# Simple Harmonic Motion

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# Basic Description

A simple harmonic motion, also called harmonic vibration or harmonic oscillation, is a type of periodic motion in physics where the restoring force on an object is directly proportional to the object's displacement from a certain point.

The following 3 animations show some examples of harmonic vibrations: Figure 1-aMass hanging on spring Figure 1-bSimple Pendulum Figure 1-cWater in tube

In Figure 1-a, a mass hanging on a light spring is pulled and released. From Hook's law in classical mechanics, we know that the more we stretch a spring, the harder it pulls back, and this restoring force is proportional to the amount of stretching. So the mass here undergoes harmonic vibration.

In Figure 1-b, a simple pendulum is formed by connecting a mass to one end of a light string, and fixing the other end to the ceiling. We can see that the more it displaces from the vertical position, the harder gravity pulls it back. When the swinging angle is small, this restoring forces is approximately proportional to the displacement. So the mass again exhibits harmonic vibration.

In Figure 1-c, a U-shaped tube with water is shaken so that water rises and falls in each of its branches. The higher water rises in one branch, the lower it falls in the other branch, and the more pressure difference is generated to push it back. This pressure difference is again proportional to the height of rising/falling. So we have got another simple harmonic motion.

The 3 vibrations above have some properties in common:

First, each one of them has an equilibrium position. At this position, if there is only one force acting on the vibrator, it should be 0; if there are multiple forces acting on the vibrator, they should cancel one another. For example, if the mass is released from a position where the pulling force of spring balances its weight, then it will not vibrate at all. The same thing happens if the pendulum is vertical when we release it, or if the water in the two branches of the tube are at the same height.

Second, when the object moves away from the equilibrium position, the system gives it a restoring force to pull it back. In simple harmonic motions, this restoring force must be proportional, or at least approximately proportional, to the object's displacement from the equilibrium position. If it's not, then we get more complicated vibrations, which we are not going to discuss here.

Based on these properties, physicists derived the equation of motion for harmonic vibrations.

# Equation of Motion Figure 2-a
Plot of position of a harmonic vibrator versus t

To get a sense of what the equation of simple harmonic motion is, we can plot the position of a harmonic vibrator against t, the time elapsed after we released it. From Figure 2-a, we can see this plot looks a lot like the sinusoidal function:

Eq. 1 $x = A\sin({\omega t} + \phi)$

It can be proven, using the two properties we mentioned above, that this function is indeed the equation of motion we want. However, this proof is long, and uses a lot of calculus. So we have postponed this proof to the Appendix of this page. Readers who are interested, and proficient in calculus, can click here to check it out. For now, let's focus on the equation of motion itself.

In Eq. 1, x is the object's displacement from equilibrium position, and t is the time elapsed. A, ω, and φ are constants determined by the nature of the system, and each of them affect different aspects of the vibration. Figure 2-b
Magnitude(red) : Magnitude(blue) = 4 : 1

The constant A is called the magnitude of harmonic vibrations. We know that the range of sin(x ) is always from -1 to 1, regardless of the variables inside. So the range of this vibration is solely determined by the factor A , which is multiplied outside the sine function. For example, in Figure 2-b the positions of two harmonic vibrators are plotted against t . The red one has a magnitude four times as large as the blue one's, and other factors are the same. We can clearly see the difference in the animation. Figure 2-c
Frequency(red) : Frequency(blue) = 3 : 1

The constant ω is called the angular frequency, or frequency, of harmonic vibrations. We already know that sin(x ) is a periodic function, and that 2 $\pi$ is its smallest complete cycle. So the larger ω is, the faster ωt will increase by 2 $\pi$ during the vibration, and the more cycles we are going to get in a time span. The influence of frequency is shown in Figure 2-c, in which the red vibrator has an angular frequency 3 times as large as the blue one's. Figure 2-d
φ(blue) = 0 ; φ(green) = 2π/3 ; φ(orange) = 4π/3

The constant φ is called the initial phase, or phase, of harmonic vibrations. It determines the vibrator's position in the 2 $\pi$ cycle at the beginning of the vibration, when t = 0. If we change the φ of a harmonic vibration, neither its magnitude nor its frequency will be affected, but its x - t graph will be shifted. For example, in Figure 2-d, 3 harmonic vibrations with same magnitude, same frequency, but different phases are plotted together. They start from different places at t = 0, and their traces are horizontal translations of one another.

