# 2.4: Harmonic Oscillator Model

## Construct Definitions and Relationships

#### In physics, many phenomena behave cyclically. Springs are the prototypical example of this.

**Elastic Object: **An object, which when deformed (perhaps by squeezing it or stretching it) returns to its original shape.

**Spring**: A coiled spiral shaped object that is constructed to be elastic over a large range of deformation. Some springs can only be stretched, but if the coils are initially separated, the spring is elastic when both stretched or compressed.

**Spring-mass**: A spring with an object whose mass is considerably greater than the mass of the spring attached to one end of the spring.

**Hanging spring-mass: **A spring-mass with one end solidly attached to a support in a manner that the spring hangs vertically with the mass at the bottom of the spring.

Before the mass is attached to the spring, the bottom of the spring (the “free end”) will be at a definite position with respect to the attached end. This is the equilibrium position of the free end without the mass attached. When the mass is attached, the free end of the spring will move to a new, lower position. This is the equilibrium position of the free end of the spring with the mass attached.

### Assumptions and Simplifications:

**For the sake of being able to perform calculations, we assume that springs have negligible mass, and always have a mass attached at one end.** These assumptions are obviously violated frequently in the real world, but without them even basic problems become very difficult. For example, finding the kinetic energy of a single mass in motion is trivial, but finding the kinetic energy of a spring alone, whose end is always moving faster than its base, becomes a daunting task, and detracts from the learning of physics principles. You'll be comforted to know that these simplifications yield realistic results in most cases involving spring-masses, since the spring's mass often *is *negligible compared to the weight at the end.

*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, \(x\).*

Hooke's Law

The force with which a spring pulls back when stretched (or pushes back when compressed) is proportional to the amount of stretch from equilibrium, provided the spring is not stretched too far. (Historically, this linear proportionality between the force and amount of stretch is referred to as Hooke’s law behavior.) We write the force with which the spring pulls back (the restoring force) as

\[F = -k x\]

where **k **is the “spring constant” or “force constant” (and is dependent on the stiffness of the particular spring). The minus sign indicates that the force is opposite in direction to the direction it was stretched or compressed.

An important point to notice in the expression for the force of the spring is that x is measured from the un-stretched position of the free end of the spring.

In order for the force to have units of newtons, the units of k must be newtons per meter. The force that you (an external agent) have to exert on the spring to stretch it a distance \(x\) is in the opposite direction to the restoring force and is equal to \(+k x\).

Hooke's Law breaks down at the extremes of a spring's motion. For example, when stretched to the point of breaking or permanent deformation, a spring's behavior will begin to deviate substantially from the linear expectation. Also, when compressed so far that it begins to touch itself, clearly the forces at play change. For these reasons, it is usually assumed that springs stretch only within a small portion of their maximum deformation.

Example: Deducing a Spring Constant

Suppose a spring stretches 13 cm when a force of 780 N is applied to it. What is it's spring constant?

Solution

To find the spring constant, model the spring as a Hooke's law spring and solve for k.

\[ F = kx \]

\[ k = \frac{F}{x} \]

Convert from cm to m: 13 cm = 0.13 m

\[ k = \frac{780 N}{0.13m} = 6000 N/m \]

The minus sign in the force equation is dropped because we are referring to the force *we* exert, not the force the spring exerts. The force is normally negative to indicate that the spring pulls opposite the direction of its deformation.

**Spring-Mass Force and Force Constant for a Hanging Mass**

A spring with a mass hanging on it acts exactly like a horizontal spring, except that the end of the spring has a different equilibrium position. There are now two forces acting on the mass: the force from the spring pulling up or pushing down and the force from the Earth always pulling down. We will not show this here, but the combination of the two forces is completely accounted for if we measure all stretches and compressions of the spring from the equilibrium point of the free end of the spring with the mass attached.

The force with which a spring-mass pulls back when stretched (or pushes back when compressed) is proportional to the amount of stretch from the equilibrium determined with the mass attached (provided the spring is not stretched too far). We will usually write the force with which the spring-mass pulls back (the restoring force) using the symbol “y” instead of with an “x”, but this is just a convention.

\[F = -k y\]

where k is the “spring constant” or “force constant” (dependent on the stiffness of the particular spring, not on the mass). The minus sign indicates that the force is opposite to the direction the spring-mass was stretched or compressed. The force that you (an external agent) have to exert on the spring-mass to stretch it a distance y is in the opposite direction to the restoring force and is equal to \(+k y\).

A critically important point to notice in the expression for the force of the spring-mass is that y is measured from the unstretched position of the free end of the spring-mass system. By measuring the stretch from this “new” equilibrium position, the effect of the force of the Earth is automatically taken into account, so does not have to be added back in. It is as if, we are in far outer space where there is no force of gravity. The only force is the “modified’ force of the spring.

### Energies of a Spring-Mass

The work I do when I pull on the mass increases the potential energy of the mass/spring system. I can calculate this work: it is simply the product of the force I apply times the distance through which I move while pulling or pushing. The only tricky part is that the force is not constant, since \(F = -kx\). That is, the force is proportional to the distance I have pushed it or pulled it. So I need to take the average force times the distance. The force varies from zero to a max of ky. The average of this is \(\dfrac{1}{2}ky\). When I multiply this average force times the distance, I get the work I do on the system:

\[W = ( \text{average force}) \times ( \text{distance}) \]

\[ = (\dfrac{1}{2}kx)(x) = \dfrac{1}{2}kx^{2}\]

Notice that while the force scales linearly with deformation, this tells us that the work required to deform the spring scales as the square of deformation.

