Chapter 11, Problem 11.26P

To determine

The normal frequencies of small oscillations and describe the motion of the oscillation in corresponding normal modes.

Answer to Problem 11.26P

The normal frequencies of small oscillations of the beads are ${\omega }_1^2=\frac{g}{2R}$ and ${\omega }_2^2=\frac{2g}{R}$. The first normal mode is $x_1\left(t\right)=A\cos \left({\omega }_{1}t-\delta \right)$ and $x_2\left(t\right)=A\cos \left({\omega }_{1}t-\delta \right)$ and the oscillation is in same phase. The second normal mode is $x_1\left(t\right)=A\cos \left({\omega }_{1}t-\delta \right)$ and $x_2\left(t\right)=-2A\cos \left({\omega }_{1}t-\delta \right)$ and the oscillation is in out of phase with different amplitude.

Explanation of Solution

Write the Lagrangian of the system

    $L=T-U$        (I)

Here, $L$ is the Lagrangian of the system, $T$ is the kinetic energy of the system and $U$ is the potential energy.

Write the Lagrangian equation of motion, if the Lagrangian of the system is function of $L=f\left(q_1,q_2,\mathrm{....},q_i\right)$

    $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)=\frac{\partial L}{\partial q}$        (II)

Write the moment of inertia of the system from parallel axis theorem

    $I=I_{hoop}+m{R}^2$        (III)

Here, $I$ is the moment of inertia of the system, $I_{hoop}$ is the moment of inertia about an axis passing through its center, $m$ is the mass of the hoop and $R$ is the distance between the center and point $A$.

The mass of the hoop is $m$ and the distance from the axis passing through its center is $R$. Therefore, the moment of inertia of the hoop is

    $I_{hoop}=m{R}^2$        (IV)

Substitute equation (IV) in (III) and solve for $I$

$\begin{array}{c}I=m{R}^2+m{R}^2\\ =2m{R}^2\end{array}$

Write the equation to find the kinetic energy of the hoop

    $T_{hoop}=\frac{1}{2}I{\left({\dot{\varphi }}_1\right)}^2$

Substitute $2m{R}^2$ for $I$ in the above equation

$\begin{array}{c}{T}_{hoop}=\frac{1}{2}\left(2m{R}^2\right){\left({\dot{\varphi }}_1\right)}^2\\ =m{R}^2{\dot{\varphi }}_1^2\end{array}$        (V)

In Figure, ${\varphi }_1$ and ${\varphi }_2$ are the angles of the position of the bead from the vertical axis.

Consider $A$ as the origin, then, the position of bead along horizontal direction to the left of $A$ is $R\sin {\varphi }_1+R\sin {\varphi }_2$ and the position of bead along vertical direction below the point $A$ is $R\cos {\varphi }_1+R\cos {\varphi }_2$.

Hence, the position of the bead can be expressed as

$r=\left(R\sin {\varphi }_1+R\sin {\varphi }_2\right)i+\left(-R\cos {\varphi }_1-R\cos {\varphi }_2\right)j$

Differentiate the above equation to find the velocity of the bead

$\dot{r}=\left({\dot{\varphi }}_{1}R\cos {\varphi }_1+{\dot{\varphi }}_{2}R\cos {\varphi }_2\right)i+\left(R\sin {\varphi }_1{\dot{\varphi }}_1+R\sin {\varphi }_2{\dot{\varphi }}_2\right)j$

Consider ${\varphi }_1$ and ${\varphi }_2$ are small angle. Therefore, $\cos {\varphi }_1\to 1$, $\cos {\varphi }_2\to 1$, $\sin {\varphi }_1\approx {\varphi }_1$ and $\sin {\varphi }_2\approx {\varphi }_2$. The above expression becomes,

$\dot{r}=\left({\dot{\varphi }}_{1}R+{\dot{\varphi }}_{2}R\right)i+\left(R{\varphi }_1{\dot{\varphi }}_1+R{\varphi }_2{\dot{\varphi }}_2\right)j$        (VI)

Write the equation to find the kinetic energy of bead

    $T_{bead}=\frac{1}{2}m{\dot{r}}^2$

Here, $T_{bead}$ is the kinetic energy of bead, $m$ is the mass of the bead and $\dot{r}$ is the velocity of the bead.

