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PU-103 / L11 · lecture

Beats and Energy Exchange Between Modes

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

BEFORE YOU START

You should be comfortable with the concept of normal modes as independent oscillating patterns and recall that the total energy of a harmonic oscillator is proportional to the square of its amplitude.

Why this matters

Beats are the visible signature of energy sloshing back and forth between two coupled oscillators.

In the previous lecture, we established that coupled oscillators decompose into normal modes. While these modes are mathematically independent, they are physically coupled through the system's constraints. When you excite the system in a way that is not a pure normal mode, you excite multiple modes simultaneously. Because these modes have slightly different frequencies, their superposition creates a phenomenon known as beats. Understanding beats is crucial because it reveals that energy is not static within a mode; it flows between the parts of the system while each normal mode keeps its own energy, a concept that generalizes to energy exchange in quantum systems and wave packets.

The Phenomenon of Beats

Imagine two tuning forks with very similar frequencies, say 440 Hz and 442 Hz, struck simultaneously. If you listen to the combined sound, you do not hear two distinct tones. Instead, you hear a single tone that grows louder and softer periodically. This periodic variation in loudness is called a beat. The rate at which the sound swells and fades is the beat frequency. In the context of coupled oscillators, if we displace one mass and release it, we are not exciting a single normal mode. We are exciting a mixture of the two normal modes. Because the two modes oscillate at slightly different rates, they periodically align in phase (constructive interference) and periodically oppose each other (destructive interference). This alignment and opposition manifest as the transfer of energy from one part of the system to another.

Superposition of Two Modes

Consider a system of two coupled oscillators with normal frequencies \(\omega_1\) and \(\omega_2\), where \(\omega_2\) is slightly larger than \(\omega_1\). Let the difference be small, such that \(\omega_2 - \omega_1 = \Delta\omega\) and the average frequency is \(\omega_0 = (\omega_1 + \omega_2)/2\). We can write \(\omega_1 = \omega_0 - \Delta\omega/2\) and \(\omega_2 = \omega_0 + \Delta\omega/2\).

Suppose we excite the system such that both modes have the same amplitude \(A\) and start in phase at \(t=0\). The displacement of a specific coordinate \(q(t)\) is the sum of the contributions from both modes. For simplicity, assume the mode shapes are similar in this coordinate, so we can write the superposition as:

\[ q(t) = A\cos(\omega_1 t) + A\cos(\omega_2 t) \]

Using the trigonometric identity \(\cos \alpha + \cos \beta = 2\cos\left(\frac{\alpha-\beta}{2}\right)\cos\left(\frac{\alpha+\beta}{2}\right)\), we can rewrite this expression:

\[ q(t) = 2A\cos\left(\frac{\omega_1 - \omega_2}{2}t\right)\cos\left(\frac{\omega_1 + \omega_2}{2}t\right) \]

Substituting our definitions for \(\omega_0\) and \(\Delta\omega\):

\[ q(t) = 2A\cos\left(-\frac{\Delta\omega}{2}t\right)\cos(\omega_0 t) \]

Since cosine is an even function, \(\cos(-x) = \cos(x)\), the expression becomes:

\[ q(t) = 2A\cos\left(\frac{\Delta\omega}{2}t\right)\cos(\omega_0 t) \]

This equation reveals the structure of the beat. The term \(\cos(\omega_0 t)\) represents the rapid oscillation at the average frequency. The term \(2A\cos\left(\frac{\Delta\omega}{2}t\right)\) acts as a slowly varying envelope. The amplitude of the fast oscillation is not constant; it changes with time according to the slower cosine term.

Beat Frequency and Period

The "loudness" or amplitude of the motion is determined by the magnitude of the envelope function \(2A\left|\cos\left(\frac{\Delta\omega}{2}t\right)\right|\). The amplitude reaches a maximum when the cosine term is \(\pm 1\). This occurs when the argument of the cosine is an integer multiple of \(\pi\):

\[ \frac{\Delta\omega}{2}t = n\pi, \quad n = 0, 1, 2, \dots \]

The time interval between successive maxima (where the envelope peaks) corresponds to \(n\) increasing by 1. Let \(\Delta t\) be the time for \(n\) to increase by 1. Then:

\[ \frac{\Delta\omega}{2}(\Delta t) = \pi \implies \Delta t = \frac{2\pi}{\Delta\omega} \]

This time interval \(\Delta t\) is the period of the beat, \(T_{\text{beat}}\). The beat frequency \(f_{\text{beat}}\) is the reciprocal of this period:

\[ f_{\text{beat}} = \frac{1}{T_{\text{beat}}} = \frac{\Delta\omega}{2\pi} \]

Since \(\Delta\omega = \omega_2 - \omega_1\), we have:

\[ f_{\text{beat}} = \frac{\omega_2 - \omega_1}{2\pi} \]

In terms of ordinary frequencies \(f_1 = \omega_1/2\pi\) and \(f_2 = \omega_2/2\pi\), the beat frequency is simply the difference between the two frequencies:

\[ f_{\text{beat}} = f_2 - f_1 \]

It is important to note that the amplitude envelope completes a full cycle (from max to min to max) over a period of \(2\pi / (\Delta\omega/2) = 4\pi/\Delta\omega\). However, the "beats" are perceived as the peaks in amplitude, which occur twice per full cycle of the envelope cosine. Thus, the beat frequency is indeed the difference in the component frequencies.

Energy Exchange

The beat phenomenon is not just a kinematic curiosity; it is a direct manifestation of energy conservation and exchange. In a linear, undamped system, the total mechanical energy is constant. However, the way that energy is shared between the two oscillators changes with time, even though the energy in each normal mode stays constant.

