6.2 Small oscillations and coupled oscillators

Equilibrium

Consider a simple system where all forces are monogenic and there is only one (cartesian) degree of freedom.

\begin{align} L=\frac{1}{2}m\dot q^2-V(q) \tag{6.49}\end{align}

The system will be in static equilibrium if the virtual work done by the monogenic forces is zero. This will be true if the generalized force is zero:

\begin{align} Q_1=-\die{V}{q}=0 \tag{6.50}\end{align}

Such a condition implies that we have an extremum of \(V\). If we then expand the potential in a Taylor series near a point \(q_0\), where the potential is at the extremum, we have:

\begin{align} V &= V(q_0)+\die{V}{q}(q-q_0)+\frac{1}{2}\die{^2V}{^2q}(q-q_0)^2+\dots\\ &\sim \frac{1}{2}k(q-q_0)^2\\ k&\equiv \die{^2V}{^2q} \tag{6.51}\end{align}

which is the potential energy of a simple harmonic oscillator. We can always choose to define \(V\) such that it is zero at the point where the forces are zero (alternatively, the first term is a constant and does not affect the motion).

An equilibrium is “stable” if, when the system is displaced from equilibrium, it returns to that point (or oscillates around it) (e.g. a ball at the bottom of a bowl). The equilibrium is called “unstable” if a small perturbation of the system away from equilibrium will lead the system go further from the equilibrium point (e.g. a ball at the top of a hill).

The system will be in static equilibrium at \(q_0\). The nature of the equilibrium (stable or unstable) will depend on the sign of \(k\). If \(k\) is positive, then the point of equilibrium is a minimum of \(V\) and the particle will be bounded by points where \(E=V\). If \(k\) is negative, then the particle will see a potential barrier at \(q_0\), which it can only surmount if \(E>0\), otherwise, it will turn around when \(E=V\).

Simple harmonic oscillator

The simple harmonic oscillator Lagrangian is given by:

\begin{align} L=\frac{1}{2}m\dot q^2-\frac{1}{2}kq^2 \tag{6.52}\end{align}

where the system is in equilibrium at \(q=0\). Since \(L\) does not depend explicitly on time, energy is conserved, and we can solve for the integral (assuming that \(k\) and \(E\) are positive):

\begin{align} E&=\frac{1}{2}m\dot q^2+\frac{1}{2}kq^2\\ \dot q&=\sqrt{\frac{2E-kq^2}{m}}=\sqrt{\frac{2E}{m}}\sqrt{1-\frac{k}{2E}q^2}\\ \therefore t&=\sqrt{\frac{m}{2E}}\int\frac{dq}{\sqrt{1-\frac{k}{2E}q^2}}\\ t&=\sqrt{\frac{m}{2E}}\sqrt{\frac{2E}{k}}\sin^{-1}\left(\sqrt{\frac{k}{2E}}q\right)+\text{const.}\\ &=\sqrt{\frac{m}{k}}\sin^{-1}\left(\sqrt{\frac{k}{2E}}q\right)+\text{const.}\\ \therefore q(t)&=A\sin\left( \sqrt{\frac{k}{m}}t+\phi_0 \right)\\ &=A\sin\left( \omega t+\phi_0 \right) \tag{6.53}\end{align}

where we have introduced the frequency \(\omega\) and the constants \(A\) and \(\phi_0\) are determined from the initial conditions. Note that if \(k\) is positive, then \(E\) must also be positive. If \(k\) is negative, the solution is a hyperbolic sine function.

We could also solve this from the Lagrange equation of motion, by postulating a solution:

\begin{align} m\ddot q &=-kq \\ \therefore q(t) &= Ae^{-i\omega t}\\ \omega&=\sqrt{\frac{k}{m}} \tag{6.54}\end{align}

where \(q(t)\) is the real part of the expression.

