Question 5.8: Determine the normal coordinates for the coupled pendulums o......

The two linearly independent solutions to Eqs. (5.98) in the case of the coupled pendulums are

(V- \omega ^{2}_{\left(s\right)}T) \varrho^{\left(s\right)}=0, s = 1, . . . , n . (5.98)

\varrho^{\left(1\right)}= \alpha \left ( \begin{matrix} 1 \\ 1 \end{matrix} \right ) , \varrho^{\left(2\right)}=\beta \left ( \begin{matrix} 1 \\ -1 \end{matrix} \right ),     \alpha ,\beta ∈ R, (5.142)

which were obtained in Example 5.7. In order to construct the modal matrix, it is necessary that \varrho^{\left(1\right)} and \varrho^{\left(2\right)} make up an orthonormal set in the inner product defined by the kinetic energy matrix T = mI – that is,

\varrho ^{\left(1\right) ^{T} } T \varrho^{\left(1\right)}=1 \Longrightarrow m \alpha^{2} \left ( \begin{matrix} 1 & 1 \end{matrix} \right ) \left ( \begin{matrix} 1 \\ 1 \end{matrix} \right )=1 \Longrightarrow 2m\alpha^{2}=1,   (5.143a)

\varrho ^{\left(2\right) ^{T} } T \varrho^{\left(2\right)}=1 \Longrightarrow m \beta^{2} \left ( \begin{matrix} 1 & -1 \end{matrix} \right ) \left ( \begin{matrix} 1 \\ -1 \end{matrix} \right )=1 \Longrightarrow 2m\beta^{2}=1,   (5.143b)

\varrho ^{\left(1\right) ^{T} } T \varrho^{\left(2\right)}=0 \Longrightarrow m \alpha \beta \left ( \begin{matrix} 1 & 1 \end{matrix} \right ) \left ( \begin{matrix} 1 \\ -1 \end{matrix} \right )=0 \Longrightarrow m \alpha \beta\left ( \begin{matrix} 1 & -1 \end{matrix} \right ) =0.  (5.143c)

The orthogonality condition is identically satisfied and the choice α = β = (2m)^{{−1}/{2}} ensures normalisation of the vectors. An orthonormal set of solutions of (5.98) is

\varrho^{\left(1\right)}= \frac{1}{\sqrt{2m} } \left ( \begin{matrix} 1 \\ 1 \end{matrix} \right ), \varrho^{\left(2\right)}= \frac{1}{\sqrt{2m} } \left ( \begin{matrix} 1 \\ -1 \end{matrix} \right ),   (5.144)

which give rise to the modal matrix

A = \frac{1}{\sqrt{2m} } \left ( \begin{matrix} 1  & 1 \\ 1 & -1\end{matrix} \right ). (5.145)

The normal coordinates are

ζ = A^{T}Tη =\frac{m}{\sqrt{2m} }\left ( \begin{matrix} 1  & 1 \\ 1 & -1\end{matrix} \right ) \left ( \begin{matrix} \eta_{1}  \\ \eta_{2} \end{matrix} \right ),  (5.146)

that is,

ζ _{1} =\sqrt{\frac{m}{2} }\left(\eta_{1}+\eta_{2}\right) , ζ _{2} =\sqrt{\frac{m}{2} }\left(\eta_{1}-\eta_{2}\right) .(5.147)

The normal mode of frequency ω_{1}has \eta_{1} = \eta_{2}, so only ζ_{1} oscillates while ζ_{2} is identically zero; the opposite occurs with the normal mode of frequency ω_{2}.