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PU-103 / L03 · lecture

Energy in the Harmonic Oscillator

~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be comfortable with the definition of simple harmonic motion, the relationship between angular frequency \(\omega_0\), mass \(m\), and stiffness \(k\), and the basic calculus of differentiating sinusoidal functions.

Why this matters

Energy provides the most robust invariant in oscillatory systems, linking the microscopic motion of a single atom to the macroscopic propagation of waves.

In the previous lecture, we established that a mass on a spring executes simple harmonic motion (SHM) due to a linear restoring force. However, the position \(x(t)\) alone does not tell the whole story. To understand how energy is stored, transferred, and eventually transported in wave phenomena, we must examine the kinetic and potential energy components separately. This lecture demonstrates that while the individual energy forms oscillate rapidly, their sum remains constant, establishing the foundation for conservation laws in more complex coupled systems.

The Energy Picture in Plain Language

Imagine a mass attached to a spring. When the mass is at its maximum displacement (amplitude), it is momentarily at rest. All the energy in the system is stored in the stretched or compressed spring as potential energy. As the mass moves toward the equilibrium position, the spring relaxes, converting that potential energy into kinetic energy. At the exact centre, the spring is unstretched, so potential energy is zero, but the mass is moving at its maximum speed, meaning kinetic energy is at its peak. As the mass overshoots the centre, it slows down, converting kinetic energy back into potential energy. This continuous back-and-forth exchange is the heartbeat of the oscillator. Crucially, in an ideal undamped system, no energy is lost to heat or sound; it merely changes form. This conservation of total mechanical energy is the key principle that allows us to predict the motion without solving the differential equation directly.

Instantaneous Energy Components

Let us define the position of the oscillator as \(x(t) = A\cos(\omega_0 t + \varphi)\), where \(A\) is the amplitude and \(\omega_0 = \sqrt{k/m}\). The velocity is the time derivative of position: \(v(t) = \dot{x}(t) = -A\omega_0\sin(\omega_0 t + \varphi)\). The potential energy \(U\) stored in the spring is given by \(U = \tfrac12 kx^2\). Substituting the position function, we find \(U(t) = \tfrac12 k A^2 \cos^2(\omega_0 t + \varphi)\). Similarly, the kinetic energy \(T\) is \(\tfrac12 mv^2\). Substituting the velocity function, we get \(T(t) = \tfrac12 m (A\omega_0\sin(\omega_0 t + \varphi))^2\). Since \(k = m\omega_0^2\), this simplifies to \(T(t) = \tfrac12 k A^2 \sin^2(\omega_0 t + \varphi)\). Notice that both energies depend on the square of a trigonometric function. Using the identity \(\cos^2\theta = \tfrac12(1+\cos 2\theta)\) and \(\sin^2\theta = \tfrac12(1-\cos 2\theta)\), we can rewrite these expressions to reveal their time dependence more clearly.

\[ U(t) = \frac{1}{4}kA^2 \left[ 1 + \cos(2\omega_0 t + 2\varphi) \right], \qquad T(t) = \frac{1}{4}kA^2 \left[ 1 - \cos(2\omega_0 t + 2\varphi) \right] \]

This form shows that both \(T\) and \(U\) oscillate about a mean value. The term \(\tfrac14 kA^2\) is the average energy for each component, while the cosine term represents the oscillating part. Importantly, the argument of the cosine is \(2\omega_0 t\), indicating that the energy oscillates at twice the frequency of the position.

Conservation and Frequency Doubling

Adding the instantaneous kinetic and potential energies yields the total mechanical energy \(E(t) = T(t) + U(t)\). When we sum the expressions derived above, the oscillating cosine terms cancel out exactly because one has a positive sign and the other a negative sign. The result is a constant value independent of time: \(E = \tfrac12 kA^2\). This confirms the principle of conservation of energy for the undamped harmonic oscillator. The total energy is determined solely by the amplitude \(A\) and the stiffness \(k\). It is worth noting that the energy is proportional to the square of the amplitude. Doubling the amplitude quadruples the total energy. This quadratic relationship is a hallmark of linear systems and will reappear in the energy density of electromagnetic waves and the intensity of sound waves.

\[ E = T + U = \frac{1}{2}kA^2 = \frac{1}{2}m\omega_0^2 A^2 = \text{const} \]

The fact that \(T\) and \(U\) oscillate at \(2\omega_0\) has a physical interpretation. The energy exchange happens twice per cycle of the motion: once when the mass moves from the extreme to the centre (potential to kinetic) and once when it moves from the centre to the opposite extreme (kinetic to potential). This frequency doubling is a critical concept when analysing parametric resonance and energy transfer in coupled oscillators.

