= Solution
Write $w=\sqrt{1+2k}$. The <initial conditions> on the <normal mode coordinates> are $u(0)=v(0)=0$ and $\dot u(0)=\dot v(0)=k$. Solving the two <harmonic oscillator equations> and recovering the original <angles> gives the exact solution within the <harmonic approximation>:
$$
\boxed{\theta_1(t)=\frac{k}{2}\left(\sin t+\frac{\sin wt}{w}\right),\qquad
\theta_2(t)=\frac{k}{2}\left(\sin t-\frac{\sin wt}{w}\right).}
$$
Their <angular velocities> are $\dot\theta_1=k(\cos t+\cos wt)/2$ and $\dot\theta_2=k(\cos t-\cos wt)/2$, which verify all four <initial conditions>.
To expose the <beat>, define $\Delta=w-1$ and $\Omega=(w+1)/2$. Replacing $1/w$ by $1$ changes each <angle> by $O(k^2)$ uniformly in time. The sum-to-product identities then give
$$
\begin{aligned}
\theta_1&=k\sin(\Omega t)\cos(\Delta t/2)+O(k^2),&
\dot\theta_1&=k\cos(\Omega t)\cos(\Delta t/2),\\
\theta_2&=-k\cos(\Omega t)\sin(\Delta t/2)+O(k^2),&
\dot\theta_2&=k\sin(\Omega t)\sin(\Delta t/2).
\end{aligned}
$$
The displayed <angular velocities> are exact in the linear system. The <absolute and signed beat envelopes> have leading <amplitudes> $k|\cos(\Delta t/2)|$ and $k|\sin(\Delta t/2)|$. Because $\Delta=k+O(k^2)$, at $\tau=\pi/k$ we have $\Delta\tau/2=\pi/2+O(k)$ and $\Omega\tau=\tau+\pi/2+O(k)$. Hence
$$
\boxed{\begin{aligned}
\theta_1(\tau)&=O(k^2),&\dot\theta_1(\tau)&=O(k^2),\\
\theta_2(\tau)&=k\sin\tau+O(k^2),&\dot\theta_2(\tau)&=k\cos\tau+O(k^2).
\end{aligned}}
$$
The first <simple pendulum> has negligible <oscillation> <amplitude>. Its <kinetic energy> is $T_1(\tau)=O(k^4)$, while the second has $T_2(\tau)=k^2\cos^2\tau/2+O(k^3)$. Thus, to the retained order, all instantaneous <kinetic energy> belongs to the second <simple pendulum>. The fast phase can make that <kinetic energy> small: the whole initial <energy> $k^2/2$ is transferred as the second <simple pendulum>'s combined <kinetic energy> and gravitational <potential energy>, since
$$
\frac12\left(\dot\theta_2(\tau)^2+\theta_2(\tau)^2\right)=\frac{k^2}{2}+O(k^3),
\qquad V_s(\tau)=O(k^3).
$$
This is <energy transfer after an impulse in two weakly coupled oscillators>. It does not require $\cos\tau=1$.
At $2\tau$, $\Delta(2\tau)/2=\pi+O(k)$, so the <beat> has returned the <oscillation> to the first <simple pendulum>:
$$
\boxed{\begin{aligned}
\theta_1(2\tau)&=k\sin(2\tau)+O(k^2),&\dot\theta_1(2\tau)&=k\cos(2\tau)+O(k^2),\\
\theta_2(2\tau)&=O(k^2),&\dot\theta_2(2\tau)&=O(k^2).
\end{aligned}}
$$
The <energy> localization returns, although the fast <oscillation> phase need not equal its initial phase. The exact first minimum of the leading <beat> envelope is at $\pi/\Delta$, whose leading value is $\pi/k$. Keeping the slow phase $\Delta t$ is essential: expanding $\sin(wt)$ at fixed $t$ and then using $t=O(1/k)$ would lose the <energy> exchange. Although <angles> are retained only to order $k$, their leading <energies> are of order $k^2$.
\Image[/past-exam-of-the-classical-mechanics-course-of-the-university-of-oxford/2023-coupled-pendulum-energy-exchange.png]
{title=Pendulum energy exchange after an initial velocity impulse}
{description=The exact linear solution at $k=0.08$, with each pendulum's kinetic plus uncoupled gravitational potential energy divided by the initial total energy. The small remaining fraction is spring potential energy. Vertical lines mark $t=\pi/k$ and $t=2\pi/k$.}
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