Hooke's Law

Restorative force of a spring opposes the force of gravity pulling a mass downward [2].

Hooke's law is an empirical physical law describing the linear relationship between the restorative force exerted by a spring and the distance by which the spring is displaced from its equilibrium length. A spring which obeys Hooke's law is said to be Hookean. In addition to springs, Hooke's law is often a good model for arbitrary physical systems that exhibit a tendency to return to a state of equilibrium quickly after perturbation.

If a Hookean spring is compressed or extended by some displacement $x$ from equilibrium, the spring will exert a force proportional to this displacement in the opposite direction:

$$F = -kx.$$

The proportionality constant $k$, called the spring constant, is dependent on the stiffness of the spring, which in turn depends on its shape and the material from which the spring is made.

How can Hooke's law be used to determine the mass of an object?

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Given a Hookean spring of spring constant $k$, fix one end of the spring to the ceiling and the other end to the object. In equilibrium, the spring force will balance the downward force of gravity on the object, which allows computation of the mass $m$ from the displacement $x$ of the spring.

Gravity exerts a force $F_g = mg$ downward proportional to the mass $m$ of the object, which is perfectly matched by the spring force $F_s = -kx$ in equilibrium, where the negative sign indicates that the spring force acts in the opposite direction. The mass is obtained by setting these forces equal:

$$m = \frac{kx}{g}.\ _\square$$

The linear relationship of Hooke's law empirically holds only for small displacements $x$. For large deformations, a spring or other Hookean material can be permanently distorted and exhibit a nonlinear restorative force.

In one dimension, the potential energy of a spring can be obtained from Hooke's law by integration:

$$U = -\int F \,dx = \int kx\,dx = \frac12 kx^2.$$

If a spring is compressed or extended and then released, it will oscillate (indefinitely if friction is neglected; otherwise, the spring will eventually return to equilibrium). The oscillation of a mass $m$ on a spring can be derived directly from Newton's second law of motion, $F = ma$. Since acceleration is the second derivative of velocity, setting the force equal to the spring force $F = -kx$ yields

$$m\ddot{x} = ma = F = -kx.$$

So the equation of motion of the mass on the spring is

$$\ddot{x} + \frac{k}{m} x = 0,$$

where dots indicate time derivatives. The coefficient $\frac{k}{m}$ is often denoted as $\omega_0^2$, the square of the natural frequency of oscillation of the spring. This is because the general solution to this equation of motion is

$$x(t) = A \cos (\omega_0 t) + B \sin (\omega_0 t)$$

for some constants $A$ and $B$ depending on initial conditions.

The equation of motion above is often called the equation of motion of the simple harmonic oscillator, and systems obeying a similar equation of motion to the above are therefore said to exhibit simple harmonic motion.

Mass $M$ suspended by two springs of spring constant $k$.

A mass $M,$ suspended by two springs each of spring constant $k$ as in the diagram, is compressed upwards a displacement $L$ from the equilibrium length of the springs and allowed to fall under the influence of gravity. Find the subsequent displacement as a function of time.

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The mass obeys the equation of motion of the simple harmonic oscillator where the coefficient $k$ is replaced by $2k$ since there are two spring forces acting directly on the mass. The displacement $x$ therefore obeys the equation

$$x(t) = A \cos \left(\sqrt{\frac{2k}{m}} t \right) + B\sin \left(\sqrt{\frac{2k}{m}} t \right)$$

for constants $A$ and $B$ to be fixed by initial conditions. These initial conditions are

$$x(0) = L, \quad \dot{x} (0) = 0.$$

Plugging into the general solution above, these initial conditions yield

$$x(0) = A = L, \quad \dot{x} (0) = B\sqrt{\frac{2k}{m}} = 0 \implies B = 0.$$

The solution is thus uniquely fixed to be

$$x(t) = L\cos \left(\sqrt{\frac{2k}{m}}t \right).\ _\square$$

Show that a circuit with an inductor and capacitor, called an LC circuit, obeys the simple harmonic oscillator equation.

