Energy, Amplitude, and Frequency in an Ideal Mass–Spring Oscillator

Begin with Hooke's law

An ideal horizontal mass–spring system obeys Hooke's law,

$$F_x=-kx.$$

The minus sign means the force points toward equilibrium. Combining this force with Newton's second law gives simple harmonic motion with angular frequency

$$\omega=\sqrt{\frac{k}{m}}, \qquad f=\frac{\omega}{2\pi}, \qquad T=\frac{2\pi}{\omega}.$$

Frequency depends on $k$ and $m$, not on amplitude, as long as the spring remains linear and damping is negligible.

Track energy through one cycle

The total mechanical energy is

$$E=\frac12mv^2+\frac12kx^2=\frac12kA^2.$$

At either turning point, $x=\pm A$ and $v=0$, so all energy is elastic potential energy. At equilibrium, $x=0$, so all energy is kinetic and speed is greatest. This gives

$$v_{\max}=A\omega, \qquad a_{\max}=A\omega^2=\frac{kA}{m}.$$

Acceleration is largest at the turning points even though speed is zero there. Speed is largest at equilibrium even though acceleration is zero there. Keeping those locations separate prevents a common conceptual error.

Work a different example

Take $m=2.0\,\mathrm{kg}$, $k=50\,\mathrm{N/m}$, and amplitude $A=0.20\,\mathrm{m}$. Then

$$\omega=\sqrt{\frac{50}{2.0}}=5.0\,\mathrm{rad/s},$$ $$f=\frac{5.0}{2\pi}=0.796\,\mathrm{Hz},$$

and

$$E=\frac12(50)(0.20)^2=1.0\,\mathrm{J}.$$

The maximum speed is $A\omega=1.0\,\mathrm{m/s}$, and the maximum acceleration is $A\omega^2=5.0\,\mathrm{m/s^2}$.

The ideal model stops being exact when damping removes energy, the spring's force is not proportional to displacement, the spring has significant mass, or motion leaves one dimension.