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# 8.5: Simple Harmonic Motion Model

Periodic or oscillatory motion is common throughout the universe, from the smallest to the largest distance and time scales. This kind of motion is related to the forces acting between objects by Newton’s 2nd law, just as all motions are. To have oscillatory motion, there must be a restoring force that acts in a direction to cause an object to return to its equilibrium position. A particularly common type of oscillatory motion results when the magnitude of the restoring force is directly proportional to the displacement of the object from equilibrium:

$F_{restoring} = -kx \label{Eq1}$

This is the force law characteristic of a spring that is stretched or compressed from its equilibrium position. The oscillatory motion that results from this force law is known as simple harmonic motion (SHM).

Even when the force law is not as simple as Equation $$\ref{Eq1}$$ for arbitrary values of $$x$$, it turns out that for an object that oscillates about an equilibrium position, this linear law provides an accurate description for small oscillations. Thus, we can make a very strong statement: essentially every system that vibrates, does so in SHM for small amplitudes of vibration.

Because it is so common, it is worth spending some effort understanding SHM and the different ways to represent it. Another reason for focusing on SHM is that periodic wave motion is the interconnected vibrations of many, many oscillators, each vibrating in SHM.

### Simple Harmonic Motion

Our approach is to use the tools we have at our disposal, namely, Newton’s 2nd law, to analyze the motion of several different physical systems that exhibit oscillatory motion. We will look for common features of the motion and its description. Then, we will generalize the description and representation. In this process we will develop explicit mathematical expressions to represent the motion and we will see how properties of the motion such as the period of oscillation are related to the physical parameters of the particular phenomena. In this process we will revisit the energies involved in oscillating systems and gain a deeper understanding of the energy relationships.

### First System: Mass on a Spring

We consider a mass hanging on a spring. There are two forces acting on the mass: the pull upward of the spring and the gravity force of the Earth pulling down. We saw previously that if we take x to be the distance from the equilibrium position of the mass as it hangs motionless on the spring, then the net force has the form

$\sum F = -kx$

Now we apply Newton’s 2nd law: net force equals mass times acceleration

$-kx = ma , or$

$-kx= m \dfrac{d^2x}{dt^2}$

This is called a differential equation, because it involves derivatives of x. A standard way to write this equation that will be useful as we compare different systems is

$a = \dfrac{d^2}{dt^2}x(t) = -\dfrac{k}{m}x(t)$

Let’s note several things about this equation. Its solution will be a mathematical expression that gives the position x as a function of the time $$t$$. The equation says that if we differentiate this function twice, we get back the same function multiplied by the negative constant -k/m (and are both positive constants). Also, the acceleration,

$a= \dfrac{d^2}{dt^2} x(t)$

is not constant. Rather, the acceleration is proportional to the displacement from equilibrium, but with the opposite sign.

What kind of function, when differentiated twice, gives back the same function, but with a negative constant coefficient? Perhaps you remember from your calculus course which function has this property. If you do not, that is OK. What’s important is to understand the properties of the solution, not how to get the solution. There are two functions that have the property we desire. One is the sine function and the other is the cosine function. The second derivative of $$A \sin (bt)$$ with respect to $$t$$ (when A and are constants) is $$-b^2A\sin(bt).$$ That is,

$\dfrac{d^2}{dt^2}A\sin bt = -Ab^2 \sin bt \label{Eq5}$

And similarly for the cosine function.

We compare Equation $$\ref{Eq5}$$ to the equation for the mass on a spring,

$\dfrac{d^2}{dt^2}x(t) = -\dfrac{k}{m}x(t)$

we notice that they are the same if $$b^2$$ equals $$k/m$$. If we make that substitution, we get

$x(t)= A \sin \left( \sqrt{\dfrac{k}{m}t} \right)$

differentiating twice with respect to t and see if you do not get the function $$x(t)$$ back multiplied by the negative constant -k/m.

These two functions (the sine and the cosine) are solutions of the differential equation we obtained by applying Newton’s 2nd law to the mass hanging on the spring. These functions repeat every time the angle $$bt$$increases by $$2\pi$$. Thus, the time to complete one oscillation is that value of $$t$$  that satisfies the relation $$bt = 2\pi$$. This time is called  the period and is denoted by the letter $$T$$. It is equal to $$2\pi /b$$.

