IPhO 1992 · Theory problem 2

The Longitudinal Motion of a Linear Molecule

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In this problem you will analyze the longitudinal motion of a linear molecule, i.e., the motion along the molecular axis. The rotational motion and the bending of the molecule are not considered. The molecule is assumed to consist of \(N\) atoms of mass \(m_1, m_2, \ldots, m_N\), respectively. Each atom is assumed to be connected to its neighbors by a chemical bond. Each bond is approximated by a massless spring which obeys Hooke’s law with spring constants \(k_1, k_2, \ldots, k_{N-1}\). The molecule is shown in Fig. 1.

Figure 1

Fig.1. A linear molecule with \(N\) atoms.

Use the following facts when solving this problem: The longitudinal vibrational motion of a linear molecule consists of a superposition of separate vibrational motions called normal vibrations, or normal modes. In a normal mode all atoms vibrate in simple harmonic motion with the same frequency and pass through their equilibrium positions simultaneously.

Questions

1) Let \(x_i\) be the displacement of atom \(i\) from its equilibrium position. Express the force \(F_i\) acting on each atom as a function of the displacements \(x_1, x_2, \ldots, x_N\) and the spring constants \(k_1, k_2, \ldots, k_{N-1}\). What relationship is there among the forces \(F_1, F_2, \ldots, F_N\)? Using this relationship, derive a relationship between the displacements \(x_1, x_2, \ldots, x_N\) and give a physical interpretation of this relationship.

2) Analyze the motion of a diatomic molecule AB (Fig. 2). The value of the spring constant is \(k\). Derive an expression for the forces acting on atoms A and B. Determine the possible types of motion of the molecule. Determine the corresponding vibrational frequencies and interpret the result. In particular, how is it possible for the atoms to vibrate with the same frequency even though their masses are not the same?

Figure 2

Fig.2. The diatomic molecule AB

3) Analyze the motion of the triatomic molecule \(\mathrm{BA_2}\) (Fig. 3)

Figure 3

Fig.3. The triatomic molecule \(\mathrm{BA_2}\).

Express the net force on each atom as a function of its displacement only. Deduce the possible motions of the molecule and the corresponding vibrational frequencies.

4) The frequencies of the two longitudinal modes of vibration of the \(\mathrm{CO_2}\) molecule are \(3.998\times10^{13}\ \mathrm{Hz}\) and \(7.042\times10^{13}\ \mathrm{Hz}\), respectively. Determine a numerical value for the spring constant of the CO bond.

How well do you think this approximation for the bond structure of the molecule describes the vibrational motion of the real molecule?

The atomic mass of the carbon atom = \(12\ \mathrm{amu}\) and that of the oxygen atom = \(16\ \mathrm{amu}\). The atomic mass unit = \(1.660\times10^{-27}\ \mathrm{kg}\).

Official English text of the problem.

Answers

  1. The forces are \[ \boxed{ \begin{aligned} F_1&=k_1(x_2-x_1),\\ F_i&=k_i(x_{i+1}-x_i)-k_{i-1}(x_i-x_{i-1}), &&2\leq i\leq N-1,\\ F_N&=-k_{N-1}(x_N-x_{N-1}). \end{aligned}} \] They satisfy \(\sum_iF_i=0\), hence \[ \sum_i m_i x_i=C_0+C_1t. \] The center of mass moves uniformly. For internal vibrations measured in its rest frame, \[ \boxed{\sum_i m_i x_i=0.} \]

  2. For AB, \[ \boxed{F_A=k(x_B-x_A),\qquad F_B=k(x_A-x_B).} \] There is rigid translation with zero frequency and one internal vibrational mode: \[ \boxed{ f=\frac{1}{2\pi}\sqrt{k\left(\frac1{m_A}+\frac1{m_B}\right)} =\frac{1}{2\pi}\sqrt{\frac{k}{\mu}}, \qquad \mu=\frac{m_Am_B}{m_A+m_B}.} \] In this vibration, \(m_Ax_A=-m_Bx_B\): the atoms move in opposite directions, with amplitudes inversely proportional to their masses.

