= Solution
At fractional position $s$ along a straight rigid rod, the velocity is $(1-s)u+sv$. Uniform mass per unit length gives
$$
T=\frac m2\int_0^1|(1-s)u+sv|^2ds
=\boxed{\frac m6(|u|^2+u\cdot v+|v|^2)}.
$$
The string constraints are $x_1^2+y_1^2+(-l+z_1)^2=l^2$ and $x_2^2+y_2^2+(-2l+z_2)^2=4l^2$. Their first-order terms give $z_1=z_2=0$. Their second-order terms then give
$$
\boxed{z_1=(x_1^2+y_1^2)/(2l),\qquad z_2=(x_2^2+y_2^2)/(4l).}
$$
The rod constraint is $(l+x_2-x_1)^2+(y_2-y_1)^2+(-l+z_2-z_1)^2=2l^2$. Its first-order part is $2l(x_2-x_1)=0$, hence $x_2=x_1$ to first order.
The <centre of mass> rises by $(z_1+z_2)/2$. Substituting the quadratic string expansions and the linear relation between the horizontal displacements gives
$$
\boxed{V=\frac{mg}{8l}(3x_1^2+2y_1^2+y_2^2).}
$$
Vertical velocities contribute only at fourth order to the <energy>. Substituting $u=(\dot x_1,\dot y_1,0)$ and $v=(\dot x_1,\dot y_2,0)$ into the rod <energy> gives
$$
\boxed{T=\frac m6(3\dot x_1^2+\dot y_1^2+\dot y_2^2+\dot y_1\dot y_2).}
$$
The decoupled $x_1$ equation is $\ddot x_1+3gx_1/(4l)=0$. For the other two coordinates put $\kappa=3g/(2l)$; their equations are
$$
2\ddot y_1+\ddot y_2+2\kappa y_1=0,\qquad
\ddot y_1+2\ddot y_2+\kappa y_2=0.
$$
A normal-mode ansatz gives $3\omega^4-6\kappa\omega^2+2\kappa^2=0$, whose roots are $\omega^2=g(3\pm\sqrt3)/(2l)$. The corresponding displacement ratios are $y_2/y_1=1\mp\sqrt3$. Therefore the three periods are
$$
\boxed{\frac{2\pi}{\sqrt{3g/(4l)}},\qquad\frac{2\pi}{\sqrt{g(3+\sqrt3)/(2l)}},\qquad\frac{2\pi}{\sqrt{g(3-\sqrt3)/(2l)}}.}
$$
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