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According to Kirchoff's laws, the voltage around a closed loop must sum to zero. The voltage across the capacitor is $V_c = Q/C,$ where $C$ is the capacitance and $Q$ is the charge stored on the capacitor. The voltage across the inductor is $V = L\frac{dI}{dt},$ where $I$ is the current in the circuit and $L$ is the inductance. Since the current $I = \frac{dQ}{dt}$ is the time rate of change of the charge on the capacitor, Kirchhoff's loop rule reduces to

$$0 = -L \frac{dI}{dt} - \frac{Q}{C} \implies L\ddot{Q} + \frac{1}{C} Q = 0.$$

This is the simple harmonic oscillator equation where the inductance $L$ plays the role of the mass and the inverse of the capacitance $\frac{1}{C}$ plays the role of the spring constant. The linear voltage response with increasing charge stored on the capacitor is analogous to Hooke's law, with the rate of change of current through the circuit analogous to the acceleration. $_\square$

The solution to the equations of motion above, in terms of cosines and sines, will continue to oscillate forever. However, real oscillators eventually return to a stable equilibrium. This is because real oscillators feel damping forces which remove energy from the system.

Almost any physical system with a stable equilibrium state is described well by Hooke's law when it is displaced slightly from equilibrium. To see why, it is useful to consider potential energy diagrams, which graphically display the potential energy of a system in different states.

In blue, a potential energy diagram for some physical system. In red, the harmonic oscillator potential approximation about the stable equilibrium point.

The above diagram plots in blue the potential energy $U (x)$ of some system as a function of location $x$. A state of sufficiently low total energy in this system will not have enough kinetic energy to escape the potential well around the minimum of $U(x)$. The minimum $x_0$ of $U(x)$ is thus an equilibrium state. Futhermore, it is stable: if the system starts at $x_0$ and is perturbed slightly, it will tend to exert a restorative force towards $x_0$.

As a result, Taylor expanding the potential energy about the minimum yields

$$U(x) \approx U(x_0) + \left. \frac12 (x-x_0)^2 \frac{d^2 U}{dx^2} \right|_{x=x_0}.$$

The first term, $U(x_0)$, is just a constant energy shift. The quadratic term is the lowest-order term that describes how $U$ changes with $x$. This quadratic potential exactly matches the form of the spring potential $U_s = \frac12 kx^2$. Therefore, near the equilibrium point $x_0$, the behavior of physical systems can be well approximated by simple harmonic motion, i.e. a force obeying Hooke's law. Above, the quadratic approximation to a potential at its minimum is plotted in red.

For a system with multiple interacting masses, it is useful to define the reduced mass $\mu$ by

$$\frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2} + \cdots.$$

The frequency of oscillation $\omega_0$ about the minimum of a potential is then, in analogy to the Hookean spring,

$$\omega_0^2 = \sqrt{\frac{1}{\mu} \left. \frac{d^2 U}{dx^2} \right|_{x=x_0}}.$$

The Lennard-Jones 6,12 potential is used to model the interactions between two neutral atoms. The potential contains an attractive $\frac{1}{r^6}$ term for the van der Waals interaction and a repulsive $\frac{1}{r^{12}}$ term that models the exchange force due to the Pauli exclusion principle:

$$U = \epsilon \left[ \left(\frac{r_m}{r} \right)^{12} - 2\left( \frac{r_m}{r} \right)^6 \right],$$

where $\epsilon$ controls the depth of the potential well and $r_m$ is the minimum of the potential well. Find the frequency of oscillation about equilibrium for two atoms of mass $m$ whose interactions are modeled by this potential.

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The Taylor expansion of $U (r)$ about $r_m$ to second order is

$$U(r) \approx -\epsilon + 36 \frac{\epsilon}{r_m^2} (r-r_m)^2.$$

The reduced mass of the system is $\mu = \frac{m}{2}$. The frequency of oscillation about equilibrium is therefore

$$\omega_0^2 = \sqrt{\frac{2}{m} 72 \frac{\epsilon}{r_m^2}} = \frac{12}{r_m} \sqrt{\frac{\epsilon}{m}}.\ _\square$$

[1] D. Klepper and R. Kolenkow, An Introduction to Mechanics. McGraw-Hill, 1973.

[2] Image from https://upload.wikimedia.org/wikipedia/commons/5/5a/Mass-spring-system.png under Creative Commons licensing for reuse and modification.