\(v^2(x_v)\ne v^2(x_v\sqrt{2})\) in general,
\(v^2_{cir}\ne v^2_{shm}\)
so,
\(v_{cir}\ne v_{shm}\)
However,
specifically there can be \(v^2\) at \(x\) such that \(x\sqrt{2}\) is the other intersection on the \(v^2\) vs \(x\) graph along a horizontal \(v^2\). In stating,
\(cos(\theta)=\cfrac{1}{\sqrt{2}}\)
we have also specified the value for \(v^2\) for this to occur. So,
\(v^2(x_v)=v^2(x_v\sqrt{2})\)
only for a specific value of \(v\) if the solution exist. If the solution does exist then the non emitting/absorbing mode of oscillation exist. Whether the oscillation is sustained is another question.
How to detect such oscillation? If this mode of oscillating leaks in any way, it is then not sustained.
It is even oscillating?!
As \(v_{cir}\) and \(v_{shm}\) changes along its path \(\theta\) varies. Because of the \(sin(\theta)\) and \(cos(\theta)\) terms, \(v_{cir}\) leads \(v_{shm}\). At \(x=0\), the center however, \(v_{cir}=0\), \(v_{shm}\) is also zero due to the common factor that they both share, as plotted above. When \(v_{cir}\) is maximum at \(a\), \(v_{shm}\) is zero due to their phase difference.
Although both \(v_{cir}\) and \(v_{shm}\) share the same graph, they are at 90o phase difference. When one is at the extrema the other is at zero and they are both zero at \(x=0\).
And silent is the night.
