Sine Waves with Increasing Wavelength With a function such as 
$$y = \frac{1}{8}\sin(\pi x)$$
How can the function be modified so that the wavelength increases by 2 each period?
 A: If
$$
f(x)=-1+\sqrt{1+4x}
$$
then
$$
f\left(n^2+n\right)=2n
$$
Therefore, the "period" of
$$
\frac18\sin(\pi f(x))
$$
increases by $2$ since $\left(n^2+n\right)-\left(n^2-n\right)=2n$:

A: Consider $\sin(\sqrt{x})$.
It has linearly increasing periods, because each intersection with the $x$-axis is a multiple of $n^2π$.
Notice that $π(n+1)^2-πn^2=π(2n+1)$, which is the distance (half-period?) between each intersection.
We have that the difference of the (half-periods?) is a linearly increasing function, so it can be easily modified.
Modifying so that the function increases each period by $2$, and changing amplitude gives 
$\boxed{y=\displaystyle \frac{1}{8}\sin(2π\sqrt{x})}$
A: If the wavelength or time-period increases with time it is no more a harmonic sine-wave.
Differential equation of sine-wave
$$ y^{''}(x)+ (\frac{ 2 \pi}{\lambda})^2\,y(x)=0$$
has a solution 
$$ x = A \sin \frac{ 2 \pi x}{\lambda} $$
We can add wave lengths separately, but not entire time wave-forms as 
$$ \lambda  = \quad  \lambda_1+ \lambda_2 $$

as shown for wavelengths
$$ \lambda_1=2,\quad \lambda_1+ \lambda_2 = 2+1 =3 $$
By above method continuous  expansion of wave-lengths with time  is possible  ($\lambda =1,\, p=0.125$ is chosen here):
$$ y = A \sin \frac{ 2 \pi x}{\lambda (1+ p\,x)} $$ 
You can now choose amplitude and adjust the variable time-period increase rate as desired.
EDIT1;
I am not sure whether the infinite product would converge/ even work
$$ y= x ( 1 - (\frac{x}{\pi}) ^2 )( 1 - (\frac{x/2}{\pi+2}) ^2 ) ( 1 - (\frac{x/3}{\pi+4}) ^2 )( 1 - (\frac{x/4}{\pi+6}) ^2 )...$$ 
