In the coupled pendulums of Figure $4.3$ let us write the modulated frequency $\omega_{m}=\left(\omega_{2}-\omega_{1}\right) / 2$ and the average frequency $\omega_{a}=\left(\omega_{2}+\omega_{1}\right) / 2$ and assume that the spring is so weak that it stores a negligible amount of energy. Let the modulated amplitude
$$
2 a \cos \omega_{m} t \quad \text { or } \quad 2 a \sin \omega_{m} t
$$
be constant over one cycle at the average frequency $\omega_{a}$ to show that the energies of the masses may be written
$$
E_{x}=2 m a^{2} \omega_{a}^{2} \cos ^{2} \omega_{m} t
$$
and
$$
E_{y}=2 m a^{2} \omega_{a}^{2} \sin ^{2} \omega_{m} t
$$
Show that the total energy $E$ remains constant and that the energy difference at any time is
$$
E_{x}-E_{y}=E \cos \left(\omega_{2}-\omega_{1}\right) t
$$
Prove that
$$
E_{x}=\frac{E}{2}\left[1+\cos \left(\omega_{2}-\omega_{1}\right) t\right]
$$
and
$$
E_{y}=\frac{E}{2}\left[1-\cos \left(\omega_{2}-\omega_{1}\right) t\right]
$$
to show that the constant total energy is completely exchanged between the two pendulums at the beat frequency $\left(\omega_{2}-\omega_{1}\right)$.