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.
Best Answer
Just to drive home the point, your derivation does not in fact show period depends on amplitude. Let's stick with angular frequency(same deal). What we really have is $\omega(t) = \frac{\sin^{-1}(\frac{x(t)}{A})}{t} $. But now, how does x depend on t well, we then have $$\omega(t) = \frac{\sin^{-1}(\frac{A sin(\omega t )}{A})}{t} = \frac{\sin^{-1}(sin(\omega t))}{t} =\omega $$ In other words your derivation is a complete tautology