Algebraic Proof that the Amplitude does not affect Frequency

Question

In Simple Harmonic Motion, the Amplitude is defined as the maximum displacement. Consider a pendulum (SHM) or a mass on a horizontal or vertical spiring. If the amplitude increases, which means that the height or starting distance becomes greater, the frequency AND period both do not change. I am looking for an algebraic proof that can show how the amplitude doesn't actually affect the period.

Answer 1

As CyclotomicField proposed in the comments, if you model your wave as a function $f(t)$ a change of amplitude corresponds to the re-scaling by a positive number $a$. This rescaling doesn't affect the zeroes of the function at all, so the distances between the zeroes, i.e. the period, doesn't change either.

Answer 2

For the mass-spring system, the equation of motion reads:

$$ \begin{aligned} 0 &=m\ddot{x}+kx\\ &=\left(D_{t}-j\omega_{n}\right)\left(D_{t}+j\omega_{n}\right)x \end{aligned} $$

where $D_{t}$ is the time differential operator and $\omega_{n}=\sqrt{k/m}$. Simplifying the procedure a little here, we have either

$$ \begin{aligned} 0 &= \left(D_{t}-j\omega_{n}\right)x \implies x=C_{1}\cdot e^{j\omega_{n}t} \\\\ &\text{or} \\\\ 0 &= \left(D_{t}+j\omega_{n}\right)x \implies x=C_{2}\cdot e^{-j\omega_{n}t} \end{aligned} $$

Combining the two homogenous solutions:

$$ x=C_{1}\cdot e^{j\omega_{n}t} + C_{2}\cdot e^{-j\omega_{n}t} $$

The thing to notice here is that any constants $C_{1}$ and $C_{2}$ satisfy the equation of motion, their values only depend on the initial conditions and are independent of the frequency $\omega_{n}$.