So the three constants in Eq. 1 affect the x - t plot of the simple harmonic motion in different ways. Changing the magnitude A dilates the plot in vertical direction; changing the frequency ω dilates the plot in horizontal direction; and changing the initial phase φ shifts the plot in horizontal direction. As we have pointed out before, these constants are determined by the nature of the system, such as the amount of mass, the stiffness of the spring, the position where we release the mass, and so on.

# The Phase Circle

Simple harmonic motion is closely related to uniform circular motion. In fact, this relationship gives us a useful tool to analyze simple harmonic motions: the phase circle. See the following images: Figure 3-aHarmonic vibration and uniform circular motion Figure 3-bThe phase circle

In Figure 3-a, the harmonic vibration of a mass hanging on a spring at left is compared to the uniform circular motion at right. As we can see, the two objects are always at the same height. So the two types of motions here have the same y - components. In other words, simple harmonic motion can be viewed as a component of uniform circular motion. This circle in the uniform circular motion is then called the phase circle of simple harmonic motion.

We can get the same result using geometry and parametric equations. In Figure 3-b, a particle P undergoing uniform circular motion is projected onto a vertical line. The radius of the circle is A, and the position of P is determined by the following formula: $\theta = \omega t + \phi$

So the coordinate of P is: $(A\cos(\omega t + \phi), A\sin(\omega t + \phi))$

Since the point P' is the projection of P onto a vertical line, it must have the same y - coordinate as P: $y_{P'} = A\sin(\omega t + \phi)$

Notice that the equation above is identical to Eq.1, the equation of motion we gave for simple harmonic motions. So simple harmonic motion can be viewed as the y component of a uniform circular motion.

The phase circle gives us another way to explain the roles of constants A, ω, and φ in Eq.1. These constants affect the simple harmonic motion by changing the radius, angular velocity, and starting point of its corresponding uniform circular motion. The following three animations are Figure 2-b through Figure 2-d remade with the corresponding phase circles. These animations are self-explanatory, and one can clearly see how these constants affect the harmonic vibration through its phase circle. Figure 3-cMagnitude A changes the radius of the phase circle. Figure 3-dFrequency ω affects the angular velocity of the circular motion. Figure 3-eDifferent phases φ generate different points on the same phase circle.

Acknowledgement: The idea of a phase circle is inspired by this website: click here.

# Appendix: Proof for the equation of motion Figure 4-a
Position and restoring force in the spring vibrator

Here is the promised proof for the equation of simple harmonic motion: $x = A\sin(\omega t + \phi)$

This proof consists of two parts: the physics part and the math part. The physics part tells us how to get a differential equation using laws in classical mechanics, and the math part tells us how to solve that differential equation to get the equation of motion.

### The physics part

We have seen many systems that generate harmonic vibrations, and each of them has a different physical nature. For example, for the spring vibrator in Figure 1-a, we may use Hook's law and Newton's second law to solve it. However, to deal with the water-tube vibrator in Figure 1-c, we must have knowledge about fluid dynamics. What's more, there are many vibrators in electromagnetism, optics, and other areas of physics that can't possibly be covered in this article. So in this physics part we will only study a simple case: the horizontal spring vibrator. Other cases can be treated in a similar manner once we have the background knowledge.

Figure 4-a shows a horizontal spring vibrator with its equilibrium position, displacement vector, and restoring force vector. By Hook's law, the restoring forces provided by the spring are proportional to the displacement, and in opposite directions: $F = -kx$

in which F stands for the restoring force, x stands for the displacement, and k stands for the spring constant, or stiffness. This equation gives us a harmonic vibration by definition.

What's more, we have Newton's second law, which states that force is equal to mass times acceleration. Since the downward force of the mass is balanced by the reaction of the surface (which is why we chose the horizontal position), the only external force left is the restoring force provided by the spring: $F = ma$

in which m stands for the mass of the object, and a stands for its acceleration. Eliminate F from these equations: $ma = -kx$

Move -kx to the left side, and divide the equation by m to make it look neater: $a + {k \over m}x = 0$

In physics, acceleration is the second order derivative of displacement: $a = {d^2x \over dt^2}$

So the previous equation changes into:

Eq. 2 ${d^2x \over dt^2} + {k \over m}x = 0$

This is the characteristic differential equation of simple harmonic motions. No matter which physics model we use, as long as the restoring force is proportional to the displacement, we will get a differential equation similar to Eq. 2.

