What's the period of $\sin \left(x^\frac{3}{2}\right)$? 
How can one tell the period of $\sin \left(x^\frac{3}{2}\right)$? Is it $(2\pi)^\frac{2}{3}$?

 A: In physics we say such waveforms are "chirping" (although, technically, this name is given to quadratic arguments - but this is close enough).  There is no one period here, but a rate of change of a frequency.  We may speak of an average frequency over an interval, an instantaneous frequency, etc.  But asking for "the frequency" of this sine is like asking for "the value" of $y=x^{3/2}$.
A: The solutions of $$\sin((x+T)^{3/2})=\sin(x^{3/2})$$ are
$$(x+T)^{3/2}=x^{3/2}+2m\pi.$$
and from there,
$$T=(x^{3/2}+2m\pi)^{2/3}-x$$
which proves that there is no constant $T$ that works.
(The same computation with $\sin(x)$ gives $T=(x+2m\pi)-x$.)
A: Leaving aside the fact that the function is only defined for $x\ge0$, you can easily see that
$$
f'(x)=\frac{3}{2}\sqrt{x}\cos(x^{3/2})
$$
If $T>0$ is a period, we should have $f(T)=f(0)=0$ and also $f'(T)=f'(0)=0$. Then we have
$$
\sin(T^{3/2})=\cos(T^{3/2})=0
$$
which is a contradiction.
A: As people have pointed out, this function isn't periodic.  But you might be thinking of something like "what's the distance between successive maxima of this function"?  $\sin(x)$ has maxima at $\pi/2, 5\pi/2, 9\pi/2, \cdots$ -- in general $(4n+1)\pi/2$ for integer $n$.  So $\sin(x^{3/2})$ has maxima at $((4n+1)\pi/2)^{2/3}$ - call this $f(n)$.  Then the distance between the $n$th peak and the $(n+1)$st peak is 
$$f(n+1) - f(n) = ((4n+5)^{2/3} - (4n+1)^{2/3}) (\pi/2)^{2/3}) = ((n+5/4)^{2/3} - (n+1/4)^{2/3}) (2\pi)^{2/3}$$
and you can approximate that difference by a derivative to get
$$(2/3) n^{-1/3} (2\pi)^{2/3} $$
So the spacing between peaks near $f(n) \approx (2\pi)^{2/3} n^{2/3}$ is approximately  $(2/3) (2\pi)^{2/3} n^{-1/3}$.    Let $A = (2\pi)^{2/3}, B = (2/3) (2\pi)^{2/3}$.  Then the spacing between peaks near $An^{2/3}$ is approximately $Bn^{-1/3}$.  Let $x = An^{2/3}$ and so $n = (x/A)^{3/2}$; the spacing of peaks near $x$ is then approximately 
$$B((x/A)^{3/2})^{-1/3} = B(x/A)^{-1/2} = B A^{1/2} / \sqrt{x} = (4\pi/3) / \sqrt{x}$$
For example, there are two successive peaks at $(401\pi/2)^{2/3} \approx 73.48119$ and $(405\pi/2)^{2/3} \approx 73.96903$ - the distance between these is about $0.48784$. In comparison, the distance between peaks around $73.7$ (halfway between these) is predicted to be $(4\pi/3)/\sqrt{73.5} \approx 0.48793$.
So in a sense the period is approximately $(4\pi/3)/\sqrt{x}$.  This depends on $x$, so of course the function isn't periodic as usually defined. 
