A particle of mass $m$ capable of moving along the $x$-axis, is acted upon by a force $F(x) = -\frac{k}{x^3}$. At the initial time moment $t=0$, the particle is at the point $x=x_0>0$, and its velocity is $\dot{x}(0)=0$. In what time $\tau$ will the particle reach the point $x=0$?
My attempt:
We basically have to solve this second order nonlinear ODE $$\frac{d^{2}x}{dt^{2}}=-\frac{k}{mx^{3}}.$$
Using substitution $v=\frac{dx}{dt}$, we get $\frac{dv}{dt}=-\frac{k}{mx^3}$, using chain rule to rewrite $\frac{dv}{dt}=v\frac{dv}{dx}$, then separate ant integrate
$$\frac{1}{2}v^2 = \frac{k}{2mx^2} + A \implies v = \sqrt{\frac{k}{mx^2} + A}.$$
The result I got is $v = \sqrt{\frac{k}{mx^2} + A}$, where $A$ is a constant of integration. After plugging in the initial conditions we find the value of $A=-\frac{k}{2mx_0^2}$.
Next, applying the initial conditions and keeping in mind that velocity is negative due to given conditions, $$$$\begin{aligned}
&\int_0^\tau dt=-\int_{x_0}^0\frac{dx}{\sqrt{\frac k{mx^2}-\frac k{mx_0^2}}} \\
&\tau=-\int_{x_0}^0\frac{dx}{\sqrt{\frac k{mx^2}-\frac k{mx_0^2}}}
\end{aligned}
Are there any ways to simplify it further? I suppose we can make use of some integrals with similar form like $\int \frac{dx}{\sqrt{a^2-x^2}}=\arcsin(\frac{x}{a})$
Best Answer
The time taken is given by: $$ \tau = -\int_{x_0}^0\frac{dx}{\sqrt{\frac{k}{mx^2}-\frac{k}{mx_0^2}}} \\ \implies \tau = -\sqrt{\frac{m}{k}}x_0\int_{x_0}^0\frac{xdx}{\sqrt{x_0^2-x^2}}. \\ $$ We have that: $$ \frac{d}{dx}(\sqrt{a^2-x^2})=-\frac{x}{\sqrt{a^2-x^2}}. $$ Can you do the rest?