Horizontal oscillations of a spring-mass system

Consider a block of mass $m$ attached to one end of a massless spring of stiffness (force/spring) constant $k$, the other end fixed, resting on a smooth (frictionless) horizontal surface. Let $x_0$ be the mean/equilibrium position of the mass when undisturbed. If the mass is pulled through a small displacement $x$ from $x_0$ and released, the stretched (or compressed) spring exerts a restoring force $F=-kx$ proportional to the displacement, so the block oscillates back and forth about $x_0$. Newton's second law gives $m\ddot x=-kx$, so, exactly as for the general linear oscillator, $\omega=\sqrt{k/m}$, the frequency is $f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$ Hz, and the time period is $T=2\pi\sqrt{\dfrac{m}{k}}$ seconds. A useful way to remember this formula is $T=2\pi\sqrt{\dfrac{\text{inertial property}}{\text{elastic property}}}$, since mass $m$ represents the system's inertia and the spring constant $k$ its elastic stiffness; equivalently $\left|\dfrac{x}{\ddot x}\right|=\dfrac{m}{k}$, so $T=2\pi\sqrt{m/k}$ follows directly from the ratio of dis …