91Ó°ÊÓ

Given data is$F\left(x\right)=m/x^3,v=0,x=1$, at$t=0$.

The definition of velocity is the rate of change of distance. i.e. $v\left(x\right)=\frac{dx}{dt}$. It is a vector quantity.

From the given information, noticing the given equation:

$m\frac{d^{2}x}{d{t}^2}=F\left(x\right)$ ……. (1)

With denoting differentiation w.r.t the time variable, can be cast into the more convenient form by multiplying it's both sides with$v\left(x\right)=\frac{dx}{dt}$:

$m\frac{d^{2}x}{d{t}^2}\frac{dx}{dt}=F\left(x\right)\frac{dx}{dt}$

This form can be simplified by noting that the first fraction on the LHS of the previous equation is a second time derivative of position, a first-time derivative of velocity, so inserting$\frac{dv}{dt}=\frac{d}{dt}\frac{dx}{dt}=\frac{d^{2}x}{d{t}^2}$into it:

$\frac{dv}{dt}=\frac{d}{dt}\frac{dx}{dt}=\frac{d^{2}x}{d{t}^2}$

The expression by exploiting the definition of velocity$v\left(x\right)≡\frac{dx}{dt}$:

$mv\frac{dv}{dt}\frac{dx}{dt}=F\left(x\right)\frac{dx}{dt}$

After multiplying both sides of the previous equation byrole="math" localid="1664863969787" $dt$to separate the integration integral,

data-custom-editor="chemistry" $\mathrm{mvdv}=\mathrm{F}\left(\mathrm{x}\right)\mathrm{dt}$

Now integrate it by using the table integral$∫xdx=\frac{1}{2}{x}^2+$const. to demonstrate the convenient form of the equation (1):

$\frac{1}{2}m{v}^2\left(x\right)=∫F\left(x\right)dx+\text{const.}$ ……. (2)

By given the function$F\left(x\right)=m/x^3,v=0,x=1$find the term on LHS of the above equation:

$∫F\left(x\right)dx=∫\frac{m}{x^3}dx=-\frac{1}{2}\frac{m}{x^2}+\text{const}$

Since$∫\frac{1}{x^3}dx=-\frac{1}{2}\frac{1}{x^2}+$const. inserting this result into equation (2) and denoting the overall integration constant as $A$,

$\frac{1}{2}m{v}^2=-\frac{1}{2}m{x}^2+A$

The initial condition evaluates the previous equation at time$t=0$to obtain the constant$A$

$$\begin{gathered} \frac{1}{2}m\left[v\right(t=0){]}^2=-\frac{m}{2}\frac{1}{\left[x{(t=0)}^2âˆ\pounds \right.}+A \\ ⇒0-\frac{m}{2}+A \\ A=\frac{m}{2} \end{gathered}$$

Rewrite the equation as:

data-custom-editor="chemistry" $\frac{1}{2}m{v}^2=-\frac{1}{2}\frac{m}{x^2}+\frac{m}{2}$

After diving through by$\frac{m}{2}$as:

$v^2=1-\frac{1}{x^2}$

Rewrite the above expression using$v\left(x\right)=\frac{dx}{dt}$and taking the square root as:

$\frac{dx}{dt}=Â\pm \sqrt{1-\frac{1}{x^2}}$

Separate it and integrate to obtain:

$∫\frac{dx}{\sqrt{1-\frac{1}{x^2}}}=∫Â\pm dt$

To solve the given integral, multiply both the numerator and denominator by$x$

$∫\frac{dx}{\sqrt{1-\frac{1}{x^2}}}=∫\frac{xdx}{\sqrt{x^2-1}}$

The substitution$u=x^2-1⇒du=2xdx$is appropriate for solving the integral, since the factor of$x$is present in the numerator:

$∫\frac{xdx}{\sqrt{x^2-1}}=∫x\frac{du}{2x}\frac{1}{\sqrt{u}}=\frac{1}{2}∫\frac{du}{\sqrt{u}}$

Integrate the above equation using the table integral$∫\frac{dx}{\sqrt{x}}=2\sqrt{x}+$const. to obtain:

$∫\frac{xdx}{\sqrt{x^2-1}}=\frac{1}{2}∫\frac{du}{\sqrt{u}}=\frac{1}{2}2\sqrt{u}+\text{const.}=\sqrt{x^2-1}+\text{const.}$

After successfully integrated both side of the third equation, insert them into said equation:

$\sqrt{x^2-1}=Â\pm t+B$

Here, $B$denoting the integration constant. This constant by exploiting the initial condition $x(t=0)=1$again:

data-custom-editor="chemistry" $\sqrt{{\left[x\right(t=0\left)\right]}^2-1}=0+B⇒B=0$

Now squaring gives the solution of the given function:

$$\begin{gathered} \sqrt{x^2-1}=Â\pm t \\ {x}^2=t^2+1 \\ x=Â\pm \sqrt{1+t^2} \end{gathered}$$

Thus, the given function's solution is $x=Â\pm \sqrt{1+t^2}$.