Chapter 11, Problem 11.35P

(a)

To determine

The natural choice for the normal coordinates ${\xi }_1$ and ${\xi }_2$.

Answer to Problem 11.35P

The natural choice for the normal coordinates ${\xi }_1$ and ${\xi }_2$ are ${\xi }_1=\frac{{\varphi }_1+{\varphi }_2}{2}$ and ${\xi }_2=\frac{{\varphi }_1-{\varphi }_2}{2}$.

Explanation of Solution

The Lagrangian equation of motion of the two coupled pendulum of Problem 11.14 is

$\left(\begin{array}{cc}m& 0\\ 0& m\end{array}\right)\left(\begin{array}{c}{\ddot{\varphi }}_1\\ {\ddot{\varphi }}_2\end{array}\right)=\left(\begin{array}{cc}-\left(\frac{mg}{L}+k\right)& k\\ k& -\left(\frac{mg}{L}+k\right)\end{array}\right)\left(\begin{array}{c}{\varphi }_1\\ {\varphi }_2\end{array}\right)$

Write the general equation of motion

$M\ddot{\varphi }=-K\varphi$

Comparing the Lagrangian equation of motion and the general equation of motion

$M=\left(\begin{array}{cc}m& 0\\ 0& m\end{array}\right)$

$\ddot{\varphi }=\left(\begin{array}{c}{\ddot{\varphi }}_1\\ {\ddot{\varphi }}_2\end{array}\right)$

$K=\left(\begin{array}{cc}-\left(\frac{mg}{L}+k\right)& k\\ k& -\left(\frac{mg}{L}+k\right)\end{array}\right)$

$\varphi =\left(\begin{array}{c}{\varphi }_1\\ {\varphi }_2\end{array}\right)$

The $\varphi \left(t\right)$ can be written as the real part of complex solution of $z\left(t\right)$ when the time dependence if just $e^{i\omega t}$

$\varphi \left(t\right)=\mathrm{Re}z\left(t\right)$

Where, $z\left(t\right)a{e}^{i\omega t}=\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)e^{i\omega t}$

The function of this form satisfies the general equation of motion $M\ddot{\varphi }=-K\varphi$ only if the frequency $\omega$ and the column matrix $a$ satisfies the Eigen value equation, $\left(K-{\omega }^{2}M\right)a=0$.

Column matrix $a$ in the above Eigen value equation has a solution only if $\det \left(K-{\omega }^{2}M\right)=0$.

$\begin{array}{c}\left|\begin{array}{cc}\frac{mg}{L}+k-{\omega }^{2}m& -k\\ -k& \frac{mg}{L}+k-{\omega }^{2}m\end{array}\right|=0\\ \left(\frac{mg}{L}+k-{\omega }^{2}m\right)\left(\frac{mg}{L}+k-{\omega }^{2}m\right)-\left(-k\right)\left(-k\right)=0\\ {\left(\frac{mg}{L}+k-{\omega }^{2}m\right)}^2-\left(k^2\right)=0\\ \left(\frac{mg}{L}+k-{\omega }^{2}m\right)=\left(\pm k\right)\end{array}$

$\pm$ sign indicates two modes are exist in this motion of the system.

When $\pm k\to +k$. The first normal mode is

$\begin{array}{c}\left(\frac{mg}{L}+k-{\omega }^{2}m\right)=+k\\ \frac{mg}{L}-{\omega }^{2}m=0\\ {\omega }^2=\frac{L}{g}\\ {\omega }_1^2=\frac{L}{g}\end{array}$

When $\pm k\to -k$. The second normal mode is

$\begin{array}{c}\left(\frac{mg}{L}+k-{\omega }^{2}m\right)=-k\\ \frac{mg}{L}+2k-{\omega }^{2}m=0\\ {\omega }_2^2=\frac{g}{L}+\frac{2k}{m}\end{array}$

In the first normal mode, ${\omega }_1^2=\frac{g}{L}$. Then Eigen value equation becomes

$\begin{array}{c}\left(K-{\omega }^{2}M\right)a=0\\ \left(\begin{array}{cc}\frac{mg}{L}+k-\frac{g}{L}m& -k\\ -k& \frac{mg}{L}+k-\frac{g}{L}m\end{array}\right)\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)=0\\ \left(\begin{array}{cc}k& -k\\ -k& k\end{array}\right)\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)=0\\ \left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)=0\end{array}$

Writing the above matrices in equations

$a_1-a_2=0$  and $-a_1+a_2=0$

The solution to the above equation is $a_1=a_2$.

