Regular N-Gon Mass / Spring Oscillation

$\text{In the above image, the green lines represent the springs in the equilibrium position, }$

$\text{and the blue ones - once the entire polygon is stretched out.}$

Let $F(t)$ be the magnitude of the force due to one of the springs on a particle, and let $a(t)$ be the distance a particle is from its equilibrium position.

As we can see from the figure, the total force acting on a particle is $m\frac{d^2a}{dt^2}=2F(t)\cdot \cos(x)$ , where $x=\frac{1}{2}(\pi-\frac{2\pi}{N})$ .

On the other hand, if $l$ is the length of a spring, $F(t)=-k\Delta l=-k\cdot (2a(t)\cdot \cos(x))$ , again evident from the picture.

Thus we have $m\frac{d^2a}{dt^2}=2F(t)\cdot \cos(x)=-4k\cdot a(t)\cdot \cos^2(x)$ .

This is just regular harmonic oscillation:

$a(t)=A\cdot \sin(\omega t)$

$-m\omega^2 A\cdot \sin(\omega t)=-4k\cdot \cos^2(x)\cdot A\cdot \sin(\omega t)$

$\omega=\displaystyle\sqrt\frac{k}{m}\cdot 2\cos(x)$

Substituting $k=Nk_{0}$ and $m=\frac{m_{0}}{N}$ :

$\omega=\displaystyle\sqrt\frac{k_{0}}{m_{0}}\cdot 2N\cos(x)$

Finally, we have $\cos(x)=\cos(\frac{1}{2}(\pi-\frac{2\pi}{N}))=\cos(\frac{\pi}{2}-\frac{\pi}{N})=\sin(\frac{\pi}{N})\approx\frac{\pi}{N}$ for large N, and so

$\omega=\displaystyle\sqrt\frac{k_{0}}{m_{0}}\cdot 2N\frac{\pi}{N}=2\pi\displaystyle\sqrt\frac{k_{0}}{m_{0}}$ , making the answer $2\pi\approx6.283$