Simple harmonic motion : why is the period independent of amplitude even when angular velocity is related to the amplitude?

Question

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) ?

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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.

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Answer 1

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

Answer 2

It's because $x$ depends on $t$ in such a way that there is no dependence on $A$ left in the expression. $A$ and $\omega$ are constants that don't depend on time, and $x$ is a function of time; it's the position at time $t$, usually written $x(t)$.

When the force isn't Hooke's law, then you can get a relationship between $A$ and $\omega$. The reason for this is because the energy of a simple harmonic oscillator is $$ E = \frac{1}{2} m [v(t)]^2 + \frac{1}{2} k [x(t)]^2, $$ that is the equation for an ellipse in $x$-$v$ space no matter how big the amplitude/energy is. The simple harmonic osillator just rotates around that ellipse with a constant angular frequency, $\omega$. If I were to change the potential energy term to, say: $$ E = \frac{1}{2} m [v(t)]^2 + \frac{1}{4} k [x(t)]^4, $$ then the shape with a constant energy is no longer an ellipse, and the shape changes as the energy changes in such a way that amplitude and period become related.

For another example, a look at the orbits of planets and Kepler's third law — there amplitude along the major axis and period are related by: $$\frac{T^2}{A^3} = \frac{4\pi^2}{G(M+m)}$$