Thus, in the first normal mode, both the masses will oscillate with same amplitude and in phase with each other, ${\varphi }_1\left(t\right)={\varphi }_2\left(t\right)=A\cos \left({\omega }_{1}t-\delta \right)$

In the second normal mode, ${\omega }_2^2=\frac{g}{L}+\frac{2k}{m}$. Then Eigen value equation becomes

$\begin{array}{c}\left(K-{\omega }^{2}M\right)a=0\\ \left(\begin{array}{cc}\frac{mg}{L}+k-\left(\frac{g}{L}+\frac{2k}{m}\right)m& -k\\ -k& \frac{mg}{L}+k-\left(\frac{g}{L}+\frac{2k}{m}\right)m\end{array}\right)\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)=0\\ \left(\begin{array}{cc}-k& -k\\ -k& -k\end{array}\right)\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)=0\\ \left(\begin{array}{cc}-1& -1\\ -1& -1\end{array}\right)\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)=0\end{array}$

Writing the above matrices in equation

$\begin{array}{c}{a}_1+a_2=0\\ a_1=-a_2\end{array}$

The solution to the above equation is $a_1=a_2$.

Thus, in the second normal mode, both the masses will oscillate with same amplitude and are in out of phase with each other, ${\varphi }_1\left(t\right)=-{\varphi }_2\left(t\right)=A\cos \left({\omega }_{1}t-\delta \right)$.

The eigenvectors $a_1$ and $a_2$ found in problem 11.14 are

$a_1=\left(\begin{array}{c}1\\ 1\end{array}\right)$  and $a_2=\left(\begin{array}{c}1\\ -1\end{array}\right)$

The eigenvector in matrix format

$\tilde{A}=\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)$

Determinate of the above matrix is

$\begin{array}{l}\left|\tilde{A}\right|=\left|\begin{array}{cc}1& 1\\ 1& -1\end{array}\right|\\ =\left(1\right)\left(-1\right)-\left(1\right)\left(1\right)\\ =-2\end{array}$

Since the determinate of $A$ is not equal to zero. The inverse of the matrix exist.

Therefore, the two eigenvectors $a_{\left(1\right)}\text{ and }{a}_{\left(2\right)}$ in $\tilde{A}$

Therefore, the $2\times 1$ column vector in terms of the two eigenvectors $a_{\left(1\right)}\text{ and }{a}_{\left(2\right)}$ $\varphi$ can be expanded in terms of them. Therefore, for any right-hand side of $\varphi$ can be

 $\varphi ={\xi }_1{a}_{\left(1\right)}+{\xi }_2{a}_{\left(2\right)}$

Here, ${\xi }_1,{\xi }_2$ are the normal coordination.

Substitute the column matrix of $a_1\text{ and }{a}_2$  in the above equation and solving

$\begin{array}{c}\varphi =\left(\begin{array}{c}{\varphi }_1\\ {\varphi }_2\end{array}\right)={\xi }_1\left(\begin{array}{c}1\\ 1\end{array}\right)+{\xi }_2\left(\begin{array}{c}1\\ -1\end{array}\right)\\ \left(\begin{array}{c}{\varphi }_1\\ {\varphi }_2\end{array}\right)=\left(\begin{array}{c}{\xi }_1\\ {\xi }_1\end{array}\right)+\left(\begin{array}{c}{\xi }_2\\ -{\xi }_2\end{array}\right)\\ \left(\begin{array}{c}{\varphi }_1\\ {\varphi }_2\end{array}\right)=\left(\begin{array}{c}{\xi }_1+{\xi }_2\\ {\xi }_1-{\xi }_2\end{array}\right)\end{array}$

The equations from the above matrix

${\varphi }_1={\xi }_1+{\xi }_2$

And,

${\varphi }_2={\xi }_1-{\xi }_2$

Solving algebraically,

$\begin{array}{l}{\varphi }_1={\xi }_1+{\xi }_2\\ \underset{\_}{{\varphi }_2={\xi }_1-{\xi }_2}\\ {\varphi }_1+{\varphi }_2=2{\xi }_1\end{array}$

${\xi }_1=\frac{{\varphi }_1+{\varphi }_2}{2}$

Substitute $\frac{{\varphi }_1+{\varphi }_2}{2}$ for ${\xi }_1$ in ${\varphi }_1={\xi }_1+{\xi }_2$ and solving

$\begin{array}{c}{\varphi }_1=\frac{{\varphi }_1+{\varphi }_2}{2}+{\xi }_2\\ 2{\xi }_2={\varphi }_1-{\varphi }_2\\ {\xi }_2=\frac{{\varphi }_1-{\varphi }_2}{2}\end{array}$

Conclusion:

Therefore, the natural choice for the normal coordinates ${\xi }_1$ and ${\xi }_2$ are ${\xi }_1=\frac{{\varphi }_1+{\varphi }_2}{2}$ and ${\xi }_2=\frac{{\varphi }_1-{\varphi }_2}{2}$.

