What is the period of the harmonic oscillator?

Question

I am trying to find the elapsed time $T$ (or transit time) over one cycle of the harmonic oscillator

$$\ddot{x} + \omega^2x=0.$$

I worked this out to be \begin{align} T &= \int_{R}^{-R} \frac{\mathrm{d}x}{-y} + \int_{-R}^{R} \frac{\mathrm{d}x}{y}\\ &= \int_{R}^{-R} \frac{\mathrm{d}x}{-\sqrt{R^2-\omega^2x^2}} + \int_{-R}^{R} \frac{\mathrm{d}x}{\sqrt{R^2-\omega^2x^2}}\\ &= -\frac{1}{\omega} \int_{R}^{-R} \frac{\mathrm{d}x}{\sqrt{D^2-x^2}} + \frac{1}{\omega} \int_{-R}^{R} \frac{\mathrm{d}x}{\sqrt{D^2-x^2}} \end{align}

where $D=R/\omega$.

When I evaluate the definite integral, I get $T=4\arcsin(\omega)/\omega $, which makes no sense to me because $\arcsin(\omega)$ bounds $\omega$ at unity. Since $\omega$ is the angular frequency, the answer should be $T=2\pi/\omega$.

Could someone point out where I went so wrong?

Answer 1

Your setup is entirely correct, but your evaluation of the integral is not. The primitive of $\frac{1}{\sqrt{D^2 -x^2}}$ is $\text{arctan} \frac{x}{\sqrt{D^2 -x^2}}$; taking the appropriate limits yields the desired factor $\pi$, for each integral.

Edit: As @MichaelSeifert rightly pointed out, the integral can also be expressed in terms of $\text{arcsin}$, since \begin{equation} \text{arctan} \frac{x}{\sqrt{D^2 -x^2}} = \text{arcsin} \frac{x}{D}. \end{equation} As $\text{arcsin} 1 = \frac{\pi}{2}$, you get the requested answer.

Answer 2

I'm assuming you integrate once to find the first integral $$ E(x,y)=\frac12(y^2+ω^2x^2) $$ and that the points of largest amplitude are $(x,y)=(\pm R,0)$. But then the constant is $$ E=\frac12ω^2R^2 $$ and thus the equation for $y$ is $$ y=\pmω\sqrt{R^2-x^2} $$

Now use the substitution $x=R\sin u$ to solve the resulting integral for the half-period, $$ \frac T2=\int_{-R}^R\frac{dx}y=\frac1ω\int_{-\pi/2}^{\pi/2}du=\frac{\pi}{ω} $$