Topic 7.4 Energy of Simple Harmonic Oscillators: express the kinetic, potential and total energy of an oscillator, apply conservation of energy to relate speed and displacement, and find the speed at any position.

What this topic is asking

The College Board (Topic 7.4) wants you to express the kinetic, potential and total energy of an oscillator, to apply conservation of energy to relate speed and displacement, and to find the speed at any position. Because an ideal oscillator has no friction, its mechanical energy is constant, sloshing back and forth between kinetic and potential forms, and this gives a fast route to speeds without solving the motion in time.

Kinetic, potential and total energy

Because no friction acts, the total energy is conserved and equals its value at any convenient point. At the turning points the mass is momentarily at rest ($K = 0$) and the displacement is maximum, so all the energy is potential: $E = \tfrac{1}{2}kA^2$. At equilibrium the displacement is zero ($U = 0$) and the speed is maximum, so all the energy is kinetic: $E = \tfrac{1}{2}mv_{max}^2$. Equating these two expressions recovers $v_{max} = A\sqrt{k/m} = A\omega$, consistent with the kinematic result.

Energy exchange through the cycle

As the oscillator moves, energy flows continuously between kinetic and potential forms while the total stays fixed. Starting at a turning point (all potential), the mass speeds up as it falls toward equilibrium (potential converting to kinetic), reaches maximum speed at the center (all kinetic), then slows as it climbs to the far turning point (kinetic converting back to potential). The two energies oscillate at twice the frequency of the motion (each goes through two maxima per cycle), and their sum is a flat line at $E = \tfrac{1}{2}kA^2$.

Speed at any displacement

Conservation of energy gives the speed at any position directly. Setting $\tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 = \tfrac{1}{2}kA^2$ and solving for $v$:

This is the workhorse relation for SHM energy problems: it gives the speed at any displacement without needing the time. The speed is maximum at $x = 0$ ($v = \omega A$) and zero at $x = \pm A$, as expected. A useful special case: the kinetic and potential energies are equal ($K = U = E/2$) when $\tfrac{1}{2}kx^2 = \tfrac{1}{2}E$, that is at $x = \pm A/\sqrt{2}$, where the speed is $v_{max}/\sqrt{2}$.

Try this

Q1. An oscillator has total energy $0.50$ J and spring constant $100$ N/m. Calculate its amplitude. [2 points]

• Cue. $E = \tfrac{1}{2}kA^2 \Rightarrow 0.50 = \tfrac{1}{2}(100)A^2 \Rightarrow A = \sqrt{0.010} = 0.10$ m.

Q2. State the displacement (as a fraction of amplitude) at which the kinetic and potential energies of an oscillator are equal. [2 points]

• Cue. $x = A/\sqrt{2} \approx 0.71A$.