How should I find the following period ## T ## of this solution?

[Math100]

Homework Statement

Consider the differential equation $\ddot{x}+x+\frac{3}{2}\beta x\lvert x\rvert=0$
for $\beta\geq 0$ with the solution $x(t)$ and initial conditions
$x(0)=A>0, \dot{x}(0)=0$. Show that the period $T$ of this solution
is given by $T=4\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1+\beta A\sec^2\theta(1-\sin^3\theta)}}$.

Relevant Equations

None.

From the original equation of $\ddot{x}+x+\frac{3}{2}\beta x\lvert x\rvert=0$,
I've got $\ddot{x}\dot{x}+x\dot{x}+\frac{3}{2}\beta x\lvert x\rvert\dot{x}=0 \implies \frac{d}{dt}(\frac{1}{2}\dot{x}^2+\frac{1}{2}x^2+\frac{\beta}{2}x^2\lvert x\rvert)=0 \implies \frac{1}{2}\dot{x}^2+\frac{1}{2}x^2+\frac{\beta}{2}x^2\lvert x\rvert=C$ where
$C$ is the constant.

From the work shown above, how should I use the initial conditions of $x(0)=A>0$
and $\dot{x}(0)=0$ to find the given period $T$? And where does the factor of
$4$ come from in front of the integral sign of the period $T$?

 

[pasmith]

Set y = \dot x and switch to scaled polar coordinates (x,y) = (Ar \cos \theta, Ar \sin \theta). The period is then given by <br /> T = \int_0^{2\pi} \frac{1}{\dot \theta}\,d\theta. In this case, by symmetry the integral over the full period from 0 to 2\pi is four times the integral over the quarter period from 0 to \pi/2.

 

[Math100]

Set y = \dot x and switch to scaled polar coordinates (x,y) = (Ar \cos \theta, Ar \sin \theta). The period is then given by <br /> T = \int_0^{2\pi} \frac{1}{\dot \theta}\,d\theta. In this case, by symmetry the integral over the full period from 0 to 2\pi is four times the integral over the quarter period from 0 to \pi/2.

But what's $\dot\theta$ from the integral?

 

[pasmith]

My earlier hint was unhelpful; please disregard it.

Your first integral gives you an expression for |\dot x| in terms of x, <br /> |\dot x| = F(x) where F is even.

You can therefore start from <br /> T = \int_0^T\,dt = \int_A^0 -F(x)^{-1} \,dx + \int_0^{-A} -F(x)^{-1}\,dx + \int_{-A}^0 F(x)^{-1}\,dx + \int_0^A F(x)^{-1}\,dx.

 

[Gavran]

But what's $\dot\theta$ from the integral?

You can find $\dot \theta$ by using the substitution $x=A\sin\theta$, $\dot x=A\cos\theta\dot \theta$.

You do not need to discard the whole post #2. Use the next part

The period is then given by <br /> T = \int_0^{2\pi} \frac{1}{\dot \theta}\,d\theta. In this case, by symmetry the integral over the full period from 0 to 2\pi is four times the integral over the quarter period from 0 to \pi/2.

to find the expression for $T$.

 

[Math100]

You can find $\dot \theta$ by using the substitution $x=A\sin\theta$, $\dot x=A\cos\theta\dot \theta$.

You do not need to discard the whole post #2. Use the next part

to find the expression for $T$.

Thank you!