Substitute equation (VI) in the above equation to solve for $T_{bead}$

$\begin{array}{c}{T}_{bead}=\frac{1}{2}m{\left[\left({\dot{\varphi }}_{1}R+{\dot{\varphi }}_{2}R\right)i+\left(R{\varphi }_1{\dot{\varphi }}_1+R{\varphi }_2{\dot{\varphi }}_2\right)j\right]}^2\\ =\frac{1}{2}m\left[{\left({\dot{\varphi }}_{1}R+{\dot{\varphi }}_{2}R\right)}^2+{\left(R{\varphi }_1{\dot{\varphi }}_1+R{\varphi }_2{\dot{\varphi }}_2\right)}^2\right]\end{array}$

Since ${\varphi }_1$ and ${\varphi }_2$ are small angle. Therefore angles ${\dot{\varphi }}_1$ and ${\dot{\varphi }}_2$ will be small. Hence, multiplying these small angles will also give insignificant value. Ignoring ${\left(R{\varphi }_1{\dot{\varphi }}_1+R{\varphi }_2{\dot{\varphi }}_2\right)}^2$ in the above equation.

$\begin{array}{c}{T}_{bead}=\frac{1}{2}m{\left({\dot{\varphi }}_{1}R+{\dot{\varphi }}_{2}R\right)}^2\\ =\frac{1}{2}m{R}^2{\left({\dot{\varphi }}_1+{\dot{\varphi }}_2\right)}^2\end{array}$        (VII)

Write the formula to find the total kinetic energy of the system

    $T=T_{hoop}+T_{bead}$

Substitute equation (V) and (VII) in the above equation and solve for $T$

$\begin{array}{c}T=m{R}^2{\dot{\varphi }}_1^2+\frac{1}{2}m{R}^2{\left({\dot{\varphi }}_1+{\dot{\varphi }}_2\right)}^2\\ =m{R}^2{\dot{\varphi }}_1^2+\frac{1}{2}m{R}^2\left({\dot{\varphi }}_1^2+{\dot{\varphi }}_2^2+2{\dot{\varphi }}_1{\dot{\varphi }}_2\right)\end{array}$

$T=\frac{1}{2}m{R}^2\left(3{\dot{\varphi }}_1^2+{\dot{\varphi }}_2^2+2{\dot{\varphi }}_1{\dot{\varphi }}_2\right)$        (VIII)

About the point $A$, the vertical component of the position of center of mass of hoop is $R\cos {\varphi }_1$ below $A$

Write the equation for the potential energy of hoop

    $U_{hoop}=-mgR\cos {\varphi }_1$

Here, $U_{hoop}$ is the potential energy of hoop, $m$ is the mass of the hoop and $g$ is the acceleration due to gravity.

Substitute the expression of $\cos {\varphi }_1$ in the above equation and solve for $U_{hoop}$

$U_{hoop}=-mgR\left(1-\frac{1}{2}{\varphi }_1^2+\mathrm{.....}\right)$

Ignoring the higher terms the above equation becomes,

$U_{hoop}=\frac{1}{2}mgR{\varphi }_1^2$        (IX)

Form figure, the vertical component of position of the bead is $R\cos {\varphi }_1+R\cos {\varphi }_2$ below the point $A$.

Write the equation for the potential energy of bead

    $U_{bead}=-mgR\left(\cos {\varphi }_1+\cos {\varphi }_2\right)$

Substitute the expression of $\cos {\varphi }_1$ in the above equation and solve for $U_{bead}$

$U_{bead}=-mgR\left[\left(1-\frac{1}{2}{\varphi }_1^2+\mathrm{.....}\right)+\left(1-\frac{1}{2}{\varphi }_2^2+\mathrm{.....}\right)\right]$

Ignoring the higher terms the above equation becomes,

$\begin{array}{c}{U}_{bead}=-mgR\left(1-\frac{1}{2}{\varphi }_1^2+1-\frac{1}{2}{\varphi }_2^2\right)\\ =-2mgR+\frac{1}{2}mgR\left({\varphi }_1^2+{\varphi }_2^2\right)\end{array}$

In the above equation, $-2mgR$ is a constant term and as it is irrelevant over here. This term can be ignored.