  3. For \(\mathrm{BA_2}\), the normal modes are:

Motion Displacement ratios \(x_1:x_2:x_3\) Frequency
Rigid translation \(1:1:1\) \(0\)
Symmetric stretch \(1:0:-1\) \(\displaystyle f_s=\frac1{2\pi}\sqrt{\frac{k}{m_A}}\)
Asymmetric stretch \(\displaystyle 1:-\frac{2m_A}{m_B}:1\) \(\displaystyle f_a=\frac1{2\pi}\sqrt{k\left(\frac1{m_A}+\frac2{m_B}\right)}\)

In the symmetric stretch, \[ \boxed{F_1=-kx_1,\qquad F_2=0,\qquad F_3=-kx_3.} \] In the asymmetric stretch, \[ \boxed{ \begin{aligned} F_1&=-k\left(1+\frac{2m_A}{m_B}\right)x_1,\\ F_2&=-k\left(2+\frac{m_B}{m_A}\right)x_2,\\ F_3&=-k\left(1+\frac{2m_A}{m_B}\right)x_3. \end{aligned}} \] General motion is a superposition of the two vibrations and uniform translation.

  1. The two measured frequencies separately give \[ \boxed{k_s=1.676\times10^3\ \mathrm{N\,m^{-1}},\qquad k_a=1.418\times10^3\ \mathrm{N\,m^{-1}}.} \] Thus a representative approximate bond stiffness is \(k\approx1.5\times10^3\ \mathrm{N\,m^{-1}}\). No single \(k\) fits both frequencies exactly: the predicted frequency ratio is \(1.915\), whereas the observed ratio is \(1.761\). The model gives a reasonable approximation, with a discrepancy of about \(9\%\) in this ratio.

Solution

1) Forces and the center of mass

Take the same positive direction along the molecular axis for every displacement. The change in length of the spring joining atoms \(i\) and \(i+1\) is \[ \Delta \ell_i=x_{i+1}-x_i. \] The potential energy of the ideal Hookean springs is therefore \[ U=\frac12\sum_{i=1}^{N-1}k_i(x_{i+1}-x_i)^2. \] Using \(F_i=-\partial U/\partial x_i\) gives \[ F_1=k_1(x_2-x_1), \] \[ F_i=k_i(x_{i+1}-x_i)-k_{i-1}(x_i-x_{i-1}), \qquad 2\leq i\leq N-1, \] and \[ F_N=-k_{N-1}(x_N-x_{N-1}). \]

Each spring exerts equal and opposite forces on its two ends. Consequently, \[ \boxed{\sum_{i=1}^{N}F_i=0.} \] Newton’s second law then gives \[ \frac{d^2}{dt^2}\left(\sum_i m_ix_i\right)=0, \] so \[ \sum_i m_ix_i=C_0+C_1t. \] Writing \(M_{\mathrm{tot}}=\sum_i m_i\), the center-of-mass displacement is \[ X_{\mathrm{CM}}=\frac{\sum_i m_ix_i}{M_{\mathrm{tot}}}. \] Thus the center of mass has constant velocity, as expected for an isolated system with no external force.

For a nonzero-frequency normal mode, \[ x_i=a_i\cos(\omega t+\phi). \] Since \(\ddot x_i=-\omega^2x_i\), \[ 0=\sum_i m_i\ddot x_i=-\omega^2\sum_i m_ix_i. \] Hence \[ \boxed{\sum_i m_ix_i=0\qquad(\omega\ne0).} \] Internal vibration leaves the center of mass fixed. Uniform translation can be superposed independently.

