Period is independent of amplitude. (Vias.org)

But given that,

Simple harmonic motion can be defined by

$$x = A * \sin(\omega t) \tag{1}$$ where $A$ is the amplitude of oscillation, $\omega$ the angular velocity, $t$ the time, and $x$ the displacement from the mean position

And,

$$T = 2 \pi/\omega = 1/f \tag{2}$$ where, $T$ is the period of motion and $f$ is the frequency of oscillation.

Equation (1) can be rearranged to give \begin{align*} x &= A * \sin (\omega t)\\ \frac{x}{A} &= \sin(\omega ωt) \\ \arcsin\left(\frac{x}{A}\right) &= \omega t\\ \omega &=\frac{\arcsin\left(\frac{x}{A}\right)}{t} \end{align*}

Subbing this into (2) gives the following relationship between $T$ and $A$

$$T = \frac{2 \pi t}{\arcsin(x/A)} = \frac{1}{f}$$

Doesn't the fact that both $T$ and $A$ appear in the above equation show that $T$(period) is dependent on $A$(amplitude) ?

FYI:

Although a physical explanation may be useful, I am particularity interested in why deriving a relationship between $T$ and $A$ doesn't mean that they are dependent. Note there is a similar question here but that is concerned with the physics of the phenomena, and not why the maths can't be used to solve it.

This is because I have this exam where we are given a stimulus and based on that stimulus alone are meant to answer questions (i.e. the exam is expected to contain material/principles that we haven't been exposed to before, but should be able to answer given the stimulus). And as I didn't know much about simple harmonic motion, my initial reaction was to see if the formulas link $T$ and $A$.

The practice question (which I have typed out the important bits of above is) The answer is D.