$T=\dfrac{2\pi}{b} = 2\pi\sqrt{\dfrac{m}{k}}$

Note that we know the period if we know the values of the factors that appear in Newton’s 2nd law (mass and spring constant).

We can now write the differential equation in terms of $$T$$.

$\dfrac{d^2}{dt^2}x(t) = -\left(\dfrac{2\pi}{T}\right)^2 x(t)$

and the possible solutions also in terms of $$T$$:

$x(t) - A \sin\dfrac{2\pi t}{T}$

or

$x(t)= A \cos\dfrac{2\pi t}{T}$

Before pursuing the analysis of the spring mass system further, we will look at another system. Then we will generalize our results and discuss them in much more detail.

### Second System: Simple Pendulum

We consider a mass hanging on a lightweight string. The mass swings back and forth when pulled aside and released. How do we apply Newton’s 2nd law?

We first identify the objects and all the forces acting on the objects. Then, the Net Force acting on any particular object is equal to the product of the mass and acceleration of that object. In the case of our pendulum, the object of interest is the bob. (In our model, the mass of the string is negligible.) Two forces act on the bob - the tension in the string directed along the string, and the gravitational pull of the earth straight down on the bob. The vector sum of these two forces is the net force or unbalanced force. The motion is constrained to be along the arc of a circle with radius equal to the length of the string, l. The tangential component of the net force, that is, the force tangent to the path the bob takes, is the component that causes the bob to speed up or slow down along this path. (The component of the net force along the string causes the bob to move in a circle, and is dependent on the instantaneous speed. We do not need to be concerned with this radial force now.)

To proceed, we draw a force diagram (Figure 8.5.1), showing the forces acting on the bob. Applying Newton’s 2nd law along the tangential direction gives

Figure 8.5.1

$mg\sin \theta = -ma_{tangential}$

The minus sign tells us that $$a_{tangential}$$ is opposite to the direction of increasing $$\theta$$. It is useful to express $$a_{tangential}$$

in terms of $$\theta$$.Since a $$a_{tangential}$$ is the second derivative of a distance moved along the arc, and since a distance along the arc is simply the product of $$l\theta$$ , $$a_{tangential} = l d^2\theta/dt^2$$. Then,

$mg\sin \theta = -ml\dfrac{d^2\theta}{dt^2}$

and cancelling out the mass

$gsin\theta = -l\dfrac{d^2\theta}{dt^2}$

This looks almost like our equation of motion for a mass on a spring. The difference is we have a $$\sin\theta$$ in stead of $$\theta$$ on the left hand side. Perhaps for small oscillations, that is small values of $$\theta$$, we can replace $$\sin\theta$$ with $$\theta$$.

If we make this approximation (substituting for sin and then group the constants together on the left hand side) we get:

$-\dfrac{g}{l}\theta=\dfrac{d^2\theta}{dt^2}$

If we put this in standard form, we can easily compare it to the equation we got for the mass oscillating on a spring.

$simple\: pendulum: \dfrac{d^2\theta}{dt^2} = -\dfrac{g}{l}\theta(t)$

$mass \: on \: spring: a=\dfrac{d^2x}{dt^2}=-\dfrac{k}{m}x(t)$

The similarity in these two equations. Except for the name of the variable, $$\theta$$ or $$x$$, which is arbitrary, they have the identical form.

We saw before, that in terms of the period to make a complete oscillation, we could write the expression for $$x(t)$$ as:

$\dfrac{d^2}{dt^2}x(t)=-\left(\dfrac{2\pi}{T}\right)^2x(t), where$

$(\dfrac{2\pi}{T})^2=\dfrac{k}{m} \leftarrow T=2\pi\sqrt{\dfrac{m}{k}}$

Now by comparing the pendulum equation to the mass and spring equation, we see that the relation giving the period for a pendulum must be:

$(\dfrac{2\pi}{T})^2=\dfrac{g}{l} \leftarrow T=2\pi\sqrt{\dfrac{l}{g}}$

Also, since the equation for the mass on a spring and the equation for the pendulum are in fact the same equation with different constants, they must have the same solution. So the mathematical function that worked for the mass-spring, must work for the simple pendulum, too. The distinguishing feature that makes these equations similar is that the acceleration is proportional to the displacement, but with the opposite sign. This is the unique feature that leads to simple harmonic motion (SHM).

Before going any further with the analysis of SHM, it is useful to investigate its general properties. This is what we will now do.