Skip to main content
Chemistry LibreTexts

5.1: A Harmonic Oscillator Obeys Hooke's Law

The motion of two particles in space can be separated into translational, vibrational, and rotational motions. The internal motions of vibration and rotation for a two-particle system can be described by a single reduced particle with a reduced mass \(μ\) located at \(r\). For a diatomic molecule, Figure \(\PageIndex{1}\), the vector \(\vec{r}\) corresponds to the internuclear axis. The magnitude or length of \(r\) is the bond length, and the orientation of \(r\) in space gives the orientation of the internuclear axis in space. Changes in the orientation correspond to rotation of the molecule, and changes in the length correspond to vibration. The change in the bond length from the equilibrium bond length is the vibrational coordinate for a diatomic molecule.

Figure \(\PageIndex{1}\) Diagrams of the coordinate systems and relevant vectors for a) a diatomic molecule with atoms of mass \(m_1\) and \(m_2\) and b) the equivalent reduced particle of reduced mass \(\mu\)

A classical description of the vibration of a diatomic molecule is needed because the quantum mechanical description begins with replacing the classical energy with the Hamiltonian operator in the Schrödinger equation. It also is interesting to compare and contrast the classical description with the quantum mechanical picture.

The Classical Harmonic Oscillator

Simple harmonic oscillators about a potential energy minimum can be thought of as a ball rolling frictionlessly in a dish (left) or a pendulum swinging frictionlessly back and forth. The restoring forces are precisely the same in either horizontal direction.


Figure \(\PageIndex{2}\): Simple image of a ball oscillating in a potential. Image used with permission from Wikipedia.

Let's start with Newton's equation of motion

\[\vec{F}= m \vec{a} \label {5.1.1}\]

to obtain a classical description of how a diatomic molecule vibrates. In Equation \(\ref{5.1.1}\):

  • \(m\) is the mass,
  • \(\mu\) is the reduced mass of the molecule
  • \(a\) is the acceleration and is expressed as \(d^2Q/dt^2\), and
  • \(f\) is the force that pulls the molecule back to its equilibrium bond length (\(x_o\)).

If we consider the bond to behave like a spring, then this restoring force is proportional to the displacement from the equilibrium length (set to zero for convenience \(x_o=0\)); this is Hooke's Law

\[ F = - kx \label {5.1.2}\]

where \(k\) is the force constant. Hooke's Law says that the force is proportional to, but in opposite direction to, the displacement, \(x\). The force constant reflects the stiffness of the spring. The idea incorporated into the application of Hooke's Law to a diatomic molecule is that when the atoms move away from their equilibrium positions, a restoring force is produced that increases proportionally with the displacement from equilibrium. The potential energy for such a system increases quadratically with the displacement.

\[ V (x) = \dfrac {1}{2} k x^2 \label {5.1.3}\]

Hooke's Law or the harmonic (i.e. quadratic) potential given by Equation \(\ref{5.1.3}\) is an excellent approximation for the vibrational oscillations of molecules. The magnitude of the force constant \(k\) depends upon the nature of the chemical bond in molecular systems just as it depends on the nature of the spring in mechanical systems. The larger the force constant, the stiffer the spring or the stiffer the bond. Since it is the electron distribution between the two positively charged nuclei that holds them together, a double bond with more electrons has a larger force constant than a single bond, and the nuclei are held together more tightly.

A stiff bond with a large force constant is not necessarily a strong bond with a large dissociation energy.


Figure \(\PageIndex{3}\): A mass connected to a spring that follows Hooke's law is a system that, when displaced from its equilibrium position, experiences a restoring force, \(F\) that is proportional to the displacement, \(x\).

Example \(\PageIndex{1}\)

  1. Show that minus the first derivative of the harmonic potential energy function in Equation \(\ref{5.1.1}\) with respect to x is the Hooke's Law force.
  2. Show that the second derivative is the force constant, \(k\).
  3. At what value of \(x\) is the potential energy a minimum; at what value of x is the force zero?
  4. Sketch graphs to compare the potential energy and the force for a system with a large force constant to one with a small force constant.

