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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:If $x$ is measured from that equilibrium position, Newton's second law becomesThus the motion is a harmonic oscillator withGravity shifts the equilibrium extension but drops out of this oscillation frequency because the downhill force component is constant.

Viscous damping. For a damped harmonic oscillator, put $\beta =b/(2m)$. Substitution of $z=A{e}^{-pt}$ in the supplied linear ordinary differential equation givesIn 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 isHere $C,D$, or equivalently $A_{\mathrm{env}}$ and $\varphi$, 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 $D=0$:The curve oscillates about zero between exponentially shrinking envelopes $\pm A_{\mathrm{env}}{e}^{-\beta t}$, where $A_{\mathrm{env}}=|x_0|\sqrt{1+(\beta /{\omega }_d{)}^2}$ for this release. The upper panel illustrates that shape; its damping strength is chosen to make the envelope visible.

Frequency change. Viscous damping reduces the angular frequency and lengthens the oscillation period: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 $b$.

Will the block reach the stop? The given parameters giveFor $x_0=-0.260\mathrm{m}$, differentiating the release-from-rest solution givesIt 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:Since $x_{+}>0.250\mathrm{m}$ and the displacement increases continuously up to that point,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.