IPhO 1986 · Theory problem 3

Normal modes of masses joined by springs (ring and chain)

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Three particles, each of mass \(m\), are in equilibrium and joined by unstretched massless springs, each with Hooke’s Law spring constant \(k\). They are constrained to move in a circular path as indicated in Figure 3.1.

Figure 1

Figure 3.1

(i) If each mass is displaced from equilibrium by small displacements \(u_1\), \(u_2\) and \(u_3\) respectively, write down the equation of motion for each mass.

(ii) Verify that the system has simple harmonic solutions of the form

\[ u_n=a_n\cos\omega t, \]

with accelerations, \((-\omega^2u_n)\) where \(a_n\) \((n=1,2,3)\) are constant amplitudes, and \(\omega\), the angular frequency, can have 3 possible values,

\[ \omega_o\sqrt{3},\ \omega_o\sqrt{3}\text{ and }0. \quad\text{where }\omega_o^2=\frac{k}{m}. \]

(iii) The system of alternate springs and masses is extended to \(N\) particles, each mass \(m\) is joined by springs to its neighbouring masses. Initially the springs are unstretched and in equilibrium. Write down the equation of motion of the \(n\)th mass \((n=1,2...N)\) in terms of its displacement and those of the adjacent masses when the particles are displaced from equilibrium.

\[ u_n(t)=a_s\sin\left(\frac{2ns\pi}{N}+\phi\right)\cos\omega_s t, \]

are oscillatory solutions where \(s=1,2,\ldots N\), \(n=1,2,\ldots N\) and where \(\phi\) is an arbitrary phase, providing the angular frequencies are given by

\[ \omega_s=2\omega_o\sin\left(\frac{s\pi}{N}\right), \]

where \(a_s\ (s=1,......N)\) are constant amplitudes independent of \(n\).

State the range of possible frequencies for a chain containing an infinite number of masses.

(iv) Determine the ratio

\[ u_n/u_{n+1} \]

for large \(N\), in the two cases:

(a) low frequency solutions

(b) \(\omega=\omega_{\max}\), where \(\omega_{\max}\) is the maximum frequency solution.

Sketch typical graphs indicating the displacements of the particles against particle number along the chain at time \(t\) for cases (a) and (b).

(v) If one of the masses is replaced by a mass \(m'\ll m\) estimate any major change one would expect to occur to the angular frequency distribution.

Describe qualitatively the form of the frequency spectrum one would predict for a diatomic chain with alternate masses \(m\) and \(m'\) on the basis of the previous result.

Reminder

\[ \sin(A+B)=\sin A\cos B+\cos A\sin B \]

\[ \sin A+\sin B =2\sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]

\[ 2\sin^2A=1-\cos2A \]

Official English text of the problem.

Answers

  • (i) \[ \boxed{ \begin{aligned} m\ddot u_1&=k(u_2+u_3-2u_1),\\ m\ddot u_2&=k(u_3+u_1-2u_2),\\ m\ddot u_3&=k(u_1+u_2-2u_3). \end{aligned}} \]

  • (ii) \[ \boxed{\omega=\sqrt{\frac{3k}{m}},\quad \sqrt{\frac{3k}{m}},\quad 0.} \] The nonzero frequency has two independent modes satisfying \(a_1+a_2+a_3=0\). The zero mode has \(a_1=a_2=a_3\).

  • (iii) \[ \boxed{m\ddot u_n=k(u_{n+1}+u_{n-1}-2u_n),\qquad u_{n+N}=u_n,} \] \[ \boxed{\omega_s=2\sqrt{\frac{k}{m}}\sin\left(\frac{s\pi}{N}\right).} \] For an infinite chain, \[ \boxed{0\leq\omega\leq2\sqrt{\frac{k}{m}}.} \] In ordinary frequency, \(0\leq f\leq\sqrt{k/m}/\pi\).

  • (iv)
    (a) \(\boxed{u_n/u_{n+1}\simeq+1}\), away from displacement nodes. The graph is a slowly varying sinusoid sampled at the particle positions.
    (b) \(\boxed{u_n/u_{n+1}=-1}\) at the upper band edge. Displacements alternate between equal positive and negative values. This is exact for even \(N\); an odd ring approaches this behaviour locally for large \(N\).