2) The diatomic molecule AB

The spring extension is \(x_B-x_A\), so \[ m_A\ddot x_A=k(x_B-x_A),\qquad m_B\ddot x_B=-k(x_B-x_A). \]

Introduce the center-of-mass displacement and the relative displacement: \[ X=\frac{m_Ax_A+m_Bx_B}{m_A+m_B}, \qquad q=x_B-x_A. \] Adding the equations of motion gives \[ \ddot X=0. \] Subtracting the accelerations gives \[ \ddot q =-k\left(\frac1{m_A}+\frac1{m_B}\right)q. \] Therefore \[ X=X_0+Vt,\qquad q=Q\cos(\omega t+\phi), \] where \[ \boxed{\omega^2=k\left(\frac1{m_A}+\frac1{m_B}\right)=\frac{k}{\mu}}, \qquad \mu=\frac{m_Am_B}{m_A+m_B}. \]

The complete motion is \[ \boxed{ \begin{aligned} x_A&=X_0+Vt-\frac{m_B}{m_A+m_B}Q\cos(\omega t+\phi),\\ x_B&=X_0+Vt+\frac{m_A}{m_A+m_B}Q\cos(\omega t+\phi). \end{aligned}} \]

There are two independent types of motion:

  • Rigid translation: \(q=0\), so \(x_A=x_B=X_0+Vt\). The spring is undeformed and both forces vanish. This is the zero-frequency mode.
  • Internal vibration: in the center-of-mass rest frame, \[ m_Ax_A+m_Bx_B=0. \] The atoms oscillate in opposite phase with frequency \[ \boxed{f=\frac{\omega}{2\pi} =\frac1{2\pi}\sqrt{\frac{k}{\mu}}.} \]

To understand the equal frequencies despite unequal masses, eliminate the other atom’s displacement: \[ x_B=-\frac{m_A}{m_B}x_A, \qquad x_A=-\frac{m_B}{m_A}x_B. \] Then \[ F_A=-k\left(1+\frac{m_A}{m_B}\right)x_A, \qquad F_B=-k\left(1+\frac{m_B}{m_A}\right)x_B. \] Dividing each restoring-force coefficient by the corresponding mass gives the same value: \[ \frac{k(1+m_A/m_B)}{m_A} =\frac{k(1+m_B/m_A)}{m_B} =k\left(\frac1{m_A}+\frac1{m_B}\right). \] Thus both atoms have the same angular frequency. Their unequal masses are accommodated by unequal amplitudes: \[ \boxed{\frac{|x_A|_{\max}}{|x_B|_{\max}}=\frac{m_B}{m_A}.} \] The heavier atom has both the smaller displacement amplitude and the smaller acceleration amplitude.

3) The triatomic molecule \(\mathrm{BA_2}\)

Write \(m=m_A\) and \(M=m_B\). Hooke’s law gives \[ \boxed{ \begin{aligned} m\ddot x_1&=k(x_2-x_1),\\ M\ddot x_2&=k(x_1+x_3-2x_2),\\ m\ddot x_3&=k(x_2-x_3). \end{aligned}} \] Rigid translation, \(x_1=x_2=x_3=X_0+Vt\), makes all three forces zero.

For internal vibration, work in the center-of-mass rest frame: \[ m(x_1+x_3)+Mx_2=0. \] Define the average and half-difference of the endpoint displacements: \[ s=\frac{x_1+x_3}{2},\qquad d=\frac{x_1-x_3}{2}. \] Then \[ x_1=s+d,\qquad x_3=s-d,\qquad x_2=-\frac{2m}{M}s. \]

Adding and subtracting the endpoint equations gives \[ m\ddot s=k(x_2-s) =-k\left(1+\frac{2m}{M}\right)s, \] \[ m\ddot d=-kd. \] These are two independent harmonic oscillators, so they identify all internal normal modes.

Symmetric stretching mode. Set \(s=0\). Then \[ x_2=0,\qquad x_1=d,\qquad x_3=-d. \] The middle atom remains stationary, while the outer atoms move with equal amplitudes in opposite directions. Both bonds lengthen together and shorten together.

The forces, expressed using each atom’s own displacement, are \[ F_1=-kx_1,\qquad F_2=0,\qquad F_3=-kx_3. \] Therefore \[ \boxed{\omega_s^2=\frac{k}{m_A},\qquad f_s=\frac1{2\pi}\sqrt{\frac{k}{m_A}}.} \] The central mass does not enter this frequency because the two spring forces on it cancel throughout the motion.

