What will the equation of a sine wave having the axis as a parabola $y= ax^2$ be?

Question

approximate representation of my curve(in red)

Answer 1

$\newcommand{\vect}[1]{{\bf #1}}$ Define a point on the parabola through a parametric equation

$$ \vect{x}(t) = {t \choose at^2} $$

The tangent vector to the parabola is

$$ \vect{v}(t) = \frac{{\rm d}}{{\rm d}t}\vect{x}(t) = {1 \choose 2at} $$

and a perpendicular vector can be constructed from this

$$ \vect{u}(t) = {2at \choose -1} $$

moreover, a unitary vector perpendicular to the parabola is

$$ \hat{\vect{u}}(t) = \frac{1}{\sqrt{1 + 4a^2 t^2}} {2at \choose -1} $$

The point on a sine function defined over the parabola can be thus obtained as

$$ \vect{x}'(t) = \vect{x}(t) + \frac{\color{red}{A\cos B \sigma(t)}}{\sqrt{1 + 4a^2 t^2}} {2at \choose -1} $$

Below there's an example for $A = 0.2$, $B = 6$ and $a = 1$

In this last expression you could use $\sigma(t) = t$ or $\sigma(t) = s(t)$, where $s(t)$ is the arg-length along the parabola (following N74's suggestion)

$$ s(t) = \int_0^t{\rm d}t~ |\vect{v}(t)| = \frac{t}{2}\sqrt{1 + 4a^2t^2} + \frac{1}{4a}\sinh^{-1}(2at) $$

Answer 2

Note that what you have drawn is not a function as it fails the vertical line test. I know it is just a sketch but the problem is real if you try to bend the base line of the sine wave to trace the parabola. It is easy to write $y=x^2+100 \sin x$ and get the below. I picked the factor $100$ to make the sine waves visible. This doesn't capture the fact that you want the amplitude of the sine wave to be perpendicular to the parabola. If you don't mind failing the vertical line test, you can write the parabola as $y=x^2, y'=2x$ and note the arc length to a point $x$ is $s=\int_0^t\sqrt{1+y'^2}dx=\frac 14(2x\sqrt{1+4x^2}-\sinh^{-1}(2x))$. Now you can take the sine of this and add a vector perpendicular to the parabola of length the sine wave you want.

Answer 3

If we apply a transformation which maps the $x-$ axis on to the parabola by rotating points of similar distance from the origin, we get a fairly even sinuisoidal wave about the parabola.

The parametric equation is:

$$\left(t+\frac{at\cos \big(mt\sqrt{1+a^2t^2}\big)}{\sqrt{1+a^2t^2}},at^2-\frac{\cos\big(mt\sqrt{1+a^2t^2}\big)}{\sqrt{1+a^2t^2}}\right)$$ where $m$ controls the sinusoidal frequency.

See desmos implementation here.