  • (v) A single light mass produces one high-frequency mode, concentrated on that mass: \[ \boxed{\omega_{\mathrm{defect}}\simeq\sqrt{\frac{2k}{m'}} \gg2\sqrt{\frac{k}{m}}.} \] An alternating chain has an acoustic band and an optical band separated by a gap. For \(m'<m\), their exact limiting ranges are \[ \boxed{ \begin{aligned} \text{Acoustic:}\quad& 0\leq\omega\leq\sqrt{\frac{2k}{m}},\\ \text{Optical:}\quad& \sqrt{\frac{2k}{m'}} \leq\omega\leq \sqrt{2k\left(\frac1m+\frac1{m'}\right)}. \end{aligned}} \]

Solution

(i)

Measure all displacements as arc lengths in the same sense around the ring. In the small-displacement approximation, a spring’s extension is the difference between the displacements of its ends. Hooke’s law therefore gives a tangential restoring force proportional to that difference. The constraint force has no tangential component.

For particle 1, the two springs exert forces \(k(u_2-u_1)\) and \(k(u_3-u_1)\). Newton’s second law gives

\[ m\ddot u_1=k(u_2-u_1)+k(u_3-u_1). \]

Applying the same reasoning to particles 2 and 3,

\[ \boxed{ \begin{aligned} m\ddot u_1&=k(u_2+u_3-2u_1),\\ m\ddot u_2&=k(u_3+u_1-2u_2),\\ m\ddot u_3&=k(u_1+u_2-2u_3). \end{aligned}} \]

(ii)

Substitute \(u_n=a_n\cos\omega t\), so that \(\ddot u_n=-\omega^2u_n\). Define

\[ \lambda=\frac{\omega^2}{\omega_o^2}, \qquad \omega_o^2=\frac{k}{m}, \qquad S=a_1+a_2+a_3. \]

Each equation becomes

\[ \lambda a_n=3a_n-S. \]

Adding the three equations gives

\[ \lambda S=0. \]

There are two possibilities.

If \(\lambda=0\), the individual equations require

\[ a_1=a_2=a_3. \]

All masses move through the same arc length, so no spring changes its length. Thus

\[ \boxed{\omega=0.} \]

This is the rigid-rotation mode. Strictly, it is not an oscillation: its general motion is \(u_1=u_2=u_3=A+Bt\). The prescribed cosine form represents its static-displacement case.

If \(\lambda\neq0\), then \(S=0\), and a nonzero amplitude requires

\[ \lambda=3. \]

Therefore

\[ \boxed{\omega=\sqrt{3}\,\omega_o=\sqrt{\frac{3k}{m}}.} \]

The condition \(a_1+a_2+a_3=0\) leaves two independent amplitudes. For example, two independent mode shapes are

\[ (a_1,a_2,a_3)\propto(1,-1,0), \qquad (a_1,a_2,a_3)\propto(1,1,-2). \]

Consequently, the three normal modes have frequencies

\[ \boxed{\sqrt{3}\,\omega_o,\quad\sqrt{3}\,\omega_o,\quad0,} \]

with the nonzero frequency occurring twice.

(iii)

Particle \(n\) experiences the forces

\[ k(u_{n+1}-u_n) \quad\text{and}\quad -k(u_n-u_{n-1}). \]

Hence

\[ \boxed{m\ddot u_n=k(u_{n+1}+u_{n-1}-2u_n).} \]

The ring imposes periodic boundary conditions,

\[ u_0=u_N,\qquad u_{N+1}=u_1. \]

Write

\[ q=\frac{2\pi s}{N}, \qquad u_n=a_s\sin(nq+\phi)\cos\omega_s t. \]

This expression is periodic because \(Nq=2\pi s\). Using the supplied addition identity,

\[ \begin{aligned} u_{n+1}+u_{n-1} &=a_s\bigl[\sin(nq+\phi+q)+\sin(nq+\phi-q)\bigr] \cos\omega_s t\\ &=2\cos q\,u_n. \end{aligned} \]

