 Solution

For the normal modes of two spring-coupled equal pendulums, a normal mode has every coordinate oscillating at the same frequency with a fixed displacement ratio. The linearized equations here areTheir normalized displacement patterns and angular frequencies areThe spring is unstretched in the in-phase mode, and stretched or compressed in the opposite-phase mode. The numerical ratio follows from $k/m=(21/200){\omega }_0^2$.

The initial displacement is the equal sum of the two mode patterns, and the initial velocities vanish. ThereforeWith the printed ${\omega }_0=2\pi \mathrm{rad}{\mathrm{s}}^{-1}$ this is $x_2/a=\sin (2.1\pi t)\sin (0.1\pi t)$, with $t$ in seconds. The fast oscillation has frequency $1.05\mathrm{Hz}$ and the amplitude is bounded by $a|\sin (\pi t/10)|$. The first complete displacement transfer is at $t=5\mathrm{s}$, when $x_1=0$, $x_2=a$ and both velocities vanish; the original state returns at $10\mathrm{s}$. Distinguishing absolute and signed beat envelopes, the absolute beat envelope repeats every $10\mathrm{s}$, although its signed sine factor has period $20\mathrm{s}$.