User:Rogerb/sandbox

We know that $$f = - \eta l \dot{x}$$ and inputs that into the equation below.

$$E = 1/2 m l^2 \omega_0^2 a_0 $$

$$\Delta E = m l^2 \omega_0^2 a_0 \Delta a_0 $$

$$ml^2 \omega^2 a_0 \Delta a_0 = - \pi \eta l^2 a_0^2 \omega $$

$$ \frac{\Delta a_0}{a_0} = - \frac{\eta}{2m} T = - \lambda / 2 T $$,

$$ \eta = \lambda m $$

$$ml^2 \omega^2 a_0 \Delta a_0 = - 4 kla_0$$

$$\delta a_0 = - \frac{2k}{\pi lm \omega} T = - \frac{2 \mu}{\pi \omega} T$$

$$ \ddot{x} + \omega_0^2x + \mu x^3 = F \sin \omega t      $$

$$ \mu = - \frac{\omega_0^2}{6} $$

$$F = \mathcal{F} / ml$$