= Solution
\b[Equilibrium and undamped motion]. Resolve <forces> along the incline, taking positive displacement down the slope. At <equilibrium>, <Hooke's law> balances the downhill component of <weight>:
$$
k(l_{\rm eq}-l_0)=mg\sin\theta,
\qquad
\boxed{l_{\rm eq}=l_0+\frac{mg\sin\theta}{k}.}
$$
If $x$ is measured from that equilibrium position, <Newton's second law> becomes
$$
m\ddot x=mg\sin\theta-k(l_{\rm eq}+x-l_0)=-kx.
$$
Thus the motion is a <harmonic oscillator> with
$$
\boxed{\omega_0=\sqrt{\frac km}.}
$$
<Gravity> shifts the equilibrium extension but drops out of this oscillation frequency because the downhill force component is constant.
\b[Viscous damping]. For a <damped harmonic oscillator>, put $\beta=b/(2m)$. Substitution of $z=Ae^{-pt}$ in the supplied <linear ordinary differential equation> gives
$$
p^2-\frac bm p+\frac km=0,
\qquad
p=\beta\pm\sqrt{\beta^2-\omega_0^2}.
$$
In the <underdamped> regime $\beta<\omega_0$, define $\omega_d=\sqrt{\omega_0^2-\beta^2}$. The two <characteristic roots> are $p=\beta\pm i\omega_d$, and the general real <displacement> is
$$
\boxed{x(t)=e^{-\beta t}\left[C\cos(\omega_dt)+D\sin(\omega_dt)\right]
=A_{\rm env}e^{-\beta t}\cos(\omega_dt-\phi).}
$$
Here $C,D$, or equivalently $A_{\rm env}$ and $\phi$, are fixed by the <initial conditions>. For release from rest at $x_0$, the condition $\dot x(0)=0$ requires $D=\beta x_0/\omega_d$, not $D=0$:
$$
\boxed{x(t)=x_0e^{-\beta t}
\left[\cos(\omega_dt)+\frac\beta{\omega_d}\sin(\omega_dt)\right].}
$$
The curve oscillates about zero between exponentially shrinking envelopes $\pm A_{\rm env}e^{-\beta t}$, where $A_{\rm env}=|x_0|\sqrt{1+(\beta/\omega_d)^2}$ for this release. The upper panel illustrates that shape; its damping strength is chosen to make the envelope visible.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019-ia-physics-q7-damped-motion.png]
{title=An underdamped release and the first encounter with the stop for the given parameters}
\b[Frequency change]. <Viscous damping> reduces the <angular frequency> and lengthens the <oscillation period>:
$$
\boxed{\omega_d=\sqrt{\frac km-\frac{b^2}{4m^2}}<\omega_0,\qquad T_d=\frac{2\pi}{\omega_d}.}
$$
For light <damping>, $\omega_d\simeq\omega_0[1-b^2/(8mk)]$, so the frequency shift is second order in $b$, even though the envelope decay rate is first order in $b$.
\b[Will the block reach the stop?] The given parameters give
$$
\beta=0.00500\,\mathrm s^{-1},\qquad
\omega_0=\sqrt{10}=3.16227766\,\mathrm{rad\,s^{-1}},\qquad
\omega_d=3.16227371\,\mathrm{rad\,s^{-1}}.
$$
For $x_0=-0.260\,\mathrm m$, differentiating the release-from-rest solution gives
$$
\dot x(t)=-\frac{x_0\omega_0^2}{\omega_d}
e^{-\beta t}\sin(\omega_dt).
$$
It is positive throughout the first half-cycle. The first positive turning point therefore occurs at $t_+=\pi/\omega_d=0.993460\,\mathrm s$. The <turning points of an underdamped release from rest> give its unobstructed height:
$$
x_+=-x_0e^{-\beta\pi/\omega_d}
=0.260e^{-0.00496730}\,\mathrm m=0.258712\,\mathrm m.
$$
Since $x_+>0.250\,\mathrm m$ and the <displacement> increases continuously up to that point,
$$
\boxed{\text{The block hits the stop on its first downward swing.}}
$$
Solving $x(t)=0.250\,\mathrm m$ gives first contact at approximately $0.91117\,\mathrm s$. The plotted dashed continuation is the hypothetical free motion after that time; the post-impact motion is not determined by the supplied damping model.
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