= Solution
For a specified negative-feedback loop with return ratio $L(s)$, let $P$ be its number of open-right-half-plane poles. Let $N_{\mathrm{cw}}$ be the signed clockwise encirclement count of $-1$ by the standard <Nyquist stability criterion> contour. Provided the contour does not hit a singularity or the critical point, the number of right-half-plane closed-loop poles is $Z=P+N_{\mathrm{cw}}$. Thus \b[closed-loop stability requires $N_{\mathrm{cw}}=-P$ and no closed-loop pole on the imaginary axis], with the usual well-posedness and absence of hidden unstable cancellations. For an already stable return ratio, there must be no encirclement of $-1$ and no passage through it.
The open cavity itself, for $\gamma>0$, has state eigenvalue and transfer-function pole $-\gamma/2$. Its initial transient decays and its impulse response is a direct-feedthrough term plus a decaying exponential. Hence \b[the open-loop cavity is asymptotically stable and its transfer function is BIBO stable]. The right-half-plane zero at $+\gamma/2$ does not make the open cavity unstable. On the frequency axis,
$$
G(i\omega)=\frac{\omega^2-(\gamma/2)^2+i\gamma\omega}{\omega^2+(\gamma/2)^2},\qquad |G(i\omega)|=1.
$$
Its unit-circle plot passes through $-1$ at $\omega=0$. A <Nyquist stability criterion> is a feedback criterion; this passage is not by itself a test of open-plant stability.
If the intended interpretation is a unit negative-feedback loop with $L=G$, then
$$
1+G(s)=\frac{2s}{s+\gamma/2},\qquad \frac{G(s)}{1+G(s)}=\frac{s-\gamma/2}{2s}.
$$
That loop has a pole at zero, so \b[unit negative feedback is marginal rather than asymptotically stable], and the critical-point passage correctly prevents a strict Nyquist stability conclusion. Without a specified feedback interconnection, the unambiguous conclusion is the stable open cavity, not that particular closed loop.
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