Complex Harmonic Motion!

First, consider the positive half of the harmonic oscillation.

We have,

$F = -kx^3 \\ \implies ma = -kx^3 \\ \implies m v \dfrac{dv}{dx} = -kx^3 \\ \implies \displaystyle m \int_0^v v \,dv = -k \int_A^x x^3 \,dx \\ \implies \dfrac{v^2}{2} = \dfrac{k}{4m} \sqrt{A^4 - x^4} \quad {\color{#3D99F6} (*)}\\ \implies v = \sqrt{\dfrac{k}{2m} (A^4 - x^4)} \\ \implies \dfrac{dx}{dt} = \sqrt{\dfrac{k}{2m} (A^4 - x^4)} \\ \implies \displaystyle \int_0^t \,dt = \sqrt{\dfrac{2m}{k}} \int_0^A \dfrac{dx}{\sqrt{A^4-x^4}} \\ \implies \displaystyle t = \sqrt{\dfrac{2m}{k}} \int_0^A \dfrac{dx}{\sqrt{A^4-x^4}}$

The value of $t$ is equal to half time-period of the oscillation on the positive side of the x-axis. The time period for positive half oscillation is thus

$T_1 = 2 \sqrt{\dfrac{2m}{k}} \int_0^A \dfrac{dx}{\sqrt{A^4-x^4}}$

Using ${\color{#3D99F6} (*)}$ , at $x=0$ , we get $\dfrac{v^2}{2} = \dfrac{kA^4}{4m}$ .

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Now, consider the negative half of the harmonic oscillation.

We have,

$F = -4kx^5 \\ \implies ma = -4kx^5 \\ \implies m v \dfrac{dv}{dx} = -4kx^5 \\ \implies \displaystyle m \int_v^0 v \,dv = -4k \int_x^p x^5 \,dx \\ \implies \dfrac{v^2}{2} = \dfrac{2k}{3m} (p^6 - x^6) \quad {\color{#3D99F6} (**)}\\ \implies v = \sqrt{\dfrac{4k}{3m} (p^6 - x^6)} \\ \implies \dfrac{dx}{dt} = \sqrt{\dfrac{4k}{3m} (p^6 - x^6)} \\ \implies \displaystyle \int_0^t \,dt = \sqrt{\dfrac{3m}{4k}} \int_{-p}^0 \dfrac{dx}{\sqrt{p^6-x^6}} \\ \implies \displaystyle t = \sqrt{\dfrac{3m}{4k}} \int_{-p}^0 \dfrac{dx}{\sqrt{p^6-x^6}}$

The value of $t$ is equal to half time-period of the oscillation on the negative side of the x-axis. The time period for negative half oscillation is thus

$T_2 = 2 \sqrt{\dfrac{3m}{4k}} \int_{-p}^0 \dfrac{dx}{\sqrt{p^6-x^6}} = \sqrt{\dfrac{3m}{k}} \int_{-p}^0 \dfrac{dx}{\sqrt{p^6-x^6}}$

Here $p$ denotes the amplitude of oscillation on the negative side of the x-axis. Using ${\color{#3D99F6} (*)}$ and ${\color{#3D99F6} (**)}$ , at $x=0$ , we get

$\dfrac{v^2}{2} = \dfrac{kA^4}{4m} = \dfrac{2k}{3m} p^6 \\ \implies p = \sqrt[6]{\dfrac{3}{8}} A^{{2}/{3}}$

Hence

$T_2 = \sqrt{\dfrac{3m}{k}} \int_{-\sqrt[6]{\frac{3}{8}} A^{{2}/{3}}}^0 \dfrac{dx}{\sqrt{\frac{3}{8} A^4 -x^6}}$

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The total time period of oscillations is thus

$T = T_1 + T_2 = 2 \sqrt{\dfrac{2m}{k}} \int_0^A \dfrac{dx}{\sqrt{A^4-x^4}} + \sqrt{\dfrac{3m}{k}} \int_{-\sqrt[6]{\frac{3}{8}} A^{{2}/{3}}}^0 \dfrac{dx}{\sqrt{\frac{3}{8} A^4 -x^6}}$

Comparing required values, we get

$a+b+c+e+f+g+j+n+q = 2+2+4+3+3+8+3+8+6 = \boxed{39}$ .