(b)

To determine

Whether the equation of motion for ${\xi }_1$ and ${\xi }_2$ will be still uncoupled when both pendulums are subject to a resistive force of magnitude $bv$.

Answer to Problem 11.35P

It is proved that the equation of motion for ${\xi }_1$ and ${\xi }_2$ will be still uncoupled when both pendulums are subject to a resistive force of magnitude $bv$.

Explanation of Solution

The Lagrangian equation of motion of the two coupled pendulum of Problem 11.14 when it is subjected to a resistive force of magnitude $bv$ is

$\left(\begin{array}{cc}m& 0\\ 0& m\end{array}\right)\left(\begin{array}{c}{\ddot{\varphi }}_1\\ {\ddot{\varphi }}_2\end{array}\right)=\left(\begin{array}{cc}-\left(\frac{mg}{L}+k\right)+\frac{b^2}{4m}& k\\ k& -\left(\frac{mg}{L}+k\right)+\frac{b^2}{4m}\end{array}\right)\left(\begin{array}{c}{\varphi }_1\\ {\varphi }_2\end{array}\right)$

Write the general equation of motion

$M\ddot{\varphi }=-K\varphi$

Comparing the Lagrangian equation of motion and the general equation of motion

$M=\left(\begin{array}{cc}m& 0\\ 0& m\end{array}\right)$

$\ddot{\varphi }=\left(\begin{array}{c}{\ddot{\varphi }}_1\\ {\ddot{\varphi }}_2\end{array}\right)$

$K=\left(\begin{array}{cc}-\left(\frac{mg}{L}+k\right)+\frac{b^2}{4m}& k\\ k& -\left(\frac{mg}{L}+k\right)+\frac{b^2}{4m}\end{array}\right)$

$\varphi =\left(\begin{array}{c}{\varphi }_1\\ {\varphi }_2\end{array}\right)$

The $\varphi \left(t\right)$ can be written as the real part of complex solution of $z\left(t\right)$ when the time dependence if just $e^{i\omega t-\beta t}$

$\varphi \left(t\right)=\mathrm{Re}z\left(t\right)$

Where, $z\left(t\right)a{e}^{i\omega t-\beta t}=\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)e^{i\omega t-\beta t}$

The function of this form satisfies the general equation of motion $M\ddot{\varphi }=-K\varphi$ only if the frequency $\omega$ and the column matrix $a$ satisfies the Eigen value equation, $\left(K-{\omega }^{2}M\right)a=0$.

Column matrix $a$ in the above Eigen value equation has a solution only if $\det \left(K-{\omega }^{2}M\right)=0$.

$\begin{array}{c}\left|\begin{array}{cc}\left(\frac{mg}{L}+k\right)-\frac{b^2}{4m}-{\omega }^{2}m& -k\\ -k& \left(\frac{mg}{L}+k\right)-\frac{b^2}{4m}-{\omega }^{2}m\end{array}\right|=0\\ {\left(\left(\frac{mg}{L}+k\right)-\frac{b^2}{4m}-{\omega }^{2}m\right)}^2-\left(-k\right)\left(-k\right)=0\\ \left(\left(\frac{mg}{L}+k\right)-\frac{b^2}{4m}-{\omega }^{2}m\right)=\pm k\end{array}$

$\pm$ sign indicates two modes are exist in this motion of the system.

When $\pm k\to +k$. The first normal mode is

$\begin{array}{c}\left(\left(\frac{mg}{L}+k\right)-\frac{b^2}{4m}-{\omega }^{2}m\right)=+k\\ \frac{mg}{L}-\frac{b^2}{4m}-{\omega }^{2}m=0\\ {\omega }^2=\frac{mg}{Lm}-\frac{b^2}{4m^2}\\ {\omega }_1^2=\frac{g}{L}-\frac{b^2}{4m^2}\end{array}$

Let $\beta =\frac{b}{2m}$. Then, the above equation becomes

${\omega }_1^2=\frac{g}{L}-{\beta }^2$

When $\pm k\to -k$. The second normal mode is

$\begin{array}{c}\left(\left(\frac{mg}{L}+k\right)-\frac{b^2}{4m}-{\omega }^{2}m\right)=-k\\ \frac{mg}{L}+2k-\frac{b^2}{4m}-{\omega }^{2}m=0\\ {\omega }^2=\frac{g}{L}+\frac{2k}{m}-\frac{b^2}{4m^2}\\ {\omega }_2^2=\frac{g}{L}+\frac{2k}{m}-{\left(\frac{b}{4m}\right)}^2\end{array}$

