= Solution
A <normal mode> of a stable undamped linear <coupled oscillator> has a single <angular frequency> and a fixed ratio of component <displacements>. All components share a sinusoidal time factor, with relative <oscillation phases> of zero or $\pi$ for real <mode shapes>. A general free motion is a <superposition> of the <normal modes>. For this system the symmetric stiffness <matrix> gives real <eigenvalues> and orthogonal <eigenvectors>; positive <spring constants> make the squared <angular frequencies> positive.
Write $m=m_1=m_2>0$ and $\kappa=k_{12}$. Assume ideal massless linear springs, negligible <damping> and positive $k_1,k_2,\kappa$. By <Hooke's law> and <Newton's second law>,
$$
\boxed{m\ddot x_1=-(k_1+\kappa)x_1+\kappa x_2,\qquad
m\ddot x_2=\kappa x_1-(k_2+\kappa)x_2.}
$$
The coupling spring extends by $x_2-x_1$, giving equal and opposite coupling <forces>.
For a <normal mode>, take $\mathbf x=\operatorname{Re}(\mathbf a e^{i\omega t})$. Nonzero <eigenvectors> require
$$
\det\begin{pmatrix}k_1+\kappa-m\omega^2&-\kappa\\-\kappa&k_2+\kappa-m\omega^2\end{pmatrix}=0.
$$
Solving this <quadratic equation> in $m\omega^2$ gives
$$
\boxed{\omega_\pm^2=\frac{k_1+k_2+2\kappa\ \pm\sqrt{(k_1-k_2)^2+4\kappa^2}}{2m}.}
$$
A corresponding <mode shape> has ratio
$$
\frac{a_2}{a_1}=\frac{k_1+\kappa-m\omega_\pm^2}{\kappa},
$$
and the general real solution is $\mathbf x=C_-\mathbf a_-\cos(\omega_-t+\phi_-)+C_+\mathbf a_+\cos(\omega_+t+\phi_+)$. The arbitrary <amplitudes> and <oscillation phases> account for four independent initial data.
\b[Weak coupling alone does not guarantee large beats.] A useful representative sketch assumes nearly equal uncoupled <angular frequencies>. In the resonant case $k_1=k_2=k$ with $\kappa\ll k$, the <mode shapes> are $(1,1)$ and $(1,-1)$, and
$$
\omega_-^2=k/m,\qquad \omega_+^2=(k+2\kappa)/m.
$$
For $x_1(0)=A$, $x_2(0)=\dot x_1(0)=\dot x_2(0)=0$, both <normal modes> have equal coefficients. With $\bar\omega=(\omega_++\omega_-)/2$ and $\Delta\omega=\omega_+-\omega_-$,
$$
\boxed{x_1=A\cos(\bar\omega t)\cos(\Delta\omega t/2),\qquad
x_2=A\sin(\bar\omega t)\sin(\Delta\omega t/2).}
$$
These rapid <oscillations> have slowly varying complementary <wave envelopes>. The initial <displacement> shifts almost entirely to the other mass at $t=\pi/\Delta\omega$, and returns after $2\pi/\Delta\omega$. Arbitrary initial data give a different <superposition>; if $|k_1-k_2|\gg\kappa$, the <normal modes> are localized mainly on individual masses, and strong exchange is suppressed.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2021-coupled-modes.png]
{title=Representative weakly coupled resonant motion: rapid displacements inside slowly exchanging envelopes}
Holding one mass fixed leaves the coupling spring attached and stretched by the other mass. Therefore the two constrained <angular frequencies> are
$$
\boxed{\omega_1=\sqrt{(k_1+\kappa)/m},\qquad \omega_2=\sqrt{(k_2+\kappa)/m}.}
$$
These differ from the <angular frequencies> obtained by physically removing the coupling spring. Ordinary <frequencies> in hertz are each corresponding $\omega/(2\pi)$.
Finally, for $k_1>k_2$ and $\kappa>0$, put $D=\sqrt{(k_1-k_2)^2+4\kappa^2}>k_1-k_2$. Then
$$
\omega_-^2-\omega_2^2=\frac{k_1-k_2-D}{2m}<0,\qquad
\omega_+^2-\omega_1^2=\frac{D-(k_1-k_2)}{2m}>0.
$$
Since $\omega_1>\omega_2$, taking positive square roots proves
$$
\boxed{\omega_-<\omega_2<\omega_1<\omega_+.}
$$
At $\kappa=0$, the outer inequalities become equalities.
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