= Solution
{c}
Measure $x_1,x_2$ as horizontal <displacements> from the respective downward <equilibrium> positions; the absolute horizontal <coordinates> are $x_1$ and $L+x_2$. Each vertical <displacement> is of second order in its small <angle>. Thus the <spring> is horizontal to first order, its extension is $x_2-x_1$, and <Hooke's law> gives horizontal <forces> $k(x_2-x_1)$ and $-k(x_2-x_1)$. Combining these with the linearized <simple pendulum> restoring <force> yields
$$
\boxed{\ddot x_1=-\frac g\ell x_1+\frac km(x_2-x_1),\qquad
\ddot x_2=-\frac g\ell x_2-\frac km(x_2-x_1).}
$$
For the <normal modes of two spring-coupled equal pendulums>, add and subtract these <ordinary differential equations>. The <normal mode coordinates> $q_+=x_1+x_2$ and $q_-=x_1-x_2$ satisfy
$$
\ddot q_++\frac g\ell q_+=0,\qquad
\ddot q_-+\left(\frac g\ell+\frac{2k}{m}\right)q_-=0.
$$
Hence the two natural <angular frequencies> and displacement patterns are
$$
\boxed{\omega_+=\sqrt{\frac g\ell},\quad(x_1,x_2)\propto(1,1);\qquad
\omega_-=\sqrt{\frac g\ell+\frac{2k}{m}},\quad(x_1,x_2)\propto(1,-1).}
$$
Ordinary <frequencies> are $\omega_+/(2\pi)$ and $\omega_-/(2\pi)$. In the first <normal mode>, both particles swing together with equal <amplitudes> and the <spring> is unextended to first order. In the second <normal mode>, the particles swing in opposite directions with equal <amplitudes>, so the <spring> adds a restoring <force> and increases the <angular frequency>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-iv-resit-pendulum-modes.png]
{title=The two normal modes: equal displacements together and equal displacements in opposite directions}
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