wo identitical pendula each of length $l$ and with bobs of mass $m$ are free to oscillate in the same plane. The bobs are joined by a spring with spring constant $k$, by looking for solutions where $x$ and $y$ vary harmonically at the same angular frequency $\omega $, form a simultaneous equation for the amplitudes of oscillation ${x}_{0}$ and ${y}_{0}$.

Considering the forces acting on each pendulum we can derive the following coupled-differential equations:

$\begin{array}{}\text{(1)}& m\ddot{x}& =k(y-x)-\frac{mgx}{\ell}\text{(2)}& m\ddot{y}& =-k(y-x)-\frac{mgy}{\ell}\end{array}$

Where $x$ and $y$ are the displacements of each of the pendulum as functions of time. If we assume they oscillate harmonically with angular frequency $\omega $ then we can write $\mathrm{\exists}\omega ,{\varphi}_{1},{\varphi}_{2}\in \mathbb{R}$:

$\begin{array}{rl}x(t)& ={x}_{0}\mathrm{cos}(\omega t+{\varphi}_{1})\\ y(t)& ={y}_{0}\mathrm{cos}(\omega t+{\varphi}_{2})\end{array}$

Substituting these solutions back into $(1)$ and $(2)$ we get:

$\begin{array}{rl}-m{\omega}^{2}{x}_{0}\mathrm{cos}(\omega t+{\varphi}_{1})& =k({x}_{0}\mathrm{cos}(\omega t+{\varphi}_{1})-{y}_{0}\mathrm{cos}(\omega t+{\varphi}_{2}))-\frac{mg}{\ell}\mathrm{cos}(\omega t+{\varphi}_{1})\\ -m{\omega}^{2}{y}_{0}\mathrm{cos}(\omega t+{\varphi}_{2})& =-k({x}_{0}\mathrm{cos}(\omega t+{\varphi}_{1})-{y}_{0}\mathrm{cos}(\omega t+{\varphi}_{2}))-\frac{mg}{\ell}\mathrm{cos}(\omega t+{\varphi}_{2})\end{array}$

However, without assuming that ${\varphi}_{1}={\varphi}_{2}$, in which case everything factors out nicely to leave a simultaneous equation in ${x}_{0}$ and ${y}_{0}$, I cannot see a way of making it linear in ${x}_{0}$ and ${y}_{0}$. So am I expected to use this assumption or is there a mathematical way of simplifying it?

If it is the former, then what would the physical justification for this assumption be?