Where the energy goes

Topic 7.4 is about energy in simple harmonic motion: how a frictionless oscillator stores a fixed total and moves it between kinetic and potential form. The total mechanical energy is $E=\tfrac12 kA^2$, set by the spring constant and the amplitude.

As the block moves, the kinetic energy $K=\tfrac12 mv^2$ and the potential energy $U=\tfrac12 kx^2$ trade back and forth, but their sum stays at $\tfrac12 kA^2$. At equilibrium the energy is all kinetic; at the turning points it is all potential.

Almost every mistake here breaks one of three habits: letting the total energy vary instead of holding it fixed, scaling the energy with the amplitude instead of its square, or swapping where the kinetic and potential energy peak.

The total mechanical energy of a frictionless oscillator is the sum of its kinetic and potential energy, $E=K+U$. With $K=\tfrac12 mv^2$, $U=\tfrac12 kx^2$, and the peak speed $v_{\max}=A\omega$ where $\omega^2=k/m$, the total comes out to the constant $E=\tfrac12 kA^2=\tfrac12 mv_{\max}^2$.

Plotted against position, the potential energy is a parabola that is zero at the center and largest at the turning points, while the kinetic energy is its mirror image, largest at the center and zero at the ends. Their sum is a flat line: the total never changes.

The total is fixed: $E=K+U=\tfrac12 kA^2$ at every position. Kinetic and potential energy exchange, but their sum holds.

Because $E=\tfrac12 kA^2$, the total energy depends on the square of the amplitude. Doubling the amplitude multiplies the energy by $2^2=4$; tripling it multiplies by $9$; halving it cuts the energy to a quarter.

So a small increase in amplitude makes a much larger increase in energy. The common slip is to scale the energy in direct proportion to $A$, so a doubled amplitude is expected to double the energy. To double the energy instead, the amplitude must grow by $\sqrt2$, not by $2$.

Energy goes as $A^2$: whatever factor the amplitude changes by, the energy changes by that factor squared. Double $A$ and $E$ quadruples.

The kinetic energy is largest where the speed is largest, which is at equilibrium $x=0$. There the displacement is zero, so the potential energy is zero and all the energy is kinetic.

The potential energy is largest where the displacement is largest, at the turning points $x=\pm A$. There the speed is zero, so the kinetic energy is zero and all the energy is potential. The two forms reach their maxima at opposite places and never peak together.

Kinetic energy peaks at equilibrium, where the speed is greatest; potential energy peaks at the turning points, where the displacement is greatest.

At a general position the energy divides between the two forms. The potential energy is $U=\tfrac12 kx^2$, and the kinetic energy is whatever is left of the constant total, $K=E-U=\tfrac12 kA^2-\tfrac12 kx^2$.

This subtraction is the fastest way to read the split: knowing the total and one form gives the other at once, with no need to find the position or the speed. Halfway out in energy terms, where $K=U$, the displacement is $x=A/\sqrt2$, not $A/2$.

At any instant $K=E-U$. The total is fixed, so one form is always the constant total minus the other.

Tracking energy through an oscillation trips people up in three predictable ways.

• Letting the total energy varyTreating the total as rising and falling with the block. Fix: with no friction $E=\tfrac12 kA^2$ is constant; only the split between $K$ and $U$ changes.
• Scaling energy with $A$ instead of $A^2$Expecting a doubled amplitude to double the energy. Fix: $E=\tfrac12 kA^2$, so doubling $A$ quadruples $E$.
• Swapping where each form peaksPutting the kinetic maximum at the turning points or the potential maximum at the center. Fix: kinetic energy peaks at equilibrium, potential at the turning points.

Hold the total fixed, scale energy by the square of the amplitude, and tie kinetic energy to the center and potential energy to the ends. The whole topic comes down to those three habits.