= Solution
Measure each rod angle from the downward vertical, as in the held diagram. The upper mass rises by $2l(1-\cos\theta_1)$; the lower mass rises by $2l(1-\cos\theta_1)+l(1-\cos\theta_2)$. With the hanging equilibrium as the zero of <gravitational potential energy>,
$$
U=mgl[4(1-\cos\theta_1)+(1-\cos\theta_2)]\simeq\boxed{\frac{mgl}{2}(4\theta_1^2+\theta_2^2).}
$$
The retained expression is quadratic in the small displacements. Both masses contribute to the upper rod's gravitational term.
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