(a) \(\mathrm{With} \omega^{2}=g / l\) and \(K=k / m\) the system of differential
equations is
\\[
\begin{array}{l}
\theta_{1}^{\prime \prime}+\omega^{2}
\theta_{1}=-K\left(\theta_{1}-\theta_{2}\right) \\
\theta_{2}^{\prime \prime}+\omega^{2}
\theta_{2}=K\left(\theta_{1}-\theta_{2}\right).
\end{array}
\\]
Denoting the Laplace transform of \(\theta(t)\) by \(\Theta(s)\) we have that the
Laplace transform of the system is
\\[
\begin{array}{l}
\left(s^{2}+\omega^{2}\right) \Theta_{1}(s)=-K \Theta_{1}(s)+K \Theta_{2}(s)+s
\theta_{0} \\
\left(s^{2}+\omega^{2}\right) \Theta_{2}(s)=K \Theta_{1}(s)-K \Theta_{2}(s)+s
\psi_{0}.
\end{array}
\\]
If we add the two equations, we get
\\[
\Theta_{1}(s)+\Theta_{2}(s)=\left(\theta_{0}+\psi_{0}\right)
\frac{s}{s^{2}+\omega^{2}}
\\]
which implies
\\[
\theta_{1}(t)+\theta_{2}(t)=\left(\theta_{0}+\psi_{0}\right) \cos \omega t.
\\]
Now solving
\\[
\begin{array}{c}
\left(s^{2}+\omega^{2}+K\right) \Theta_{1}(s)-K \Theta_{2}(s)=s \theta_{0} \\
-k \Theta_{1}(s)+\left(s^{2}+\omega^{2}+K\right) \Theta_{2}(s)=s \psi_{0}
\end{array}
\\]
gives
\\[
\left[\left(s^{2}+\omega^{2}+K\right)^{2}-K^{2}\right]
\Theta_{1}(s)=s\left(s^{2}+\omega^{2}+K\right) \theta_{0}+K s \psi_{0}.
\\]
Factoring the difference of two squares and using partial fractions we get
\\[
\Theta_{1}(s)=\frac{s\left(s^{2}+\omega^{2}+K\right) \theta_{0}+K s
\psi_{0}}{\left(s^{2}+\omega^{2}\right)\left(s^{2}+\omega^{2}+2
K\right)}=\frac{\theta_{0}+\psi_{0}}{2}
\frac{s}{s^{2}+\omega^{2}}+\frac{\theta_{0}-\psi_{0}}{2}
\frac{s}{s^{2}+\omega^{2}+2 K},
\\]
so \(\theta_{1}(t)=\frac{\theta_{0}+\psi_{0}}{2} \cos \omega
t+\frac{\theta_{0}-\psi_{0}}{2} \cos \sqrt{\omega^{2}+2 K} t.\)
Then from \(\theta_{2}(t)=-\theta_{1}(t)+\left(\theta_{0}+\psi_{0}\right)\) cos
\(\omega t\) we get
\\[
\theta_{2}(t)=\frac{\theta_{0}+\psi_{0}}{2} \cos \omega
t-\frac{\theta_{0}-\psi_{0}}{2} \cos \sqrt{\omega^{2}+2 K} t.
\\]
(b) With the initial conditions \(\theta_{1}(0)=\theta_{0},
\theta_{1}^{\prime}(0)=0, \theta_{2}(0)=\theta_{0}, \theta_{2}^{\prime}(0)=0\)
we have
\\[
\theta_{1}(t)=\theta_{0} \cos \omega t, \quad \theta_{2}(t)=\theta_{0} \cos
\omega t.
\\]
Physically this means that both pendulums swing in the same direction as if
they were free since the spring exerts no influence on the motion
\(\left(\theta_{1}(t) \text { and } \theta_{2}(t) \text { are free of }
K\right)\).
With the initial conditions \(\theta_{1}(0)=\theta_{0},
\theta_{1}^{\prime}(0)=0, \theta_{2}(0)=-\theta_{0}, \theta_{2}^{\prime}(0)=0\)
we have
\\[
\theta_{1}(t)=\theta_{0} \cos \sqrt{\omega^{2}+2 K} t, \quad
\theta_{2}(t)=-\theta_{0} \cos \sqrt{\omega^{2}+2 K} t.
\\]
Physically this means that both pendulums swing in the opposite directions,
stretching and compressing the spring. The amplitude of both displacements is
\(\left|\theta_{0}\right| .\) Moreover, \(\theta_{1}(t)=\theta_{0}\) and
\(\theta_{2}(t)=-\theta_{0}\) at precisely
the same times. At these times the spring is stretched to its maximum.
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