Many oscillators

Suppose that we have \(n\) degrees of freedom \(q_i\) and a Lagrangian of the form:

\begin{align} L&=T-V\\ T&=\frac{1}{2}\sum_{i,j} T_{ij}\dot{q}_i\dot q_j \tag{6.55}\end{align}

Suppose that the point (\(q_{0,1}\), … ,\(q_{0,n}\)) corresponds to a static equilibrium:

\begin{align} \left(\die{V}{q_i}\right)\biggr\rvert_0=0 \tag{6.56}\end{align}

Consider a point, \(q_i'\), that is slightly displaced from the equilibrium by a distance, \(\eta_i\):

\begin{align} q_i'=q_{0,i}+\eta_i \tag{6.57}\end{align}

The potential at \(q'_i\) can be expanded in a Taylor series near that point:

\begin{align} V(q_1',\dots ,q_n')&=V(q_{0,1}, \dots , q_{0,n})+\sum_i\left(\die{V}{q_i}\right)\biggr\rvert_0\eta_i+\frac{1}{2}\sum_{i,j}\left(\die{V}{q_i}\die{V}{q_j}\right)\biggr\rvert_0\eta_i\eta_j+\dots\\ &\sim\frac{1}{2}\sum_{i,j}\left(\die{V}{q_i}\die{V}{q_j}\right)\biggr\rvert_0\eta_i\eta_j\\ &=\frac{1}{2}\sum_{i,j} V_{ij}\eta_i\eta_j \tag{6.58}\end{align}

where we have dropped the first two terms, and introduced \(V_{ij}\) as the second order derivatives evaluated at the equilibrium. The first term is a constant and does not influence the motion, and the second is zero at the equilibrium. The kinetic energy can also be written in terms of the small displacements from equilibrium, \(\eta_i\):

\begin{align} \eta_i&=q'_i-q_{0,i}\\ \therefore \dot\eta_i&=\dot q_i'\\ \therefore T&=\frac{1}{2}\sum_{i,j} T_{ij}\dot\eta_i\dot\eta_j \tag{6.59}\end{align}

and the Lagrangian, near equilibrium, can be written as:

\begin{align} L=\frac{1}{2}\sum_{i,j} T_{ij}\dot\eta_i\dot\eta_j-\frac{1}{2}\sum_{i,j}V_{ij}\eta_i\eta_j \tag{6.60}\end{align}

Taking the Lagrange equation for \(\eta_i\), we obtain:

\begin{align} \sum_{j}\left(T_{ij}\ddot\eta_j+V_{ij}\eta_j\right)=0 \tag{6.61}\end{align}

where we have used the fact that \(V_{ij}=V_{ji}\). Again, we can postulate a solution of the form:

\begin{align} \eta_i(t)=Ca_ie^{-i\omega t} \tag{6.62}\end{align}

where the \(a_i\) correspond to each \(\eta_i\), \(C\) is an overall (complex) constant, and only the real part is used to represent the motion. This solution is called a “normal mode”, as all coordinates oscillate with the same frequency. If we substitute this back into the Lagrange equation:

\begin{align} \sum_{j}\left(-\omega^2 T_{ij}\eta_j+V_{ij}\eta_j\right)=0\\ \sum_{j}a_j(V_{ij}-\omega^2T_{ij})=0 \tag{6.63}\end{align}

Consider this as a matrix equation:

\begin{align} \left(\begin{array}{ccc} V_{11}-\omega^2T_{11}&\cdots &V_{1n}-\omega^2T_{1n} \\ \vdots &\ddots &\vdots \\ V_{n1}-\omega^2T_{n1}&\cdots &V_{nn}-\omega^2T_{nn} \\ \end{array}\right) \left( \begin{array}{c} a_1\\ \vdots \\ a_n\\ \end{array} \right)& =0 \tag{6.64}\end{align}

We can re-write it as:

\begin{align} V\vec{a}-\omega^2T\vec{a}&=0\\ V\vec{a}&=\omega^2T\vec{a}\\ T^{-1}V\vec{a}&=\omega^2\vec{a} \tag{6.65}\end{align}

where we have treated \(T\) and \(V\) as matrices. The last step required \(T\) to be invertible. Note that \(V\) is a symmetric matrix. You should recognize that this is a characteristic eigenvalue equation (\(\vec{a}\) is an eigenvector and \(\omega^2\) is the corresponding eigenvalue).

The particular solution that we postulated corresponds to the “normal modes” of the system. These normal modes have characteristic frequencies and characteristic displacement vectors.