The classical equation of motion for a one-dimensional simple harmonic oscillator with a particle of mass \(m\) attached to a spring having spring constant \(k\) is

\[ m \dfrac{d^2x(t)}{dt^2} = -kx(t) \label{5.1.4a}\]

which can be written in the standard wave equation form:

\[ \dfrac{d^2x(t)}{dt^2} + \dfrac{k}{m}x(t) = 0 \label{5.1.4b}\]

Equation \(\ref{5.1.4}\) is a linear second-order differential equation that can be solved by the standard method of factoring and integrating. The resulting solution to Equation \(\ref{5.1.4}\) is

\[ x(t) = x_o \sin (\omega t + \phi) \label{5.1.5}\]


\[\omega = \sqrt{\dfrac{k}{m}} \label{5.1.6}\]

and the momentum \(p = mv\) has time dependence

\[ p =mx_o \omega \cos (\omega t + \phi) \label{5.1.7}\]

Figure \(\PageIndex{4}\) is a graph showing the displacement of the bond from its equilibrium length as a function of time. Such motion is called harmonic.



Figure \(\PageIndex{4}\): Solution to the Harmonic Oscillator. Displacement (y-axis) is plotted as a function of time.

Example \(\PageIndex{2}\)

Substitute the following functions into Equation \(\ref{5.1.4b}\) to show that they are both possible solutions to the classical equation of motion.

\[x(t) = x_0 e^{i \omega t} \]


\[ x(t) = x_0 e^{-i \omega t}\]


\[ \omega = \sqrt {\dfrac {k}{\mu}}\]

Note that the Greek symbol \(\omega\) for frequency represents the angular frequency \(2π\nu\).

Example \(\PageIndex{3}\)

Show that sine and cosine functions also are solutions to Equation \(\ref{5.1.4b}\).

Example \(\PageIndex{4}\)

Identify what happens to the frequency of the motion as the force constant increases in one case and as the mass increases in another case. If the force constant is increased 9-fold and the mass is increased by 4-fold, by what factor does the frequency change?

The energy of the vibration is the sum of the kinetic energy and the potential energy. The momentum associated with the harmonic oscillator is

\[P_x = m \dfrac {dx}{dt} \label {5.1.8}\]

so the energy can be written as

\[ E = T + V = \dfrac {P^2_x}{2 m} + \dfrac {k}{2} x^2 \label {5.1.9}\]

Example \(\PageIndex{5}\)

What happens to the frequency of the oscillation as the vibration is excited with more and more energy? What happens to the maximum amplitude of the vibration as it is excited with more and more energy?

Example \(\PageIndex{6}\)

If a molecular vibration is excited by collision with another molecule and is given a total energy \(E_{hit}\) as a result, what is the maximum amplitude of the oscillation? Is there any constraint on the magnitude of energy that can be introduced?

Figure \(\PageIndex{5}\): The potential \(V(x)\) for a harmonic oscillator. Vertical axis energy; horizontal axis displacement \(x\). The potential energy \(V(x) = ½ kx^2\) is shown in red. Image used with permission from Wikipedia.

The energy of the harmonic oscillator is equal to the maximum potential energy stored in the spring when \(x = \pm A\), called the turning points. The variations of Kinetic Energy (\(T\)) and \(V\) with the displacement \(x\) is plotted in Figure \(\PageIndex{6}\). The total energy (Equitation \(\ref{5.1.9}\)) is continuously being shifted between potential energy stored in the spring and kinetic energy of the mass.

Figure \(\PageIndex{6}\): Kinetic energy and potential energy versus time for a simple harmonic oscillator with \(\phi = 0\). The Yellow curve is Kinetic energy and the blue is the potential. Notice the total energy is constant. Image used with permission from Wikipedia.