### The math part

Now we have got Eq. 2: ${d^2x \over dt^2} + {k \over m}x = 0$

This is a second order differential equation, which states the relationship between the function x(t) and its second derivative. In this section we are going to solve it to get the equation of simple harmonic motion that we have long wanted.

Acknowledgement: The following approach to solve Eq. 2 comes from James Stewart's Second-Order Linear Differential Equations. To read this article, click here.

It’s not hard to think of some likely candidates for particular solutions of Eq. 2 if we state the equation verbally. We are looking for a function such that its second derivative plus a constant times itself is equal to 0. We know that the exponential function x = ert (where r is a constant) has the property that its derivative is a constant multiple of itself: ${d \over dt}e^{rt} = re^{rt}$, and: ${d^2 \over dt^2}e^{rt} = r^2e^{rt}$

If we substitute x = ert into Eq. 2, we will get: $r^2e^{rt} + {k \over m}e^{rt} = 0$

Divide the equation above by ert: $r^2 + {k \over m} = 0$

Solve for r: $r = \pm \sqrt{-{k \over m}}$

Notice that in our physics model both k and m must be positive. So the expression inside the square root sign is negative, which makes r a complex number. We can use the imaginary unit $i = \sqrt {-1}$, and do the substitution $\omega = \sqrt{k \over m}$ to make the equation above look neater: $r = \pm \sqrt{-{k \over m}} = \pm \sqrt{-1} \cdot \sqrt{k \over m} = \pm i \omega$

Recall that we assumed x = ert. So we have got two different solutions for Eq. 2: $x_1 = e^{i \omega t}$ , $x_2 = e^{ - i \omega t}$

According to the linear differential equation theory, any linear combination of them is also a solution for Eq. 2: $x_1' = {{x_1 + x_2} \over 2} = {{e^{i \omega t} + e^{-i \omega t}} \over 2}$ , $x_2' = {{x_1 - x_2} \over 2i} = {{e^{i \omega t} - e^{-i \omega t}} \over 2i}$

Now, applying the Euler's formula, which states that: $\cos \theta = {{e^{i \theta} + e^{-i \theta}} \over 2}$ , $\sin \theta = {{e^{i \theta} - e^{-i \theta}} \over 2i}$

We will get: $x_1' = \cos \omega t$ , $x_2' = \sin \omega t$

Surprisingly, we have these two simple and unexpected solutions. To make sure that they are indeed the solutions we want, we can substitute them back into Eq. 2, and recall that ω2 = k / m: ${d^2x_1' \over dt^2} + {k \over m}x_1' = -\omega^2 \cos \omega t + {k \over m} \cos \omega t = 0$ ${d^2x_2' \over dt^2} + {k \over m}x_2' = -\omega^2 \sin \omega t + {k \over m} \sin \omega t = 0$

So we see that they indeed satisfy the original differential equation. Since any linear combination of them also satisfies the equation, the general solution of Eq. 2 must be:

Eq. 3 $x = a \sin \omega t + b \cos \omega t$

in which a and b are arbitrary constants. Eq. 3 is already a satisfactory solution, but it's not in the form we want. We can apply the following trigonometric identity to it: $a \sin \theta + b \cos \theta = {\sqrt {a^2 + b^2}} \sin(\theta + \arctan{b \over a})$

which gives us: $x = {\sqrt {a^2 + b^2}}\sin(\omega t + \arctan{b \over a})$

Let A = √(a2 + b2) and φ = arctan(b/a) , then we will get Eq. 1: $x = A\sin(\omega t + \phi)$

So we have proved that the equation of simple harmonic motion is indeed a sinusoidal function. What's more, we know that the frequency of this vibrator is determined by the following formula: $\omega = \sqrt{k \over m}$

The other two factors, A and φ, are determined by its initial velocity and the position where we release the mass.