Question 5.7: Two equal pendulums perform small oscillations coupled by a ......

Let us assume that the equilibrium configuration is \theta_{1} = \theta_{2}= 0, in which the spring has its natural length d_{0}.We shall use as coordinates the small horizontal displacements \eta_{1} and \eta_{2} shown in Fig. 5.7, which are related to the small angles \theta_{1} and \theta_{2} by

\eta_{1}= l \sin\theta_{1} \approx  l \theta_{1}, \eta_{2}= l \sin\theta_{2} \approx  l \theta_{2}.  (5.105)

The kinetic energy is

T =\frac{ml^{2}}{2}\dot{\theta} ^{   2}_{1}+ \frac{ml^{2}}{2}\dot{\theta} ^{   2}_{2} \approx \frac{m}{2} \dot{\eta}^{   2}_{1}+\frac{m}{2} \dot{\eta}^{   2}_{2},  (5.106)

while the potential energy is written

V = \frac{k}{2}\left[\left(d_{0}+\eta_{2}-\eta_{1}\right)-d_{0}\right] ^{2}+mg \left[l\left(1- \cos \theta_{1}\right)+l\left(1- \cos \theta_{2}\right)\right].  (5.107)

Using \cos \theta \approx  1- {\theta^{2}}/{2} and (5.105) we get

V = \frac{k}{2}\left(\eta_{2}-\eta_{1}\right)^{2}+ \frac{mg}{2l}\left(\eta^{  2}_{1}-\eta^{  2}_{2}\right),  (5.108)

so the Lagrangian appropriate to the description of small oscillations is

L =\frac{m}{2}\left(\dot{\eta}^{2}_{1}+\dot{\eta}^{2}_{2}\right)-\frac{1}{2}\left\{\frac{mg}{l}\left(\eta^{   2}_{1}+\eta^{   2}_{2}\right)+k \left(\eta_{2}-\eta_{1}\right)^{2}\right\} .  (5.109)

Comparing this with the standard form (5.84) we have

L=\frac{1}{2} \sum\limits_{kl}{T_{kl}\dot{\eta }_{k}\dot{\eta }_{l} }-\frac{1}{2} \sum\limits_{kl}{V_{kl}\eta _{k}\eta _{l} },  (5.84)

T=\begin{pmatrix} m & 0 \\ 0 & m \end{pmatrix}, V=\begin{pmatrix} \frac{mg}{l}+k & -k \\ -k & \frac{mg}{l}+k \end{pmatrix}.  (5.110)

Warning In the standard form (5.84), the off-diagonal terms appear in duplicate. For example, in the second double sum over k and l there is one term for k = 1, l = 2 and another for k = 2, l = 1, which are equal to each other because V_{12} = V_{21}. Thus, the coefficient of \eta_{1}\eta_{2} is V_{12}+V_{21}= 2V_{12}. On the other hand, the coefficient of \eta_{1}\eta_{2} in the expression within brackets in (5.109) is −2k, hence the identification V_{12} = −k. This kind of attention is needed when comparing the standard form (5.84) with a specific Lagrangian such as (5.109), otherwise errors by a factor of two will be made, which will damage all subsequent results.

The frequencies of the normal modes are the solutions of

det\left(V-\omega ^{2}T\right)= det\begin{pmatrix} \frac{mg}{l}+k-m\omega ^{2} & -k \\ -k & \frac{mg}{l}+k-m\omega ^{2} \end{pmatrix} =0,  (5.111)

that is,

\left(\frac{mg}{l}+k-m\omega ^{2}\right)^{2}-k^{2}=0 \Longrightarrow  \frac{mg}{l}+k-m\omega ^{2}=\pm k.  (5.112)

The characteristic frequencies are

\omega_{1}= \left(\frac{g}{l}\right)^{{1}/{2}} , \omega_{2}= \left(\frac{g}{l}+\frac{2k}{m}\right)^{{1}/{2}}, (5.113)

and the corresponding eigenvectors \varrho ^{(s)} are such that

\left(V-\omega ^{2}_{1}T\right)\varrho^{\left(1\right)}=0 \Longrightarrow \begin{pmatrix} k & -k \\ -k & k \end{pmatrix} \left ( \begin{matrix} \varrho^{\left(1\right)}_{1} \\ \varrho^{\left(1\right)}_{2} \end{matrix} \right ) =0 \Longrightarrow \varrho^{\left(1\right)}=\left ( \begin{matrix} \alpha \\ \alpha \end{matrix} \right ),

