Energy in oscillatory motion

A mass $m$ on a spring of constant $k$ moves as $x(t)=A\cos(\omega_0t+\phi)$, with $\omega_0=\sqrt{k/m}$ fixed by the physical setup and $A,\phi$ fixed by how the motion was started. Differentiating gives the velocity $\dot x(t)=-A\omega_0\sin(\omega_0t+\phi)$, and from that the kinetic energy is

where: $m$ is the oscillating mass, $A$ the amplitude, $\omega_0=\sqrt{k/m}$ the angular frequency, and $\phi$ the phase constant fixed by initial conditions. The potential energy stored in the spring comes from $U=\tfrac12 kx^2$, valid because the restoring force is linear, $F=-kx=-dU/dx$:

Since $k=m\omega_0^2$, both expressions share the same prefactor $\tfrac12 kA^2$, so $K$ and $U$ are just that constant amount split between $\sin^2$ and $\cos^2$. Adding them and using $\sin^2\theta+\cos^2\theta=1$ removes the time dependence entirely — the two terms are always complementary shares of one fixed total. That total,

is the headline result: the total mechanical energy of an undamped oscillator is fixed the instant the amplitude is fixed, and it never depends on time, phase, or where in the cycle you happen to look. Two consequences are worth carrying forward. First, $E\propto A^2$ — doubling the amplitude quadruples the energy, the same scaling that reappears later as "intensity is proportional to amplitude squared" for waves. Second, $K$ and $U$ each oscillate at twice the frequency of $x(t)$ itself, because $\sin^2\theta$ and $\cos^2\theta$ complete two full wobbles for every one cycle of $\cos\theta$. $K$ peaks (equal to $E$) at the equilibrium position $x=0$, where the mass moves fastest; $U$ peaks (equal to $E$) at the turning points $x=\pm A$, where the mass is momentarily at rest.

There is one more useful number hiding in these formulas: the time-averaged kinetic and potential energies over a full cycle. Because $\sin^2\theta$ and $\cos^2\theta$ each average to $\tfrac12$ over any whole number of periods, $\langle K\rangle=\langle U\rangle=\tfrac12E=\tfrac14kA^2$. So although $K$ and $U$ individually swing all the way from $0$ up to $E$ and back every half-period, on average each carries exactly half of the total energy — a distinction worth keeping straight when a question asks for an average value rather than the value at one particular instant.

Problem. A $0.20$ kg mass on a spring with $k=80$ N/m oscillates with amplitude $A=0.10$ m. Find (a) the total energy, (b) the speed at $x=0.06$ m, (c) confirm $K+U=E$ there, and (d) the position where $K=U$.

Step 1. Total energy from the amplitude alone: $E=\tfrac12kA^2=\tfrac12(80)(0.10)^2=\tfrac12(80)(0.01)=0.40$ J. This number is fixed for the whole motion, regardless of where the mass currently is.

Step 2. At $x=0.06$ m, the potential energy is $U=\tfrac12kx^2=\tfrac12(80)(0.06)^2=\tfrac12(80)(0.0036)=0.144$ J. By conservation, the kinetic energy there must be $K=E-U=0.40-0.144=0.256$ J, and since $K=\tfrac12mv^2$, $v=\sqrt{2K/m}=\sqrt{2(0.256)/0.20}=\sqrt{2.56}=1.6$ m/s.

Step 3. Check: $K+U=0.256+0.144=0.40$ J $=E$, confirming energy conservation at this arbitrary intermediate point, not just at the extremes.

Step 4. For $K=U$, each must equal $E/2=0.20$ J. Setting $U=\tfrac12kx^2=E/2=\tfrac12\left(\tfrac12kA^2\right)$ gives $x^2=A^2/2$, so $x=\pm A/\sqrt2=\pm0.10/\sqrt2\approx\pm0.0707$ m. The mass has speed $1.6$ m/s at $x=0.06$ m, and $K=U$ at $x\approx\pm0.071$ m — always at $A/\sqrt2$, about $70.7\%$ of the amplitude, whatever the values of $m$, $k$, or $A$.

Competitive stretch

Problem. For the same spring-mass system ($m=0.20$ kg, $k=80$ N/m, $A=0.10$ m), find the maximum speed $v_{\max}$ directly from the energy relation, and check it against $v_{\max}=A\omega_0$.

Step 1. At the equilibrium position all the energy is kinetic, so $K_{\max}=E=0.40$ J. From $K_{\max}=\tfrac12mv_{\max}^2$, $v_{\max}=\sqrt{2E/m}=\sqrt{2(0.40)/0.20}=\sqrt{4.0}=2.0$ m/s.

Step 2. Cross-check using $\omega_0=\sqrt{k/m}=\sqrt{80/0.20}=\sqrt{400}=20$ rad/s, so $v_{\max}=A\omega_0=0.10\times20=2.0$ m/s. Both routes agree: $v_{\max}=2.0$ m/s, confirming that $E=\tfrac12kA^2$ and $E=\tfrac12mv_{\max}^2$ describe the same total energy from two different vantage points in the cycle.

• Treating $K$ and $U$ as each individually constant — only their sum $E$ is constant; both $K(t)$ and $U(t)$ oscillate with time, at twice the frequency of $x(t)$.
• Forgetting $U=\tfrac12kx^2$ is measured from equilibrium, so at $x=0$ the energy is entirely kinetic, not half-and-half by some assumed symmetry.
• Double-counting energy — $\tfrac12kA^2$ already equals $K_{\max}=\tfrac12mv_{\max}^2$; they are the same total computed two ways, not two separate energies to add together.
• Guessing that $K=U$ happens at $x=A/2$; the correct crossing is $x=A/\sqrt2$, found by explicitly solving $\tfrac12kx^2=\tfrac12\left(\tfrac12kA^2\right)$ rather than assuming the midpoint.

Ask the tutor: if the kinetic-potential trade-off or the $A/\sqrt2$ crossing point still feels shaky, ask the AI tutor to plot $K(t)$ and $U(t)$ together over one full period or to walk through another numerical energy-conservation example.