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I have been given the following question and I am unsure if I am missing an assumption or if I am misunderstanding something else:

Two identitical pendula each of length $\ell$ 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{align} m\ddot{x}&=k(y-x)-\frac{mgx}{\ell} \tag{1}\\ m\ddot{y}&=-k(y-x)-\frac{mgy}{\ell} \tag{2} \end{align}

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 $\exists \omega, \phi_{1},\phi_{2} \in \mathbb{R}$:

\begin{align} x(t) &= x_{0}\cos(\omega t + \phi_{1}) \\ y(t) &= y_{0}\cos(\omega t + \phi_{2}) \end{align}

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

\begin{align} -m\omega^{2}x_{0}\cos(\omega t + \phi_{1})&=k(x_{0}\cos(\omega t + \phi_{1}) - y_{0}\cos(\omega t + \phi_{2})) - \frac{mg}{\ell}\cos(\omega t + \phi_{1}) \\ -m\omega^{2}y_{0}\cos(\omega t + \phi_{2}) &= -k(x_{0}\cos(\omega t + \phi_{1}) -y_{0}\cos(\omega t + \phi_{2}))-\frac{mg}{\ell}\cos(\omega t + \phi_{2}) \end{align}

However, without assuming that $\phi_{1}=\phi_{2}$, 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 (despite it not being mentioned) or is there a mathematical way of simplifying it?

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

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If you add the two equations the terms in $k$ disappear and the terms in $\cos$ distribute. You are left for an equation in $x+y$. Then subtract the two equations and again the cosine terms distribute, leaving an equation in $x-y$ with a different frequency. You never need to assume $\phi_1=\phi_2$ In fact, you can set one to zero, which just says you start your timer when that pendulum is at angle zero. You will eventually discover that the other is zero as well.

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Thank you Ross; I should probably have clarified, I am aware of solving coupled ODEs by using normal modes (in this case $q_{1}=\frac{1}{\sqrt{2}}(x+y)$, $q_{2}=\frac{1}{\sqrt{2}}(x-y)$), but the question seems to imply that it wants a simultaneous equation in $x_0$ and $y_0$ instead without using a change of variable? Or have I just read something else into it? – Shaktal Feb 12 '14 at 17:05
The equations are already linear in $x_0,y_0$. The $\cos$ terms would be considered constants. It is easiest if you solve the first for $y_0\cos(\omega t+\phi_2)$, then plug that into the second and get an equation in $x_0$ which you can solve. – Ross Millikan Feb 12 '14 at 17:13
Ahh I see now; thank you!! – Shaktal Feb 12 '14 at 17:14

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