The equation of motion therefore becomes

\[ -m\omega_s^2u_n =2k(\cos q-1)u_n =-4k\sin^2\left(\frac q2\right)u_n. \]

For a nontrivial mode,

\[ \omega_s^2 =4\frac{k}{m}\sin^2\left(\frac{\pi s}{N}\right). \]

Since \(\sin(\pi s/N)\geq0\) for the stated range \(s=1,\ldots,N\),

\[ \boxed{\omega_s=2\omega_o\sin\left(\frac{\pi s}{N}\right).} \]

As the chain becomes infinite, the allowed wave numbers become continuous. The sine factor takes every value between zero and one, giving

\[ \boxed{0\leq\omega\leq2\omega_o=2\sqrt{\frac{k}{m}}.} \]

Zero is the rigid-displacement endpoint; genuine oscillations have \(\omega>0\). Since \(f=\omega/(2\pi)\),

\[ \boxed{0\leq f\leq\frac1\pi\sqrt{\frac{k}{m}}.} \]

With SI values of \(k\) and \(m\), angular frequencies are in \(\mathrm{rad\,s^{-1}}\), and ordinary frequencies are in \(\mathrm{Hz}\).

(iv)

At times and positions where the denominator is nonzero, the common time factor cancels:

\[ \frac{u_n}{u_{n+1}} = \frac{\sin(nq+\phi)}{\sin(nq+\phi+q)}. \]

The same expression applies directly to the ratio of modal amplitudes.

(a) Low frequency

Choose the representative of \(q\) close to zero. Then

\[ \omega=2\omega_o\left|\sin\frac q2\right| \simeq\omega_o|q|, \]

so low frequency means \(|q|\ll1\). Writing \(\theta=nq+\phi\),

\[ \frac{u_n}{u_{n+1}} =\frac{1}{\cos q+\cot\theta\sin q} \simeq1-q\cot\theta. \]

Thus, away from nodes,

\[ \boxed{\frac{u_n}{u_{n+1}}\simeq+1.} \]

Neighbouring masses have almost equal displacements and move approximately in phase. Near a node, the displacements themselves are small, so their ratio need not be close to one.

For the requested sketch, plot a sinusoid with wavelength \(L\gg1\), measured in particle spacings, and mark its values at integer particle numbers. A representative profile is

\[ u_n=C(t)\sin\left(\frac{2\pi(n-n_0)}{L}\right). \]

Its key values are:

\(n-n_0\) \(0\) \(L/4\) \(L/2\) \(3L/4\) \(L\)
\(u_n/C(t)\) \(0\) \(+1\) \(0\) \(-1\) \(0\)

There are many particles between successive nodes, and neighbouring points usually have very similar heights.

(b) Maximum frequency

At the upper band edge, \(q=\pi\), giving

\[ \omega_{\max}=2\omega_o. \]

The displacement is

\[ u_n=a_s\sin(n\pi+\phi)\cos\omega_{\max}t =(-1)^n a_s\sin\phi\cos\omega_{\max}t. \]

Consequently,

\[ \boxed{u_{n+1}=-u_n,\qquad \frac{u_n}{u_{n+1}}=-1.} \]

Every neighbouring pair moves in opposite directions with equal amplitudes.

The sketch consists of alternating points at \(+C(t)\) and \(-C(t)\):

\(n-n_0\) \(0\) \(1\) \(2\) \(3\) \(4\)
\(u_n/C(t)\) \(+1\) \(-1\) \(+1\) \(-1\) \(+1\)

The spatial period is two particle spacings. If neighbouring points are connected as a visual guide, they form a zigzag.

This pattern fits an even ring exactly. For odd finite \(N\), the largest allowed wave number is \(\pi\pm\pi/N\), and

\[ \omega_{\max} =2\omega_o\cos\left(\frac{\pi}{2N}\right). \]

Its displacement pattern alternates in sign with a slowly varying envelope. For large \(N\), the neighbouring-displacement ratio is approximately \(-1\) away from envelope nodes.

