But,
v2(xv)≠v2(xv√2) in general,
v2cir≠v2shm
so,
vcir≠vshm
However,
specifically there can be v2 at x such that x√2 is the other intersection on the v2 vs x graph along a horizontal v2. In stating,
cos(θ)=1√2
we have also specified the value for v2 for this to occur. So,
v2(xv)=v2(xv√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 vcir and vshm changes along its path θ varies. Because of the sin(θ) and cos(θ) terms, vcir leads vshm. At x=0, the center however, vcir=0, vshm is also zero due to the common factor that they both share, as plotted above. When vcir is maximum at a, vshm is zero due to their phase difference.
Although both vcir and vshm 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.