gathered} K^2{\Theta }_1\left(s\right)-K+s^2+{{\omega }^2}^2{\Theta }_1\left(s\right)=-s{\theta }_{0}K+s^2+{\omega }^2-s{\psi }_{0}K \\ K+s^2+{\omega }^2-K^2{\Theta }_1\left(s\right)=s{\theta }_{0}K+s^2+{\omega }^2+s{\psi }_{0}K \\ {s}^2+{\omega }^{2}2K+s^2+{\omega }^2{\Theta }_1\left(s\right)=s{\theta }_{0}K+s^2+{\omega }^2+s{\psi }_{0}K \\ {\Theta }_1\left(s\right)=\frac{sK{\theta }_0+s^3{\theta }_0+sK{\psi }_0+s{\omega }^2{\theta }_0}{\left(s^2+{\omega }^2\right)\left(s^2+{\omega }^2+2K\right)} \end{gathered}$$

Decompose the fraction $\frac{sK{\theta }_0+s^3{\theta }_0+sK{\psi }_0+s{\omega }^2{\theta }_0}{\left(s^2+{\omega }^2\right)\left(s^2+{\omega }^2+2K\right)}$using the partial fraction technique.

$$\begin{gathered} \star \frac{sK{\theta }_0+s^3{\theta }_0+sK{\psi }_0+s{\omega }^2{\theta }_0}{\left(s^2+{\omega }^2\right)\left(s^2+{\omega }^2+2K\right)}=\frac{As+B}{s^2+{\omega }^2}+\frac{Cs+D}{s^2+{\omega }^2+2K}----\left(7\right) \\ =\frac{(As+B)s^2+{\omega }^2+2K+(Cs+D)s^2+{\omega }^2}{\left(s^2+{\omega }^2\right)\left(s^2+{\omega }^2+2K\right)} \end{gathered}$$

Since the fractions in above equation have the same denominators, it takes their numerators to be equal.

$$\begin{gathered} sK{\theta }_0+s^3{\theta }_0+sK{\psi }_0+s{\omega }^2{\theta }_0=(As+B)s^2+{\omega }^2+2K+(Cs+D)s^2+{\omega }^2 \\ {s}^3{\theta }_0+K{\theta }_0+K{\psi }_0+{\omega }^2{\theta }_{0}s=(A+C)s^3+\left(A{\omega }^2+2KA+C{\omega }^2\right)s+(B+D)s^2+B{\omega }^2+2KB+D{\omega }^2 \end{gathered}$$

Both side comparing the coefficient

Comparing coefficient on both sides, we get

$$\begin{gathered} A+C={\theta }_0\hspace{0.17em} \\ A{\omega }^2+2KA+C{\omega }^2=K{\theta }_0+K{\psi }_0+{\omega }^2{\theta }_0 \end{gathered}$$

$$\begin{gathered} B+D=0 \\ \underset{¯}{B{\omega }^2+2KB+D{\omega }^2=0} \end{gathered}$$

Based on the previous two systems, we get

$A=\frac{1}{2}\left({\psi }_0+{\theta }_0\right),B=0,C=-\frac{1}{2}\left({\psi }_0-{\theta }_0\right)\text{and}D=0$

Substituting A, B, C and D in the equation (7), we get.

$\frac{sK{\theta }_0+s^3{\theta }_0+sK{\psi }_0+s{\omega }^2{\theta }_0}{\left(s^2+{\omega }^2\right)\left(s^2+{\omega }^2+2K\right)}=\frac{1}{2}\left({\psi }_0+{\theta }_0\right)\frac{s}{s^2+{\omega }^2}-\frac{1}{2}\left({\psi }_0-{\theta }_0\right)\frac{s}{s^2+{\omega }^2+2K}$

The equation $(*)$ becomes.

localid="1668612043479" ${\Theta }_1\left(s\right)=\frac{1}{2}\left({\psi }_0+{\theta }_0\right)\frac{s}{s^2+{\omega }^2}-\frac{1}{2}\left({\psi }_0-{\theta }_0\right)\frac{s}{s^2+{\omega }^2+2K}$

Take inverse Laplace transform of both side.

localid="1668612061404" $\star {\mathcal{L}}^{-1}\left\{{\Theta }_1\left(s\right)\right\}=\frac{1}{2}\left({\psi }_0+{\theta }_0\right){\mathcal{L}}^{-1}\frac{s}{s^2+{\omega }^2}-\frac{1}{2}\left({\psi }_0-{\theta }_0\right){\mathcal{L}}^{-1}\frac{s}{s^2+{\omega }^2+2K}$

localid="1668612073894" $\stackrel{(**)}{\to }\frac{{\theta }_0+{\psi }_0}{2}\cos \omega t+\frac{1}{2}\left({\theta }_0-{\psi }_0\right)\cos \sqrt{{\omega }^2+2K}t$

localid="1668612084841" $$\begin{gathered} (**) \\ {\mathcal{L}}^{-1}\frac{s}{s^2+k^2}=\cos kt \end{gathered}$$

localid="1668612094375" ${\theta }_1\left(t\right)={\mathcal{L}}^{-1}\left\{{\Theta }_1\left(s\right)\right\}=\frac{{\theta }_0+{\psi }_0}{2}\cos \omega t+\frac{1}{2}\left({\theta }_0-{\psi }_0\right)\cos \sqrt{{\omega }^2+2K}t$

Using the differential equation from the system, we have

localid="1668612108989" $\begin{array}{rcl}{\theta }_1^{\text{'}\text{'}}+{\omega }^2{\theta }_1& =& -K\left({\theta }_1-{\theta }_2\right)\\ {\theta }_1^{\text{'}\text{'}}+{\omega }^2{\theta }_1& =& K{\theta }_2-K{\theta }_1\\ K{\theta }_2& =& {\theta }_1^{\text{'}\text{'}}+{\omega }^2+K{\theta }_1----\left(8\right)\\ & & \end{array}$