The above equation becomes,

$U_{bead}=\frac{1}{2}mgR\left({\varphi }_1^2+{\varphi }_2^2\right)$        (X)

Write the formula to find the total potential energy of the system

    $U=U_{bead}+U_{hoop}$

Substitute (IX) and (X) in the above equation to solve for $U$

$\begin{array}{c}U=\frac{1}{2}mgR\left({\varphi }_1^2+{\varphi }_2^2\right)+\frac{1}{2}mgR{\varphi }_1^2\\ =\frac{1}{2}mgR\left(2{\varphi }_1^2+{\varphi }_2^2\right)\end{array}$        (XI)

Substitute (VIII) and (XI) in equation (I)

$\begin{array}{c}L=T-U\\ =\frac{1}{2}m{R}^2\left(3{\dot{\varphi }}_1^2+{\dot{\varphi }}_2^2+2{\dot{\varphi }}_1{\dot{\varphi }}_2\right)-\frac{1}{2}mgR\left(2{\varphi }_1^2+{\varphi }_2^2\right)\end{array}$        (XII)

Write the expression of Lagrangian equations of motion in term of ${\varphi }_1$ direction

    $\frac{d}{dt}\left(\frac{\partial L}{\partial {\dot{\varphi }}_1}\right)=\frac{\partial L}{\partial {\varphi }_1}$

Substitute (XII) in the above equation and solve

$\begin{array}{c}\frac{d}{dt}\left(\frac{\partial \left[\frac{1}{2}m{R}^2\left(3{\dot{\varphi }}_1^2+{\dot{\varphi }}_2^2+2{\dot{\varphi }}_1{\dot{\varphi }}_2\right)-\frac{1}{2}mgR\left(2{\varphi }_1^2+{\varphi }_2^2\right)\right]}{\partial {\dot{\varphi }}_1}\right)=\frac{\partial \left[\frac{1}{2}m{R}^2\left(3{\dot{\varphi }}_1^2+{\dot{\varphi }}_2^2+2{\dot{\varphi }}_1{\dot{\varphi }}_2\right)-\frac{1}{2}mgR\left(2{\varphi }_1^2+{\varphi }_2^2\right)\right]}{\partial {\varphi }_1}\\ 3m{R}^2{\ddot{\varphi }}_1+m{R}^2{\ddot{\varphi }}_2=\frac{1}{2}mgR\left(4{\varphi }_1\right)\\ 3m{R}^2{\ddot{\varphi }}_1+m{R}^2{\ddot{\varphi }}_2=-2mgR{\varphi }_1\end{array}$

Dividing $m{R}^2$ on both sides and solving

$3{\ddot{\varphi }}_1+{\ddot{\varphi }}_2=-2\frac{g}{R}{\varphi }_1$        (XIII)

Write the expression of Lagrangian equations of motion in term of ${\varphi }_2$ direction

    $\frac{d}{dt}\left(\frac{\partial L}{\partial {\dot{\varphi }}_2}\right)=\frac{\partial L}{\partial {\varphi }_2}$

Substitute (XII) in the above equation and solve

$\begin{array}{c}\frac{d}{dt}\left(\frac{\partial \left[\frac{1}{2}m{R}^2\left(3{\dot{\varphi }}_1^2+{\dot{\varphi }}_2^2+2{\dot{\varphi }}_1{\dot{\varphi }}_2\right)-\frac{1}{2}mgR\left(2{\varphi }_1^2+{\varphi }_2^2\right)\right]}{\partial {\dot{\varphi }}_2}\right)=\frac{\partial \left[\frac{1}{2}m{R}^2\left(3{\dot{\varphi }}_1^2+{\dot{\varphi }}_2^2+2{\dot{\varphi }}_1{\dot{\varphi }}_2\right)-\frac{1}{2}mgR\left(2{\varphi }_1^2+{\varphi }_2^2\right)\right]}{\partial {\varphi }_2}\\ m{R}^2{\ddot{\varphi }}_1+m{R}^2{\ddot{\varphi }}_2=-mgR{\varphi }_2\end{array}$

