Two simple pendulums with equal lengths $(L=0\mathrm{.}956m)$ and masses $(m=0\mathrm{.}426kg)$ are separated by a distance $D=0\mathrm{.}43m.$ The masses are connected to

each other by a spring with un$-$compressed length $D$ and spring constant $k=8\mathrm{.}35\frac{N}{m}.$ This is illustrated in the image below, where ${}_1$ measures the angular

displacement of the left pendulum and ${}_2$ measures the angular displacement of the right pendulum.

When the masses are set into simple harmonic motion, the spring and pendulum work as a coupled harmonic oscillator. Solving Newton's Laws for the two

pendulum gives the coupled harmonic oscillator equation:

$\frac{d^2}{d{t}^2}{}_1=-\frac{g}{L}{}_1-\frac{k}{m}({}_1-{}_2)$

$\frac{d^2}{d{t}^2}{}_2=-\frac{g}{L}{}_2-\frac{k}{m}({}_2-{}_1)$

The two natural modes of the system are when the pendulum are moving in the same direction, ${}_a={}_1+{}_2,$ and when they are moving in opposite directions

${}_b={}_1-{}_2$

The two natural $($angular$)$ frequencies of the system are ${}_a$ and ${}_b.$ What is numerical value of the larger $$ for the system?

$s^{-1}$