(v)

First replace just one mass by \(m'\ll m\). A rapid vibration of this light mass causes comparatively little motion of its heavy neighbours. To leading order, treat those neighbours as fixed.

Displacing the light mass by \(x\) produces a restoring force \(-kx\) from each spring. Therefore

\[ m'\ddot x\simeq-2kx, \]

giving

\[ \boxed{\omega_{\mathrm{defect}}\simeq\sqrt{\frac{2k}{m'}}.} \]

Relative to the original upper band edge,

\[ \frac{\omega_{\mathrm{defect}}}{2\omega_o} \simeq\sqrt{\frac{m}{2m'}}\gg1. \]

The approximation is self-consistent. For a neighbouring heavy mass with amplitude \(u_1\), the leading high-frequency balance is

\[ -m\omega_{\mathrm{defect}}^2u_1\simeq kx, \]

so

\[ \frac{u_1}{x}\simeq-\frac{m'}{2m}. \]

Its motion is small and opposite to that of the light mass. Motion farther away is smaller still: the high-frequency mode is localized around the defect.

Thus one mode moves far above the original frequency band. The other frequencies are perturbed but remain on the scale \(\sqrt{k/m}\); in an infinite chain the extended-mode band remains \(0\leq\omega\leq2\omega_o\). The zero-frequency rigid-displacement mode also remains.

For an alternating chain, the repeated light masses give a whole high-frequency band. The spectrum separates into:

  • An acoustic band, containing low-frequency collective motion. At long wavelengths, neighbouring heavy and light masses move approximately together.
  • An optical band, in which the light masses oscillate rapidly, with much smaller motion of the heavy masses. At long wavelengths, the two types of mass move in opposite directions.

The band edges can be derived directly. Let \(x_j\) be the displacement of heavy mass \(j\), and \(y_j\) that of the light mass between \(x_j\) and \(x_{j+1}\). Then

\[ \begin{aligned} m\ddot x_j&=k(y_{j-1}+y_j-2x_j),\\ m'\ddot y_j&=k(x_j+x_{j+1}-2y_j). \end{aligned} \]

For a mode with phase change \(Q\) per two-particle cell, use

\[ \begin{aligned} x_j&=X\cos(jQ+\phi)\cos\omega t,\\ y_j&=Y\cos\left((j+\tfrac12)Q+\phi\right)\cos\omega t. \end{aligned} \]

The amplitude equations are

\[ \begin{aligned} (2k-m\omega^2)X&=2k\cos(Q/2)Y,\\ (2k-m'\omega^2)Y&=2k\cos(Q/2)X. \end{aligned} \]

A nonzero solution requires

\[ (2k-m\omega^2)(2k-m'\omega^2) =4k^2\cos^2(Q/2), \]

or

\[ mm'\omega^4-2k(m+m')\omega^2 +4k^2\sin^2(Q/2)=0. \]

Its two roots are

\[ \omega_\pm^2 = k\left(\frac1m+\frac1{m'}\right) \pm k\sqrt{ \left(\frac1m+\frac1{m'}\right)^2 -\frac{4}{mm'}\sin^2(Q/2) }. \]

As \(|Q|\) increases from \(0\) to \(\pi\), the lower root increases and the upper root decreases. For \(m'<m\), their ranges are therefore

\[ \boxed{ 0\leq\omega_-\leq\sqrt{\frac{2k}{m}}, } \]

and

\[ \boxed{ \sqrt{\frac{2k}{m'}} \leq\omega_+\leq \sqrt{2k\left(\frac1m+\frac1{m'}\right)}. } \]

There are no modes in the gap

\[ \boxed{ \sqrt{\frac{2k}{m}} <\omega< \sqrt{\frac{2k}{m'}}. } \]

For \(m'\ll m\), the optical band is narrow relative to its high frequency, approximately \(\sqrt{2k/m'}\), and is widely separated from the acoustic band. A finite periodic chain has discrete frequencies within these bands; they become continuous in the infinite-chain limit.

The solution was checked against the official solution.