First we find the second derivative of localid="1668612121388" ${\theta }_1.$

localid="1668612132830" ${\theta }_1^{\text{'}}\left(t\right)=-\frac{{\theta }_0+{\psi }_0}{2}\omega \sin \omega t-\frac{1}{2}\left({\theta }_0-{\psi }_0\right)\sqrt{{\omega }^2+2K}\sin \sqrt{{\omega }^2+2K}t$

localid="1668612141719" ${\theta }_1^{\text{'}\text{'}}\left(t\right)=-\frac{{\theta }_0+{\psi }_0}{2}{\omega }^2\cos \omega t-\frac{1}{2}\left({\theta }_0-{\psi }_0\right){\omega }^2+2K\cos \sqrt{{\omega }^2+2K}t$

Plugging into equation (8), we have

localid="1668612151413" ${\theta }_2=\frac{1}{K}-\frac{{\theta }_0+{\psi }_0}{2}{\omega }^2\cos \omega t-\frac{1}{2}\left({\theta }_0-{\psi }_0\right){\omega }^2+2K\cos \sqrt{{\omega }^2+2K}t$localid="1668612163926" $+\frac{{\omega }^2+K}{K}\frac{{\theta }_0+{\psi }_0}{2}\cos \omega t+\frac{1}{2}\left({\theta }_0-{\psi }_0\right)\cos \sqrt{{\omega }^2+2K}t$

Simplifying, we have

localid="1668612172964" ${\theta }_2=-\frac{{\theta }_0+{\psi }_0}{2K}{\omega }^2\cos \omega t-\frac{1}{2K}\left({\theta }_0-{\psi }_0\right){\omega }^2+2K\cos \sqrt{{\omega }^2+2K}t$

localid="1668612182285" $+\frac{1}{2K}\left({\theta }_0+{\psi }_0\right){\omega }^2+K\cos \omega t+\frac{1}{2K}\left({\theta }_0-{\psi }_0\right){\omega }^2+K\cos \sqrt{{\omega }^2+2K}t$

localid="1668612193247" ${\theta }_2\left(t\right)=\frac{{\theta }_0+{\psi }_0}{2}\cos \omega t-\frac{{\theta }_0-{\psi }_0}{2}\cos \sqrt{{\omega }^2+2K}t$

To find the initial condition

(b) With the initial conditions ${\theta }_1\left(0\right)={\theta }_0,{\theta }_1^{\text{'}}\left(0\right)=0,{\theta }_2\left(0\right)={\theta }_0,{\theta }_2^{\text{'}}\left(0\right)=0$ we have

${\theta }_1\left(t\right)=\frac{{\theta }_0+{\theta }_0}{2}\cos \omega t+\frac{1}{2}\left({\theta }_0-{\theta }_0\right)\cos \sqrt{{\omega }^2+2K}t$

${\theta }_2\left(t\right)=\left(\frac{{\theta }_0+{\theta }_0}{2}\right)\cos \omega t-\left(\frac{{\theta }_0-{\theta }_0}{2}\right)\cos \sqrt{{\omega }^2+2K}t$

Simplifying, we have

${\theta }_1\left(t\right)={\theta }_0\cos \omega t$

${\theta }_2\left(t\right)={\theta }_0\cos \omega t$

With the initial conditions $\${\theta }_1\left(0\right)={\theta }_0,{\theta }_1^{\text{'}}\left(0\right)=0,{\theta }_2\left(0\right)=-{\theta }_0,{\theta }_2^{\text{'}}\left(0\right)=0$we have

${\theta }_1\left(t\right)=\frac{{\theta }_0-{\theta }_0}{2}\cos \omega t+\frac{1}{2}\left({\theta }_0+{\theta }_0\right)\cos \sqrt{{\omega }^2+2K}t$

${\theta }_2\left(t\right)=\left(\frac{{\theta }_0-{\theta }_0}{2}\right)\cos \omega t-\left(\frac{{\theta }_0+{\theta }_0}{2}\right)\cos \sqrt{{\omega }^2+2K}t$

Simplifying, we have

$$\begin{gathered} {\theta }_1\left(t\right)={\theta }_0\cos \sqrt{{\omega }^2+2K}t \\ {\theta }_2\left(t\right)=-{\theta }_0\cos \sqrt{{\omega }^2+2K}t \end{gathered}$$

The equations are the same, but have opposite sign. That means they move away at the same time and then towards each other at the same time.

Final proof

$$\begin{gathered} \left(a\right){\theta }_1\left(t\right)=\frac{{\theta }_0+{\psi }_0}{2}\cos \omega t+\frac{1}{2}\left({\theta }_0-{\psi }_0\right)\cos \sqrt{{\omega }^2+2K}t \\ {\theta }_2\left(t\right)=\frac{{\theta }_0+{\psi }_0}{2}\cos \omega t-\frac{{\theta }_0-{\psi }_0}{2}\cos \sqrt{{\omega }^2+2K}t \end{gathered}$$

(b)

$\begin{array}{rcl}{\theta }_1\left(t\right)& =& {\theta }_0\cos \omega t\\ {\theta }_2\left(t\right)& =& {\theta }_0\cos \omega t\\ {\theta }_1\left(t\right)& =& {\theta }_0\cos \sqrt{{\omega }^2+2K}t\\ {\theta }_2\left(t\right)& =& -{\theta }_0\cos \sqrt{{\omega }^2+2K}t\end{array}$