Potential Energy of a spring-mass

The expression in the preceding paragraph represents the work I did on the system, so by energy conservation the potential energy of the spring must have changed by that same amount. It doesn’t matter how the spring got stretched (that is, whether it was extended or compressed). If stretched by an amount *x*, it must have a PE relative to the equilibrium position (with mass attached) of

\[\text{PE spring-mass} = \dfrac{1}{2}kx^{2}\]

The **change in the spring-mass potential energy, ∆PE _{ spring-mass} **can be found using the

*Energy-Interaction Model*in the standard way. Let’s assume in this example that I compress or stretch the spring further from its equilibrium position than when it was in its initial position, y

_{i}, to a final position y

_{f}. Therefore, I am doing positive work on the spring-mass system. That is, I am adding energy to the spring-mass system. (The work I do on the spring-mass system will be a positive number of joules.) Thus, its energy must increase. If I do not change its KE, i.e., do not change its velocity (the final velocity is equal to its initial velocity), the only energy system that changes is the PE

_{ spring-mass}energy system.

Kinetic energy of a spring-mass

The kinetic energy of the spring-mass system is the same as for any mass that has a speed, v.

\[\text{KE spring-mass} = \dfrac{1}{2}mv^{2}\]

where m is the mass of the object attached to the spring and v is the speed of the mass.

The change in the KE spring-mass is calculated the same way as the change of any energy:

\[\Delta KE = KE_{f} - KE_{i} = \dfrac{1}{2}m(v_{f}^{2} - v_{i}^{2}) \]

Example: Applying Energy Conservation to a Spring-Mass

A spring of negligible mass and a spring constant of 120 N/m is fixed to a wall, free to oscillate. On the other end, a ball with a mass of 1.5 kg is attached. The spring-mass is then stretched 0.4 m, and released. What is the top speed of the attached ball?

**Solution**

Upon stretching the spring, some energy is stored in the springs' bonds as potential energy. This potential energy is released when the spring is allowed to oscillate. The maximum speed is accomplished when the spring returns to its equilibrium position, and all energy is kinetic energy.

First, calculate the amount of potential energy the spring initially stores:

\[ PE_{i} = \frac{1}{2} k x_{i}^{2} = \frac{1}{2} (120 N/m)(0.4m)^{2} = 9.6 J \]

When the spring reaches it's maximum speed, all energy which was potential is then kinetic. Use this to find the mass' top speed:

\[ PE_{max} = KE_{max} = \frac{1}{2}mv_{max}^{2} \]

\[ v_{max} = \sqrt{ \frac{2KE}{m} } = \sqrt{ \frac{2 \times 9.6 J}{1.5 kg} } = 3.58 m/s \]

### Spring-Mass Systems: a Uni

versal MotionMost physical systems that vibrate back and forth do so like a hanging spring-mass, particularly when the amplitude of vibration is not too large. Because this motion is so common, it is worth looking at it a little closer.

The maximum value of the PE of this spring-mass system occurs when the mass is at its extreme positions and its speed is zero. Conversely, the KE of the spring-mass system is a maximum when the mass is at its equilibrium position (y = 0). We know from the last chapter that in the absence of friction, ∆E_{tot} = ∆KE + ∆PE = 0, or in words, E_{tot} is a constant. Looking at E_{tot} at any particular time in its cycle of vibration, the energy is still going to be equal to the same total value. Written symbolically, this becomes:

\[E_{tot} = PE + KE = constant = PE_{max} = KE_{max}\]

The graph below shows the kinetic energy K and potential energy U of a spring-mass system as a function of the position from equilibrium. The kinetic energy as a function of y, K(y), is just the difference between the maximum energy, Emax, and U(y). This is just another way of expressing conservation of energy for this system. Both PE(y) and KE(y) are parabolas, centered about y = 0. The shape is due to the dependence on the square of y in the expression of the PE.

Although we cannot show it now without investigating the time behavior of the motion of the mass, it turns out that the *time-average* (that is, the average over time) of the PE and KE are the same, and are consequently both equal to one-half the total energy.

\[\text{avg } KE_{spring-mass} = \text{ avg } PE_{spring-mass} = \dfrac{1}{2}E_{total} \]

The fact that the time-average potential and kinetic energies are the same has a profound implication for the model of matter that we are about to develop in Chapter 3. This result is true only for a potential energy that depends on the square of the variable. It is precisely the fact that the potential energy is *quadratic* with respect to position that makes a spring-mass system so special, so universal, and so important.

The lowest point on the PE(*y*) curve is frequently called the potential energy minimum. Why is the value of the position variable for which the potential energy is a minimum significant? Consider what happens as energy is removed from a system which is oscillating about the equilibrium value of its position (e.g., a real spring-mass system, because of friction, gradually transfers mechanical energy to thermal energy). The amplitude (maximum extent) of the oscillations decreases until eventually, when all mechanical energy has been transferred to thermal systems, the system comes to rest at the equilibrium position, the position of the potential minimum.

Now consider *any* physical system that oscillates and which will settle down to a stable position as energy is transferred to thermal systems. That stable equilibrium position represents the physical state where potential energy is smallest. (All of the PE and KE have been transferred to thermal systems.) The potential energy must increase as y increases (either positively or negatively) away from zero, the equilibrium position. Now nature seems to prefer smooth changes of things like potential energies. The simplest smooth mathematical function that increases for both positive and negative y is y^{2}. When we look at real physical systems, it turns out that sufficiently close to the minimum, the potential energy always “looks parabolic”! This is a result with far reaching consequences. It implies that ** any oscillating system behaves just like our simple spring-mass system, at least for small amplitudes of oscillation! **What’s important is not that the spring-mass system is so special itself; it is, rather, that the behavior of the spring-mass system represents a

*truly universal behavior of any oscillating system.*We will use this property shortly when we model the real atoms of liquids and solids as “oscillating masses and springs.”

### Contributors

Authors of Phys7A (UC Davis Physics Department)