Dividing $m{R}^2$ on both sides and solving

${\ddot{\varphi }}_1+{\ddot{\varphi }}_2=-\frac{g}{R}{\varphi }_2$        (XIV)

The equation of the motion of the mass along $x$ and $y$ -coordinate can be expressed as,

$\left(\begin{array}{cc}3& 1\\ 1& 1\end{array}\right)\left(\begin{array}{c}{\ddot{\varphi }}_1\\ {\ddot{\varphi }}_2\end{array}\right)=-\left(\frac{g}{R}\right)\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}{\varphi }_1\\ {\varphi }_2\end{array}\right)$

Write the general equation of motion

    $M\ddot{\varphi }=-K\varphi$

Comparing the above equation of motion with the general equation of the motion

$M=\left(\begin{array}{cc}3& 1\\ 1& 1\end{array}\right)$, $K=\left(\frac{g}{R}\right)\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)$, $\ddot{\varphi }=\left(\begin{array}{c}{\ddot{\varphi }}_1\\ {\ddot{\varphi }}_2\end{array}\right)$ and $\varphi =\left(\begin{array}{c}{\varphi }_1\\ {\varphi }_2\end{array}\right)$

If the frequency $\omega$ and column matrix $a$ satisfied the eigenvalue equation $\left(K-{\omega }^{2}M\right)a=0$. And as $a$ is a constant it can be ignored. The equation becomes,

  $\det \left(K-{\omega }^{2}M\right)=0$

Substitute the matrices in the above equation

$\begin{array}{c}K-{\omega }^{2}M=\left(\frac{g}{R}\right)\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)-{\omega }^2\left(\begin{array}{cc}3& 1\\ 1& 1\end{array}\right)\\ =\left(\frac{g}{R}\right)\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)-\left(\begin{array}{cc}3{\omega }^2& {\omega }^2\\ {\omega }^2& {\omega }^2\end{array}\right)\end{array}$

$K-{\omega }^{2}M=\left(\begin{array}{cc}2\left(\frac{g}{R}\right)-3{\omega }^2& -{\omega }^2\\ -{\omega }^2& \left(\frac{g}{R}\right)-{\omega }^2\end{array}\right)$        (XV)

The determinant of $\left(K-{\omega }^{2}M\right)$ is

$\begin{array}{c}\det \left(K-{\omega }^{2}M\right)=0\\ \left|\begin{array}{cc}2\left(\frac{g}{R}\right)-3{\omega }^2& -{\omega }^2\\ -{\omega }^2& \left(\frac{g}{R}\right)-{\omega }^2\end{array}\right|=0\\ \left(2\left(\frac{g}{R}\right)-3{\omega }^2\right)\left(\left(\frac{g}{R}\right)-{\omega }^2\right)-{\omega }^4=0\\ 2{\omega }^4-5\left(\frac{g}{R}\right){\omega }^2+2{\left(\frac{g}{R}\right)}^2=0\\ 2{\omega }^4-4\left(\frac{g}{R}\right){\omega }^2-\left(\frac{g}{R}\right){\omega }^2+2{\left(\frac{g}{R}\right)}^2=0\end{array}$

Factorize the above expression and solving,

$\left(2{\omega }^2-\left(\frac{g}{R}\right)\right)\left({\omega }^2-2\left(\frac{g}{R}\right)\right)=0$

Hence,

$\begin{array}{c}\left(2{\omega }^2-\left(\frac{g}{R}\right)\right)=0\\ {\omega }^2=\frac{g}{2R}\end{array}$

 And

$\begin{array}{c}\left({\omega }^2-2\left(\frac{g}{R}\right)\right)=0\\ {\omega }^2=\frac{2g}{R}\end{array}$

Therefore, the normal frequency for the small oscillation of the beads are ${\omega }_1^2=\frac{g}{2R}$ and ${\omega }_2^2=\frac{2g}{R}$.

