Energy in SHM

A pendulum bob swinging past the bottom of its arc is moving fastest of all — pure kinetic energy, no height gained. A moment later, at the top of its swing, it hangs motionless for an instant — pure potential energy, no speed at all. Between these two extremes, at every point on the way, energy is quietly changing form: kinetic becoming potential, potential becoming kinetic, back and forth, forever (in the ideal case). What never changes is the total. This article makes that trade precise for simple harmonic motion. Starting from the SHM kinematics you already know — $y=A\sin(\omega t+\phi_0)$ and $v=\omega\sqrt{A^2-y^2}$ — we derive expressions for kinetic energy $K$ and potential energy $U$ at any displacement $y$ , show that their sum $E=U+K$ is a constant equal to $\tfrac12m\omega^2A^2$ , and read off what that constant equals at the two special points: the mean position and the extremes. The result is one of the cleanest conservation statements in all of mechanics, and a favourite source of numerical problems.

The core insight

In SHM, kinetic energy $K=\tfrac12m\omega^2(A^2-y^2)$ and potential energy $U=\tfrac12m\omega^2y^2$ continuously exchange into each other, but their sum $E=\tfrac12m\omega^2A^2$ stays constant at every instant and every displacement.

The core idea

Take a particle of mass $m$ executing SHM with amplitude $A$ and angular frequency $\omega$ , so that $y=A\sin(\omega t+\phi_0)$ and $v=\omega\sqrt{A^2-y^2}$ .

Kinetic energy follows directly from $K=\tfrac12mv^2$ once $v$ is written in terms of displacement:

K = \frac12 mv^2 = \frac12 m\omega^2(A^2-y^2)

where: $m$ is the mass (kg), $\omega$ is the angular frequency (rad/s), $A$ is the amplitude (m), and $y$ is the instantaneous displacement (m) from the mean position.

Potential energy is the elastic energy stored against the restoring force $F=-ky=-m\omega^2y$ . Integrating the work done against this force from $0$ to $y$ gives $U=\int_0^y ky\,dy=\tfrac12ky^2$ , and since $k=m\omega^2$ for SHM:

U = \frac12 ky^2 = \frac12 m\omega^2y^2

where: $k$ is the effective force constant (N/m), equal to $m\omega^2$ . Adding the two, the $y^2$ terms cancel and only $A^2$ survives:

E = U+K = \frac12m\omega^2y^2+\frac12m\omega^2(A^2-y^2)=\frac12m\omega^2A^2

This total is independent of both $y$ and $t$ — it is the same number throughout the motion. That is conservation of mechanical energy, and it holds because the SHM restoring force is conservative (no friction or damping assumed). At the two special points the split is as extreme as it gets. At the mean position ( $y=0$ ) all the energy is kinetic: $K=K_{max}=\tfrac12m\omega^2A^2=E$ , and speed is maximum. At the extremes ( $y=\pm A$ ) all the energy is potential: $U=U_{max}=\tfrac12m\omega^2A^2=E$ , and the particle is momentarily at rest — though its acceleration, and so the restoring force on it, is at its own maximum there. Plotted against $y$ (from $-A$ to $+A$ ), $U(y)$ is an upward parabola with its minimum at $y=0$ , and $K(y)$ is a downward parabola with its maximum at $y=0$ — mirror images of each other about the horizontal line $E$ . The two curves cross exactly where $K=U$ , at $y=\pm A/\sqrt2$ .

Figure

Kinetic potential and total energy plotted against displacement for a particle in SHM

Substituting the time dependence $y=A\sin(\omega t+\phi_0)$ gives $K(t)=\tfrac12m\omega^2A^2\cos^2(\omega t+\phi_0)$ and $U(t)=\tfrac12m\omega^2A^2\sin^2(\omega t+\phi_0)$ , so both $K$ and $U$ oscillate at twice the frequency of the displacement itself (since $\sin^2\theta$ and $\cos^2\theta$ complete two cycles for every one cycle of $\sin\theta$ ), even though the particle's own period is unchanged. Averaged over a full cycle, $\langle\sin^2\theta\rangle=\langle\cos^2\theta\rangle=\tfrac12$ , so $\langle K\rangle=\langle U\rangle=E/2$ — on average the energy is split evenly, even though at any single instant it rarely is.

Derivation ATotal mechanical energy in SHM is constant

The goal is to show, starting only from the SHM displacement and velocity relations and the definitions of kinetic and potential energy, that the sum $E=K+U$ for a particle oscillating with amplitude $A$ and angular frequency $\omega$ does not depend on the displacement $y$ or the time $t$ , and to identify what that constant value equals.

Step 1. Write kinetic energy in terms of displacement. From $v=\omega\sqrt{A^2-y^2}$ , $K=\tfrac12mv^2=\tfrac12m\omega^2(A^2-y^2)$ . This already shows $K$ is largest when $y=0$ and falls to zero at $y=\pm A$ .

Step 2. Write potential energy in terms of displacement. The restoring force is $F=-ky=-m\omega^2y$ , so the work done by an external agent to stretch the system quasi-statically from $0$ to $y$ is $U=\int_0^y ky\,dy=\tfrac12ky^2=\tfrac12m\omega^2y^2$ , using $k=m\omega^2$ .

Step 3. Add the two expressions. $E=K+U=\tfrac12m\omega^2(A^2-y^2)+\tfrac12m\omega^2y^2$ . The two $y^2$ terms are equal and opposite in sign, so they cancel exactly, leaving $E=\tfrac12m\omega^2A^2$ — a quantity built only from the fixed constants $m$ , $\omega$ and $A$ , with no $y$ remaining.