Asymmetric stretching mode. Set \(d=0\). Then \[ x_1=x_3=s,\qquad x_2=-\frac{2m}{M}s. \] The outer atoms move together, and the middle atom moves oppositely. One bond lengthens while the other shortens.

For either outer atom, \[ F_1=k(x_2-x_1) =-k\left(1+\frac{2m}{M}\right)x_1, \] \[ F_3=-k\left(1+\frac{2m}{M}\right)x_3. \] For the middle atom, use \(x_1+x_3=-(M/m)x_2\): \[ F_2=k(x_1+x_3-2x_2) =-k\left(2+\frac{M}{m}\right)x_2. \] All three equations give \[ \boxed{\omega_a^2=k\left(\frac1{m_A}+\frac2{m_B}\right),\qquad f_a=\frac1{2\pi}\sqrt{k\left(\frac1{m_A}+\frac2{m_B}\right)}.} \]

These single-displacement force laws apply to the indicated normal modes. A general superposition obeys the original coupled force equations.

The complete motion can be written \[ \begin{aligned} \begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} ={}&(X_0+Vt)\begin{pmatrix}1\\1\\1\end{pmatrix}\\ &+A_s\cos(\omega_st+\phi_s) \begin{pmatrix}1\\0\\-1\end{pmatrix}\\ &+A_a\cos(\omega_at+\phi_a) \begin{pmatrix}1\\-2m_A/m_B\\1\end{pmatrix}. \end{aligned} \] Thus there are two longitudinal vibrational modes and one translational mode.

4) Numerical bond stiffness and assessment of the model

For \(\mathrm{CO_2}\), the outer atoms are oxygen and the central atom is carbon: \[ m_A=m_{\mathrm O} =16(1.660\times10^{-27}) =2.656\times10^{-26}\ \mathrm{kg}, \] \[ m_B=m_{\mathrm C} =12(1.660\times10^{-27}) =1.992\times10^{-26}\ \mathrm{kg}. \] Since \(f_a>f_s\), assign \[ f_s=3.998\times10^{13}\ \mathrm{Hz},\qquad f_a=7.042\times10^{13}\ \mathrm{Hz}. \]

From the symmetric mode, \[ k_s=m_{\mathrm O}(2\pi f_s)^2, \] so \[ \boxed{ k_s=(2.656\times10^{-26}) \left[2\pi(3.998\times10^{13})\right]^2 =1.676\times10^3\ \mathrm{N\,m^{-1}}.} \]

From the asymmetric mode, \[ k_a=\frac{(2\pi f_a)^2}{1/m_{\mathrm O}+2/m_{\mathrm C}}, \] so \[ \boxed{k_a=1.418\times10^3\ \mathrm{N\,m^{-1}}.} \]

These estimates place the bond stiffness near \[ \boxed{k\approx1.5\times10^3\ \mathrm{N\,m^{-1}}.} \] Their difference shows that the model cannot reproduce both measured frequencies with a single spring constant.

A direct consistency test, independent of \(k\), is the frequency ratio. The model predicts \[ \frac{f_a}{f_s} =\sqrt{1+\frac{2m_{\mathrm O}}{m_{\mathrm C}}} =\sqrt{1+\frac{32}{12}} =\sqrt{\frac{11}{3}} =1.915. \] The measured ratio is \[ \left(\frac{f_a}{f_s}\right)_{\mathrm{measured}} =\frac{7.042}{3.998} =1.761. \] For example, fitting the lower frequency predicts \[ f_a=1.91485(3.998\times10^{13}) =7.656\times10^{13}\ \mathrm{Hz}, \] which is \(8.71\%\) above the measured upper frequency.

The ideal-spring model therefore gives a reasonable approximate description of the two stretching motions and their frequency scale. Its quantitative accuracy is limited: the real molecular restoring forces can include coupling between the two bond extensions, and departures from Hooke’s law, which this model omits.

The solution was checked against the official solution.