$t=0$ to the approximate end time $t=t_s$ when $n_{m}^{(n)}(t)$ begins to be stable and $n_{m}^{(s)}(t_s)\rightarrow 0$ (the associated dynamical behaviors at $t_s$ asymptotically approach those at $t=\infty$). A slower process will accumulate a higher noise contribution $n_{m,f}$, which can be the time average of $n_{m}^{(n)}(t)$ ($t>t_s$) for the finally oscillating ones in Fig. 2(b). At the BS coupling resonance $\Delta=\omega_m$, the coefficient of the BS (squeezing) coupling term in Eqs. (2)-(3) takes the form $J \Gamma_{b(s)}(\kappa t)$. The parameter defined as $J=(g/\omega_m)\cdot (E/\kappa)$ decides how fast the cooling of a specific OMS will be; see the illustration in Fig. 2(a). Were there no quantum noises, an uncoupled system with $J=0$ would take the longest time scale $1/\gamma_m$ ($\gamma_m\ll \kappa$) to reach its stability; cf. the solution $\hat{b}(t)=e^{-\gamma_m t}\hat{b}$ to Eq. (3) when $E, \hat{\xi}_m(t)=0$. With the drive intensity $E$ fixed as in Fig. 3(a), a lower sideband resolution $\omega_m/\kappa$ corresponding to a higher $J$ makes the cooling faster. It is also shown in Fig. 2(b) that, once the parameter $J$ is fixed, the OMSs with different sideband resolution $\omega_m/\kappa$ will evolve to the stabilized phase almost together.

Apart from speeding up the cooling of an OMS, increasing $J$ will make 