Phase Relationship and Energy Flow

Examining the expressions for \(T(t)\) and \(U(t)\), we observe that they are exactly out of phase. When \(U\) is at its maximum, \(T\) is at its minimum (zero), and vice versa. This anti-correlation ensures that the sum remains constant. In a more general context, such as a damped oscillator or a driven system, this perfect cancellation no longer holds. Energy is then exchanged with the environment or the driving force. Understanding the ideal case first allows us to quantify the deviations caused by damping. The rate of energy transfer between kinetic and potential forms is given by the time derivative of \(T\) (or \(U\)). This rate is proportional to \(\sin(2\omega_0 t + 2\varphi)\), so the power is greatest midway, at \(x = \pm A/\sqrt{2}\), and is zero both at the equilibrium position (where the force vanishes) and at the turning points (where the velocity vanishes). This insight is crucial for understanding how energy flows in wave media, where local kinetic and potential energies continuously exchange as the wave propagates.

The derivation

◆ DERIVATION EMBED · D-058 · not re-derived here

Energy Exchange and Conservation in SHM

For a one-dimensional undamped simple harmonic oscillator of mass \(m\) and stiffness \(k\), executing \(x(t)=A\cos(\omega_0 t+\varphi)\) with \(\omega_0=\sqrt{k/m}\), the total mechanical energy \(E=T+U\) is constant in time and equal to \(\tfrac12 kA^2\), while the kinetic part \(T\) and potential part \(U\) each oscillate about the mean value \(\tfrac14 kA^2\) at angular frequency \(2\omega_0\) — twice the oscillation frequency — exactly out of phase with one another.

\[ E=T+U=\tfrac12 kA^2=\tfrac12 m\omega_0^2 A^2=\text{const},\qquad \langle T\rangle=\langle U\rangle=\tfrac14 kA^2,\qquad f_{\text{exch}}=2f_0 \]
Open full derivation →
Tier 3 · watch out

A common misconception is that the energy of the oscillator is zero at the equilibrium position because the potential energy is zero. This is incorrect; at equilibrium, the potential energy is zero, but the kinetic energy is at its maximum, equal to the total energy \(\tfrac12 kA^2\). The total energy is never zero for a non-trivial oscillation. C

Worked example

Problem. A mass of \(2.0\ \mathrm{kg}\) is attached to a spring with stiffness \(k = 50\ \mathrm{N/m}\). It is displaced by \(0.10\ \mathrm{m}\) from equilibrium and released from rest. Calculate the total mechanical energy of the system and the maximum speed attained by the mass.

  1. Identify the given values: \(m = 2.0\ \mathrm{kg}\), \(k = 50\ \mathrm{N/m}\), amplitude \(A = 0.10\ \mathrm{m}\).
  2. Calculate the total energy using \(E = \tfrac12 kA^2\). Substituting the values: \(E = \tfrac12 (50) (0.10)^2 = 25 \times 0.01 = 0.25\ \mathrm{J}\).
  3. At maximum speed, the potential energy is zero, so all energy is kinetic: \(T_{\max} = \tfrac12 m v_{\max}^2 = E\).
  4. Solve for \(v_{\max}\): \(v_{\max} = \sqrt{2E/m} = \sqrt{2(0.25)/2.0} = \sqrt{0.25} = 0.50\ \mathrm{m/s}\).
\[ E = 0.25\ \mathrm{J}, \qquad v_{\max} = 0.50\ \mathrm{m/s} \]

Units check. Energy: \(\mathrm{N/m \cdot m^2} = \mathrm{N \cdot m} = \mathrm{J}\). Speed: \(\sqrt{\mathrm{J/kg}} = \sqrt{\mathrm{kg \cdot m^2/s^2 / kg}} = \sqrt{\mathrm{m^2/s^2}} = \mathrm{m/s}\). Both are consistent.

CHECK YOURSELF

Three quick tests

  1. What is the ratio of kinetic energy to potential energy when the displacement is half the amplitude (\(x = A/2\))?
    AnswerAt \(x = A/2\), \(U = \tfrac12 k(A/2)^2 = \tfrac18 kA^2\). Since \(E = \tfrac12 kA^2\), \(T = E - U = \tfrac12 kA^2 - \tfrac18 kA^2 = \tfrac38 kA^2\). The ratio \(T/U = (\tfrac38)/(\tfrac18) = 3\).
  2. How does the total energy change if the amplitude is tripled?
    AnswerSince \(E \propto A^2\), tripling the amplitude increases the energy by a factor of \(3^2 = 9\).
  3. At what angular frequency do the kinetic and potential energies oscillate?
    AnswerThey oscillate at \(2\omega_0\), which is twice the natural angular frequency of the oscillator.
PROBLEMS

Problem set for PU-103 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-058 · Threads: force, energy, waves