First normal mode for the normal frequency is

Substitute $\frac{g}{2R}$ for ${\omega }^2$ in equation (XV) and solving

$\begin{array}{c}K-{\omega }^{2}M=\left(\begin{array}{cc}2\left(\frac{g}{R}\right)-3\left(\frac{g}{2R}\right)& -\left(\frac{g}{2R}\right)\\ -\left(\frac{g}{2R}\right)& \left(\frac{g}{R}\right)-\left(\frac{g}{2R}\right)\end{array}\right)\\ =\left(\begin{array}{cc}\frac{g}{2R}& -\frac{g}{2R}\\ -\frac{g}{2R}& \frac{g}{2R}\end{array}\right)\end{array}$

The eigenvalue of the equation becomes

$\left(\begin{array}{cc}\frac{1}{2}& -\frac{1}{2}\\ -\frac{1}{2}& \frac{1}{2}\end{array}\right)\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)=0$

The equation becomes,

$\left(\frac{1}{2}\right)a_1-\left(\frac{1}{2}\right)a_2=0$

And,

$-\left(\frac{1}{2}\right)a_1+\left(\frac{1}{2}\right)a_2=0$

Solving the above equation gives,

$a_1=a_2$

These implies equation can be written in the form of $a_1=a_2=A{e}^{-i\delta }$.

Therefore, the complex column $z\left(t\right)$ is

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

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

$\begin{array}{l}x\left(t\right)=\left(\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right)\\ =\left(\begin{array}{c}A\\ A\end{array}\right)\cos \left({\omega }_{1}t-\delta \right)\end{array}$

Thus, the first normal mode is $x_1\left(t\right)=A\cos \left({\omega }_{1}t-\delta \right)$ and $x_2\left(t\right)=A\cos \left({\omega }_{1}t-\delta \right)$. The magnitude and the sign are the same. Therefore, the oscillation is in same phase.

Second normal mode for the normal frequency is

Substitute $\frac{2g}{R}$ for ${\omega }^2$ in equation (XV) and solving

$\begin{array}{c}K-{\omega }^{2}M=\left(\begin{array}{cc}2\left(\frac{g}{R}\right)-3\left(\frac{2g}{R}\right)& -\left(\frac{2g}{R}\right)\\ -\left(\frac{2g}{R}\right)& \left(\frac{g}{R}\right)-\left(\frac{2g}{R}\right)\end{array}\right)\\ =\left(\begin{array}{cc}-\frac{4g}{R}& -\frac{2g}{2R}\\ -\frac{2g}{2R}& -\frac{g}{2R}\end{array}\right)\end{array}$

The eigenvalue of the equation becomes

$\left(\begin{array}{cc}-4& -2\\ -2& -1\end{array}\right)\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)=0$

The equation becomes,

$-4a_1-2a_2=0$

And,

$-2a_1-a_2=0$

Solving the above equation gives,

$a_1=-\frac{a_2}{2}$

These implies equation can be written in the form of $a_1=-\frac{a_2}{2}=A{e}^{-i\delta }$.

Therefore, the complex column $z\left(t\right)$ is

$\begin{array}{c}z\left(t\right)=\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)e^{i{\omega }_{z}t}\\ =\left(\begin{array}{c}A\\ -2A\end{array}\right)e^{i{\omega }_{z}t-\delta }\end{array}$

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

$\begin{array}{l}x\left(t\right)=\left(\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right)\\ =\left(\begin{array}{c}A\\ -2A\end{array}\right)\cos \left({\omega }_{1}t-\delta \right)\end{array}$

Thus, the second normal mode is $x_1\left(t\right)=A\cos \left({\omega }_{1}t-\delta \right)$ and $x_2\left(t\right)=-2A\cos \left({\omega }_{1}t-\delta \right)$. The magnitude and the sign are the different. Therefore, the oscillation is in out of phase with different amplitude.

Conclusion:

The normal frequencies of small oscillations of the beads are ${\omega }_1^2=\frac{g}{2R}$ and ${\omega }_2^2=\frac{2g}{R}$. The first normal mode is $x_1\left(t\right)=A\cos \left({\omega }_{1}t-\delta \right)$ and $x_2\left(t\right)=A\cos \left({\omega }_{1}t-\delta \right)$ and the oscillation is in same phase. The second normal mode is $x_1\left(t\right)=A\cos \left({\omega }_{1}t-\delta \right)$ and $x_2\left(t\right)=-2A\cos \left({\omega }_{1}t-\delta \right)$ and the oscillation is in out of phase with different amplitude.