Why $\omega$ in $x = \cos(\omega t + \alpha)$ , $\omega$ isn't considered an arbitrary constant?

Question

We know the SHM differential equation is of second-order $$\dfrac{d^2 x}{dt^2} = -{\omega}^2 x$$ . So, the solution of this equation must contain two arbitrary constants. And also we know that $x = \cos(\omega t + \alpha)$ is readily the solution of the above equation as it contains two arbitrary constants : $A(\text{amplitude}) \quad \& \quad \alpha(\text{phase constant})$. But isn't $\omega$ also a constant? Why isn't it counted as another parameter or arbitrary constant?

Answer 1

It's not arbitrary - its value comes directly from the defining differential equation. A function (for example) $x(t)=A\cos(\omega't+\alpha)$ with $\omega'\neq\omega$ is not a solution to the differential equation, so $\omega$ is not arbitrary.

Answer 2

Because it is already given, it is not arbitrary.

Answer 3

Let us look at a simpler example:

When you have the equation:

$2 x + y + z = 5$.

Then the set of solutions (over the reals) is $S_5=\{( (5 - s - t)/2 ,s,t ) \colon s ,\, t \in \mathbb{R}\}$.

The set of all solutions is parametrized by two real parameters, or you have two constants to choose.

Now, if you modify this to $2 x + y + z = a$ where $a$ is some constant.

Then the set of all solutions of the equation given some $a$, for example $a=5$ is $S_a=\{( (a - s - t)/2 ,s,t ) \colon s ,\, t \in \mathbb{R}\}$.

There role that $a$ and $s,t$ play is different. The former is part of what is given, and not a free parameter. It can still be any real number, so in that sense one could say it is arbitrary, but the situation is:

Let us pick some $a$ and fix it. Then, let us look at the equation for this fixed $a$.