Let $\beta =\frac{b}{2m}$. Then, the above equation becomes

${\omega }_2^2=\frac{g}{L}+\frac{2k}{m}-{\beta }^2$

In the first normal mode, ${\omega }_1^2=\frac{g}{L}-{\beta }^2$. Then Eigen value equation becomes

$\begin{array}{c}\left(K-{\omega }^{2}M\right)a=0\\ \left(\begin{array}{cc}\frac{mg}{L}+k-\frac{b^2}{4m}\left(\frac{g}{L}-\frac{b^2}{4m^2}\right)m& -k{L}^2\\ -k{L}^2& \frac{mg}{L}+k-\frac{b^2}{4m}\left(\frac{g}{L}-\frac{b^2}{4m^2}\right)m\end{array}\right)\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)=0\\ \left(\begin{array}{cc}k& -k\\ -k& k\end{array}\right)\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)=0\\ \left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)=0\end{array}$

Writing the above matrices in equations

$a_1-a_2=0$  and $-a_1+a_2=0$

The solution to the above equation is $a_1=a_2$.

Thus, in the first normal mode, both the masses will oscillate with same amplitude and in phase with each other, ${\varphi }_1\left(t\right)={\varphi }_2\left(t\right)=A\cos \left({\omega }_{1}t-\delta \right)$.

In the second normal mode, ${\omega }_2^2=\frac{g}{L}+\frac{2k}{m}-\frac{b^2}{4m^2}$. Then Eigen value equation becomes

$\begin{array}{c}\left(K-{\omega }^{2}M\right)a=0\\ \left(\begin{array}{cc}\frac{mg}{L}+k-\frac{b^2}{4m}\left(\frac{g}{L}+\frac{2k}{m}-\frac{b^2}{4m^2}\right)m& -k{L}^2\\ -k{L}^2& \frac{mg}{L}+k-\frac{b^2}{4m}\left(\frac{g}{L}+\frac{2k}{m}-\frac{b^2}{4m^2}\right)m\end{array}\right)\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)=0\\ \left(\begin{array}{cc}k& k\\ k& k\end{array}\right)\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)=0\\ \left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)=0\end{array}$

Writing the above matrices in equation

$\begin{array}{c}{a}_1+a_2=0\\ a_1=-a_2\end{array}$

The solution to the above equation is $a_1=a_2$.

Thus, in the second normal mode, both the masses will oscillate with same amplitude and are in out of phase with each other, ${\varphi }_1\left(t\right)=-{\varphi }_2\left(t\right)=A\cos \left({\omega }_{1}t-\delta \right)$.

The eigenvectors $a_1$ and $a_2$ found in problem 11.14 are

$a_1=\left(\begin{array}{c}1\\ 1\end{array}\right)$  and $a_2=\left(\begin{array}{c}1\\ -1\end{array}\right)$

The eigenvector in matrix format

$\tilde{A}=\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)$

Determinate of the above matrix is

$\begin{array}{l}\left|\tilde{A}\right|=\left|\begin{array}{cc}1& 1\\ 1& -1\end{array}\right|\\ =\left(1\right)\left(-1\right)-\left(1\right)\left(1\right)\\ =-2\end{array}$

Since the determinate of $A$ is not equal to zero. The inverse of the matrix exist.

Therefore, the two eigenvectors $a_{\left(1\right)}\text{ and }{a}_{\left(2\right)}$ in $\tilde{A}$

Therefore, the $2\times 1$ column vector in terms of the two eigenvectors $a_{\left(1\right)}\text{ and }{a}_{\left(2\right)}$ $\varphi$ can be expanded in terms of them. Therefore, for any right-hand side of $\varphi$ can be

 $\varphi ={\xi }_1{a}_{\left(1\right)}+{\xi }_2{a}_{\left(2\right)}$

Here, ${\xi }_1,{\xi }_2$ are the normal coordination.