Example 6-2

Find the linear normal modes of vibration for three equal masse, \(m\) connected by two springs with spring constant \(k\)

Since we only consider the linear modes, the masses are constrained to move along a single direction and we thus have 3 degrees of freedom. The position of the three masses are labeled as \(\eta_i\) and correspond to their displacement from the rest position. The Lagrangian is thus:

\begin{align*} L&=\frac{1}{2}m(\dot\eta_1^2+\dot\eta_2^2+\dot\eta_3^2)-\frac{1}{2}k\left((\eta_2-\eta_1)^2+(\eta_3-\eta_2)^2 \right)\nonumber\\ &=\frac{1}{2}m(\dot\eta_1^2+\dot\eta_2^2+\dot\eta_3^2)-\frac{1}{2}k(2\eta_2^2-2\eta_2\eta_1+\eta_1^2+\eta_3^3-2\eta_2\eta_3) \end{align*}

The matrices are thus:

\begin{align} T&= \left( \begin{array}{ccc} m &0 &0\\ 0 &m &0\\ 0 &0 &m\\ \end{array} \right) \\ T^{-1}&= \left( \begin{array}{ccc} \frac{1}{m} &0 &0\\ 0 &\frac{1}{m} &0\\ 0 &0 &\frac{1}{m}\\ \end{array} \right) \\ V&= \left( \begin{array}{ccc} k &-k &0\\ -k &2k &-k\\ 0 &-k &k\\ \end{array} \right) \end{align}

Hence, we wish to find the eigenvalues and vectors of:

\begin{align*} \left( \begin{array}{ccc} \frac{k}{m} &-\frac{k}{m} &0\\ -\frac{k}{m} &\frac{2k}{m} &-\frac{k}{m}\\ 0 &-\frac{k}{m} &\frac{k}{m}\\ \end{array} \right) \end{align*}

The three eigenvalues are:

\begin{align*} \omega_1^2&=\frac{k}{m}\\ \omega_2^2&=\frac{3k}{2m}\\ \omega_3^2&=0 \end{align*}

and the three corresponding eigen vectors are:

\begin{align*} \vec{a}_1&=\left(\begin{array}{c} -1 \\0 \\1\\ \end{array} \right)\nonumber\\ \vec{a}_2&=\left(\begin{array}{c} 1 \\ -2 \\1\\ \end{array} \right)\nonumber\\ \vec{a}_3&=\left(\begin{array}{c} 1 \\ 1 \\1\\ \end{array} \right) \end{align*}

These solutions correspond to “equilibrium” solutions, or “normal modes”. The first one corresponds to the middle mass being fixed and the two masses on the sides moving at the same frequency in opposite directions. The second one corresponds to the two masses on the side moving in one direction and the center mass moving in the opposite direction. The last solution has a frequency of zero and all three masses moving in the same direction. This is in fact related to the total momentum of the system being conserved, and suggests that there is a better choice of coordinates that leads to only two degrees of freedom.

Problem 6-1: Two masses connected by a spring

Two masses \(m_1\) and \(m_2\) are connected by a spring with spring constant \(k\).
a)Write the Lagrangian for this system and show that it can be reduced to a single degree of freedom
b)Plot the effective potential for the system, and show that all orbits are bound
c)Determine the shape of the orbit, and plot it.

Problem 6-2: The Coulomb force

Two masses \(m_1\) and \(m_2\) with charges \(q_1\) and \(q_2\) interact through the Coulomb force.
a)Write the Lagrangian for this system and show that it can be reduced to a single degree of freedom
b)Plot the effective potential for the system when the force is attractive and describe the possible orbits
c)Plot the effective potential for the system when the force is repulsive and show that only hyperbolae are possible orbits
d)Plot the effective potential for the system when the force is repulsive but is constrained to act within a radius \(r<R\) (for example, the Coulomb force from a nucleus that is screened by electrons at a large distance) and describe the possible orbits.

Problem 6-3: Kepler’s problem

a) Show that the analytic treatment presented in this chapter is consistent with Kepler’s three laws
b) Show that for elliptic orbits, the major axis only depends on the energy of the system. c)Make a plot showing how the effective potential depends on angular momentum (show several curves for different values of angular momentum and comment).

Problem 6-4: Simple Harmonic Oscillator

a) Discuss the solutions for the simple harmonic oscillator for the cases when \(E\) and/or \(k\) are negative. If the solution exists, plot the position as a function of time.

Problem 6-5: Coupled oscillators

a) Show that in the case illustrated in example 6-2, the total momentum is conserved. In particular, show that the Lagrangian can be written with one less degree of freedom.