Let \(E_1(t)\) and \(E_2(t)\) be the energies associated with mode 1 and mode 2, respectively. For a harmonic oscillator, energy is proportional to the square of the amplitude. If we look at the superposition derived earlier, the instantaneous energy of the system can be decomposed into contributions from each mode. More rigorously, using the normal coordinates \(Q_1\) and \(Q_2\), the energy in each mode is:

\[ E_1(t) = \frac{1}{2}k_1 Q_1^2(t) + \frac{1}{2}m_1 \dot{Q}_1^2(t) \]
\[ E_2(t) = \frac{1}{2}k_2 Q_2^2(t) + \frac{1}{2}m_2 \dot{Q}_2^2(t) \]

Because the normal modes are independent, each mode energy \(E_1\) and \(E_2\) is constant in time (within a mode, energy only swaps between kinetic and potential forms). The initial conditions fix the mode amplitudes \(A_1\) and \(A_2\). If we start with one mass displaced, both modes are excited, and it is the energy of each oscillator that changes: energy flows from one mass to the other. When the two modes are in phase, their motions reinforce each other in one part of the system, concentrating energy there. When they are out of phase, they cancel in one part and reinforce in the other, shifting the energy to the other part.

The rate of this energy exchange is governed by the beat frequency. The energy of a given oscillator is greatest when its beat envelope is at a peak and smallest when the envelope is at a trough. Thus, the energy sloshes back and forth between the two oscillators at the beat frequency, while \(E_1\) and \(E_2\) stay fixed.

The derivation

◆ DERIVATION EMBED · D-033 · not re-derived here

Coupled Oscillators and Normal Modes

For a mechanical system near a stable equilibrium, the linearized equations of motion \( \mathbf{M}\ddot{\mathbf{q}} = -\mathbf{K}\mathbf{q} \), with symmetric positive-definite mass matrix \( \mathbf{M} \) and symmetric positive-semidefinite stiffness matrix \( \mathbf{K} \), are decoupled by the generalized eigenvalue problem \( \mathbf{K}\mathbf{a}^{(r)} = \omega_r^2\,\mathbf{M}\mathbf{a}^{(r)} \): the eigenvectors \( \mathbf{a}^{(r)} \) are the normal modes (rigid spatial patterns that oscillate in unison) and the eigenvalues \( \omega_r^2 \) are the squared normal frequencies, so every free motion is a superposition \( \mathbf{q}(t)=\sum_r \mathbf{a}^{(r)}\,(A_r\cos\omega_r t + B_r\sin\omega_r t) \).

\[ \mathbf{K}\,\mathbf{a}^{(r)}=\omega_r^2\,\mathbf{M}\,\mathbf{a}^{(r)},\qquad \mathbf{q}(t)=\sum_{r=1}^{n}\mathbf{a}^{(r)}\big(A_r\cos\omega_r t+B_r\sin\omega_r t\big),\qquad \big(\mathbf{a}^{(r)}\big)^{\mathsf T}\mathbf{M}\,\mathbf{a}^{(s)}=\delta_{rs} \]
Open full derivation →
Tier 3 ยท watch out

A common misconception is that the beat frequency is \(\omega_2 - \omega_1\). This is the angular beat frequency. The physical beat frequency in Hertz is \((\omega_2 - \omega_1)/2\pi\). Furthermore, the envelope function \(\cos(\Delta\omega t / 2)\) has a period of \(4\pi/\Delta\omega\), but the amplitude maxima occur every \(2\pi/\Delta\omega\). Confusing the period of the envelope with the period of the beat leads to a factor-of-two error in calculating the rate of energy exchange. C

Worked example

Problem. Two coupled pendulums have normal frequencies of \(f_1 = 1.00\ \mathrm{Hz}\) and \(f_2 = 1.02\ \mathrm{Hz}\). They are released from rest with one pendulum displaced. Calculate the beat frequency and the time interval between successive maxima of the amplitude envelope.

  1. Identify the two normal frequencies: \(f_1 = 1.00\ \mathrm{Hz}\) and \(f_2 = 1.02\ \mathrm{Hz}\).
  2. Calculate the beat frequency using the difference formula: \(f_{\text{beat}} = |f_2 - f_1|\).
  3. Calculate the time interval between successive amplitude maxima, which is the period of the beat \(T_{\text{beat}} = 1/f_{\text{beat}}\).
\[ f_{\text{beat}} = 1.02\ \mathrm{Hz} - 1.00\ \mathrm{Hz} = 0.02\ \mathrm{Hz} \]
\[ T_{\text{beat}} = \frac{1}{0.02\ \mathrm{Hz}} = 50\ \mathrm{s} \]

Units check. Frequency is in \(\mathrm{Hz}\) (\(\mathrm{s^{-1}}\)). The difference of frequencies is in \(\mathrm{Hz}\). The reciprocal of frequency is in seconds (\(\mathrm{s}\)). The result is consistent.

CHECK YOURSELF

Three quick tests

  1. If two modes have frequencies 100 Hz and 105 Hz, what is the beat frequency?
    Answer5 Hz.
  2. Does the total energy of the coupled system change during the beat phenomenon?
    AnswerNo, it is conserved; only its distribution between the two oscillators changes; the energy in each normal mode is constant.
  3. If the frequency difference \(\Delta f\) is very small, is the beat period long or short?
    AnswerLong, since \(T_{\text{beat}} = 1/\Delta f\).
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-033 · Threads: force, energy, waves