Substitute the column matrix of $a_1\text{ and }{a}_2$  in the above equation and solving

$\begin{array}{c}\varphi =\left(\begin{array}{c}{\varphi }_1\\ {\varphi }_2\end{array}\right)={\xi }_1\left(\begin{array}{c}1\\ 1\end{array}\right)+{\xi }_2\left(\begin{array}{c}1\\ -1\end{array}\right)\\ \left(\begin{array}{c}{\varphi }_1\\ {\varphi }_2\end{array}\right)=\left(\begin{array}{c}{\xi }_1\\ {\xi }_1\end{array}\right)+\left(\begin{array}{c}{\xi }_2\\ -{\xi }_2\end{array}\right)\\ \left(\begin{array}{c}{\varphi }_1\\ {\varphi }_2\end{array}\right)=\left(\begin{array}{c}{\xi }_1+{\xi }_2\\ {\xi }_1-{\xi }_2\end{array}\right)\end{array}$

The equations from the above matrix

${\varphi }_1={\xi }_1+{\xi }_2$

And,

${\varphi }_2={\xi }_1-{\xi }_2$

Solving algebraically,

$\begin{array}{l}{\varphi }_1={\xi }_1+{\xi }_2\\ \underset{\_}{{\varphi }_2={\xi }_1-{\xi }_2}\\ {\varphi }_1+{\varphi }_2=2{\xi }_1\end{array}$

${\xi }_1=\frac{{\varphi }_1+{\varphi }_2}{2}$

Substitute $\frac{{\varphi }_1+{\varphi }_2}{2}$ for ${\xi }_1$ in ${\varphi }_1={\xi }_1+{\xi }_2$ and solving

$\begin{array}{c}{\varphi }_1=\frac{{\varphi }_1+{\varphi }_2}{2}+{\xi }_2\\ 2{\xi }_2={\varphi }_1-{\varphi }_2\\ {\xi }_2=\frac{{\varphi }_1-{\varphi }_2}{2}\end{array}$

The natural choice of ${\xi }_1$ and ${\xi }_2$ is same even when the resistive force is added to the system. This proves that ${\xi }_1$ and ${\xi }_2$ are still uncoupled.

Conclusion:

It is proved that the equation of motion for ${\xi }_1$ and ${\xi }_2$ will be still uncoupled when both pendulums are subject to a resistive force of magnitude $bv$.

(c)

To determine

The motion of the two pendulums for the two modes and describe.

Answer to Problem 11.35P

The equation of the motion of the pendulum for the first mode is ${\varphi }_1\left(t\right)={\varphi }_2\left(t\right)=A{e}^{-\beta t}\cos \left(\omega t-\delta \right)$. The two masses are oscillating with same amplitude and in phase. The equation of the motion of the pendulum for the second mode is ${\varphi }_1\left(t\right)=-{\varphi }_2\left(t\right)=A{e}^{-\beta t}\cos \left(\omega t-\delta \right)$. The two masses are oscillating with same amplitude and out of phase.

Explanation of Solution

From part (b), the eigenvalue equations can be written in the form of $a_1=a_2=A{e}^{-i\delta }$.

The complex column $z\left(t\right)$ can be expressed as

$\begin{array}{c}z\left(t\right)=\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)e^{i\omega t-\beta t}\\ =A\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)e^{i\left(\omega -\delta \right)t-\beta t}\end{array}$

The real form of $z\left(t\right)$ will be corresponding to the actual motion of the mass. Therefore, the real column $\varphi \left(t\right)=\mathrm{Re}z\left(t\right)$.

$\begin{array}{c}\varphi \left(t\right)=\left(\begin{array}{c}{\varphi }_1\left(t\right)\\ {\varphi }_2\left(t\right)\end{array}\right)\\ =A\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)e^{-\beta t}\cos \left(\omega t-\delta \right)\end{array}$

Therefore, the equation of the motion of the pendulum for the first mode is ${\varphi }_1\left(t\right)={\varphi }_2\left(t\right)=A{e}^{-\beta t}\cos \left(\omega t-\delta \right)$. The two masses are oscillating with same amplitude and in phase.

For second mode,

$\varphi \left(t\right)=A\left(\begin{array}{c}1\\ -1\end{array}\right)e^{-\beta t}\cos \left(\omega t-\delta \right)$

Therefore, the equation of the motion of the pendulum for the second mode is ${\varphi }_1\left(t\right)=-{\varphi }_2\left(t\right)=A{e}^{-\beta t}\cos \left(\omega t-\delta \right)$. The two masses are oscillating with same amplitude and out of phase.

Conclusion:

The equation of the motion of the pendulum for the first mode is ${\varphi }_1\left(t\right)={\varphi }_2\left(t\right)=A{e}^{-\beta t}\cos \left(\omega t-\delta \right)$. The two masses are oscillating with same amplitude and in phase. The equation of the motion of the pendulum for the second mode is ${\varphi }_1\left(t\right)=-{\varphi }_2\left(t\right)=A{e}^{-\beta t}\cos \left(\omega t-\delta \right)$. The two masses are oscillating with same amplitude and out of phase.