\left(V-\omega ^{2}_{2}T\right)\varrho^{\left(2\right)}=0 \Longrightarrow \begin{pmatrix} -k & -k \\ -k & -k \end{pmatrix} \left ( \begin{matrix} \varrho^{\left(2\right)}_{1} \\ \varrho^{\left(2\right)}_{2} \end{matrix} \right ) =0 \Longrightarrow \varrho^{\left(2\right)}=\left ( \begin{matrix} \beta \\ -\beta \end{matrix} \right ),

where α and β are non-zero real numbers. The mode with frequency \omega _{1} corresponds to an oscillatory motion of the pendulums with \eta_{1} = \eta_{2}. In this case, it all happens as if the spring did not exist since it remains with its natural length all the time, which explains the value of \omega_{1} given by (5.113): the pendulums oscillate in unison with the frequency of an individual simple pendulum. The second normal mode consists in an oscillatory motion of the pendulums with frequency \omega_{2}, but now \eta_{1} = -\eta_{2} and the pendulums oscillate in opposite directions, as depicted in Fig. 5.8.

The two linearly independent real solutions can be chosen as

\varrho^{\left(1\right)}=\left ( \begin{matrix} 1 \\ 1 \end{matrix} \right ), \varrho^{\left(2\right)}=\left ( \begin{matrix} 1 \\ -1 \end{matrix} \right )  (5.114)

and the general solution (5.104) takes the form

\eta _{k}\left(t\right)=\sum\limits_{s=1}^{n}{C^{\left(s\right) }\varrho^{\left(s\right)}_{k}\cos \left(\omega _{s}t+\phi_{s} \right) } ,  k = 1, . . . , n . (5.104)

\eta _{1}\left(t\right)=\sum\limits_{s=1}^{2}{C^{\left(s\right) }\varrho^{\left(s\right)}_{1}\cos \left(\omega _{s}t+\phi_{s} \right) } = C^{\left(1\right) }\cos \left(\omega _{1}t+\phi_{1} \right)+C^{\left(2\right) } \cos \left(\omega _{2}t+\phi_{2} \right),  (5.115)

\eta _{2}\left(t\right)=\sum\limits_{s=1}^{2}{C^{\left(s\right) }\varrho^{\left(s\right)}_{2}\cos \left(\omega _{s}t+\phi_{s} \right) } = C^{\left(1\right) }\cos \left(\omega _{1}t+\phi_{1} \right)-C^{\left(2\right) } \cos \left(\omega _{2}t+\phi_{2} \right).  (5.116)

From the initial conditions \eta_{1}\left( 0\right) = 0, \eta_{2}\left( 0\right) = a, \dot{\eta_{1}} \left( 0\right) =\dot{\eta_{2}}\left( 0\right) = 0, we infer

C^{\left(1\right) }\cos \phi_{1} +C^{\left(2\right) }\cos \phi_{2} =0,  (5.117a)

C^{\left(1\right) }\cos \phi_{1}  -C^{\left(2\right) }\cos \phi_{2} =a,  (5.117b)

-\omega _{1}C^{\left(1\right) }\sin \phi_{1} -\omega _{2}C^{\left(2\right) }\sin \phi_{2}=0,  (5.117c)

-\omega _{1}C^{\left(1\right) }\sin \phi_{1} +\omega _{2}C^{\left(2\right) }\sin \phi_{2}=0.  (5.117d)

By adding the first two and the last two of Eqs. (5.117) we deduce

C^{\left(1\right) }\cos \phi_{1}= -C^{\left(2\right) }\cos \phi_{2}= {a}/{2}, (5.118a)

C^{\left(1\right) }\sin \phi_{1} =C^{\left(2\right) }\sin \phi_{2}= 0.  (5.118b)

Neither C^{\left(1\right) }=0 nor C^{\left(2\right) }=0 is admissible because this would give rise to a contradiction with (5.118a). Therefore, \sin \phi_{1}=\sin \phi_{2}=0, whose simplest solution is \phi_{1}=\phi_{2}=0, by reason of which C^{\left(1\right) }=-C^{\left(2\right) }={a}/{2}. Finally, the solution to the equations of motion is written

\eta_{1}\left( t\right)= \frac{a}{2} \left(\cos \omega _{1}t- \cos \omega _{2}t\right),   (5.119a)

\eta_{2}\left( t\right)= \frac{a}{2} \left(\cos \omega _{1}t+ \cos \omega _{2}t\right),   (5.119b)

with \omega _{1} and \omega _{2} given by (5.113).