Step 4. Confirm it is constant in time too. Substituting $y=A\sin(\omega t+\phi_0)$ and $v=A\omega\cos(\omega t+\phi_0)$ gives $K(t)+U(t)=\tfrac12m\omega^2A^2[\cos^2(\omega t+\phi_0)+\sin^2(\omega t+\phi_0)]=\tfrac12m\omega^2A^2$ , using $\sin^2\theta+\cos^2\theta=1$ . The same constant appears from the time-based route as from the displacement-based route, confirming $E$ is genuinely conserved throughout the motion, not just an algebraic coincidence at one instant.

Worked example

A body of mass $0.5\ \text{kg}$ executes SHM with amplitude $0.04\ \text{m}$ and angular frequency $\omega=10\ \text{rad/s}$ . Find (a) the total energy, (b) the KE and PE when the displacement is $0.02\ \text{m}$ , (c) the displacement at which KE equals PE, and (d) the maximum speed.

Step 1. Total energy: $E=\tfrac12m\omega^2A^2=\tfrac12\times0.5\times10^2\times(0.04)^2=25\times0.0016=0.04\ \text{J}$ .

Step 2. At $y=0.02\ \text{m}$ : $U=\tfrac12m\omega^2y^2=25\times(0.02)^2=25\times0.0004=0.01\ \text{J}$ . By conservation, $K=E-U=0.04-0.01=0.03\ \text{J}$ — no need to recompute $K$ from scratch once $E$ and $U$ are known.

Step 3. Setting $K=U$ : $\tfrac12m\omega^2(A^2-y^2)=\tfrac12m\omega^2y^2 \Rightarrow A^2-y^2=y^2 \Rightarrow y=\dfrac{A}{\sqrt2}=\dfrac{0.04}{1.414}=0.0283\ \text{m}$ . This point lies at about $71\%$ of the amplitude, not halfway — a common source of error.

Step 4. Maximum speed occurs at $y=0$ , where all the energy is kinetic: $\tfrac12mv_{max}^2=E \Rightarrow v_{max}=\sqrt{2E/m}=\sqrt{2\times0.04/0.5}=\sqrt{0.16}=0.4\ \text{m/s}$ .

KE and PE are equal at $y=\pm0.0283\ \text{m}$ , and the maximum speed is $0.4\ \text{m/s}$ , both consistent with the total energy of $0.04\ \text{J}$ found in Step 1.

Common traps

Believing energy is lost as the particle travels from an extreme to the mean position — in ideal (undamped) SHM, total mechanical energy $E$ is exactly conserved at every instant; only the $K$ – $U$ split changes.
Writing $E=\tfrac12m\omega^2y^2$ , using the instantaneous displacement instead of the amplitude — total energy must use $A$ , since $E$ is constant while $y$ varies.
Assuming $K$ and $U$ oscillate at the same frequency as the displacement — they actually vary at twice the frequency, since both involve $\sin^2$ or $\cos^2$ terms.
Forgetting that at $y=\pm A$ the particle has zero velocity but maximum acceleration — zero KE does not mean zero force acting on it.
Trying to use $E=\tfrac12kA^2$ for a general SHM problem without first checking that $k=m\omega^2$ is actually known or given — for a pendulum or any non-spring oscillator, express $E$ through $m$ and $\omega$ instead, since there may be no physical spring constant to quote.

Ask the tutor: Want to see the full $U$ – $K$ – $E$ derivation worked through with a spring-and-mass example instead of a pendulum, or try a harder energy-split problem — ask your AI tutor for a step-by-step walkthrough.

Why does the total energy depend on A squared rather than just A?

Because $E=\tfrac12m\omega^2A^2$ comes from summing $K=\tfrac12m\omega^2(A^2-y^2)$ and $U=\tfrac12m\omega^2y^2$ , and both terms carry a squared length — kinetic energy from $v^2$ , potential energy from $y^2$ . Squaring the amplitude is not a special rule for SHM; it is the same pattern as spring PE ( $\tfrac12kx^2$ ) or gravitational free-fall KE ( $\tfrac12mv^2$ with $v^2\propto h$ ), where energy always scales with the square of a length or speed. Doubling $A$ quadruples $E$ , which is exactly why larger-amplitude oscillations demand disproportionately more energy to set up.

Does the total energy E change if I know the displacement y at some instant?

No — that is the entire point of conservation. $E=\tfrac12m\omega^2A^2$ depends only on the fixed quantities $m$ , $\omega$ and $A$ , none of which change as the particle moves. Knowing $y$ at a particular instant only tells you how $E$ is currently split between $K$ and $U$ (via $U=\tfrac12m\omega^2y^2$ and $K=E-U$ ); it never changes the value of $E$ itself. If a problem gives you $E$ and asks for $K$ or $U$ at some $y$ , always compute the one you're missing as a subtraction from $E$ , rather than rederiving both from scratch.

Is this the same total energy used for wave intensity, or something different?

Something different, and it is worth keeping the two apart. This article is about the mechanical energy of a single oscillating particle or body — a pendulum bob, a mass on a spring — and it stays fixed at $\tfrac12m\omega^2A^2$ for that one oscillator. Wave intensity, by contrast, describes energy carried per unit area per unit time by a whole travelling disturbance made of many such oscillating particles, and depends additionally on the wave's speed and the medium's density. The SHM energy result derived here is a necessary building block for that later idea, but it is not itself a statement about waves or intensity, which stays out of scope for this topic.

Energy in SHM

The core idea

Worked example

Common traps

Before this

Where this leads