 Solution

At fractional position $s$ along a straight rigid rod, the velocity is $(1-s)u+sv$. Uniform mass per unit length givesThe 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 giveThe 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 givesVertical 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 givesThe decoupled $x_1$ equation is ${\ddot{x}}_1+3g{x}_1/(4l)=0$. For the other two coordinates put $\kappa =3g/(2l)$; their equations areA normal-mode ansatz gives $3{\omega }^4-6\kappa {\omega }^2+2{\kappa }^2=0$, whose roots are ${\omega }^2=g(3\pm \sqrt{3})/(2l)$. The corresponding displacement ratios are $y_2/y_1=1\mp \sqrt{3}$. Therefore the three periods are