Not-So-Simple Harmonic Motion

_Note: _ The integral you have shown is incorrect , you have reversed the limits.

Solution :

$F =-kx^3$

$\Rightarrow a = \frac{-k}{m} x^3$

$\Rightarrow \frac{v \text{dv}}{\text{dx}} = \frac{-k}{m} x^3$

$\Rightarrow \displaystyle \int_{0}^{v} v \text{dv} = \int_{{10}^{-1}}^{x} \frac{-k}{m} x^3 \text{dx}$

$\Rightarrow \frac{v^2}{2} = \frac{k}{4m} ({10}^{-4} - x^4)$ .

Let us consider the motion from $+0.1 m$ to $-0.1 m$ , of course velocity is negative,

Say the time period is $T$ , then the time taken in the above motion is clearly $\frac{T}{2}$ .

$\Rightarrow v = - \sqrt{\frac{k}{2m}} \sqrt{{(10}^{-4} - x^4)}$

$\Rightarrow \frac{\text{dx}}{\text{dt}} = - \sqrt{\frac{k}{2m}} \sqrt{({10}^{-4} - x^4)}$

$\Rightarrow \displaystyle \int_{0}^{\frac{T}{2}} \text{dt} = -\sqrt{\frac{2m}{k}} \int_{{10}^{-1}}^{-{10}^{-1}} \frac{\text{dx}}{\sqrt{{10}^{-4} - x^4}}$

Substitute $t = -10x$ ,

$\Rightarrow \frac{T}{2} = \displaystyle \sqrt{\frac{2m}{k}}\int_{-1}^{1} \frac{10}{\sqrt{1 - t^4}} \text{dt}$

Hence, $T = 20 \sqrt{\frac{2m}{k}} \times 2.622 \approx 2.345 s$