5.5: Harmonic Motion

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Roy Simpson
Cosumnes River College

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Learning Objectives

Build a model for simple harmonic motion.
Build a model for damped harmonic motion.

Harmonic motion is a form of periodic motion, but there are factors to consider that differentiate the two types. Applications involving general periodic motion cycle through their periods with no outside interference. This type of motion is what we have dealt with up to this point in Trigonometry. In terms of sinusoidal functions, we think of this as oscillatory motion. Examples of periodic motion include temperature measurements throughout the day, the elevation of a rider on a Ferris wheel, and sound waves. In all of these examples, the motion is periodic. Still, there is no physical force requiring the measured quantity to tend back to the midline of the periodic motion.

Harmonic motion, on the other hand, requires a restorative force. As a reminder from Physics, a force can be considered a push or pull. Thus, an object goes through harmonic motion if it oscillates (much like periodic motion) but is constantly being pulled by a force back to the midline of the oscillatory motion. Harmonic motion is crucial in Differential Equations, Engineering, and Physics. Harmonic motion includes the motion of springs, gravitational force, and magnetic force.

Harmonic motion can further be broken into two major types - simple harmonic motion (also called undamped harmonic motion) and damped harmonic motion.¹

Simple Harmonic Motion

The "prototype" motion, called simple harmonic motion, involves a restorative force but assumes that the motion will continue forever. This is not the most realistic type of harmonic motion; however, it's a necessary starting point to understand the robust forms.

Imagine a weighted object hanging on a spring. When that object is not disturbed, we say that the object is at rest or in equilibrium. If the object is pulled down and then released, the spring's force (called the spring's restorative force) pulls the object back toward equilibrium, and harmonic motion begins. The restorative force is directly proportional to the object's displacement from its equilibrium point.²

Interactive Element: Simple Harmonic Motion

Observation: The simple harmonic motion of the object is sinusoidal.

Theorem: Simple Harmonic Motion

The motion of a particle undergoing simple harmonic motion is modeled using a sinusoidal function. It can be written as either\[ s(t) = A \, \sin\left( \omega t + \phi \right) \quad \text{or} \quad s(t) = A \, \cos\left( \omega t + \phi \right), \nonumber \]where the frequency of the motion is $ f = \frac{1}{\text{Period}} = \frac{\omega}{2\pi} $ cycles per unit time.

The position of a particle relative to the ground is often unimportant when building a model for simple harmonic motion. This is because we are only concerned with modeling the particle's motion - not the height. Moreover, all our previous training with graphs of sinusoidal functions still holds here. The only difference is that we are replacing $ B $ with $ \omega $, and $ C $ with $ \phi $. This is mainly because $ \omega $ and $ \phi $ are the typical notations used in Physics.

Finally, the sinusoidal nature of these phenomena observed in physics is just that - an observation. We have not, and cannot, prove this result in Trigonometry. You do, however, prove this result in Differential Equations.

Example $\PageIndex{1}$

The displacement of a mass suspended by a spring is modeled by the function\[ s(t) = 1.5\sin\left( 6\pi t \right), \nonumber \]where $ s(t) $ is measured in cm and $ t $ in seconds.

Find the amplitude, period, and frequency of the motion of the mass.
Sketch a graph of the displacement of the mass.

Solutions

From our previous experience, the amplitude is $ |1.5| = 1.5 $, the period is $ \frac{2\pi}{\omega} = \frac{2\pi}{6\pi} = \frac{1}{3} $ seconds per cycle, and the frequency is $ f = \frac{1}{\text{Period}} = 3 $ cycles per second.
$7.5.1 Example.png$

Damped Harmonic Motion

In reality, a pendulum does not swing back and forth forever, and an object on a spring does not bounce up and down forever. Eventually, the pendulum stops swinging, the object stops bouncing, and both return to equilibrium. Periodic motion in which a frictional force or damping factor acts is known as damped harmonic motion.

Interactive Element: Damped Harmonic Motion

Observation: The quasi-sinusoidal harmonic motion of the object is being damped.

In Physics, various formulas account for the damping factor on the moving object. Some of these are Calculus-based formulas that involve derivatives. We will use damping functions based on decreasing exponential functions (as these are the most common damping functions).

Theorem: Damped Harmonic Motion

The motion of a particle undergoing damped harmonic motion is modeled using a quasi-sinusoidal function. The displacement of the particle can be written as either\[ s(t) = A \, e^{−ct} \sin\left( \omega t + \phi \right) \quad \text{or} \quad s(t) = A \, e^{−ct} \cos\left( \omega t + \phi \right),\]where $c$ is called the damping factor, $\frac{2\pi}{\omega}$ is called the quasi-period, and $ \frac{1}{\text{Quasi-Period}} = \frac{\omega}{2\pi} $ is called the quasi-frequency.

The words quasi-period and quasi-frequency reflect the fact that these models are not true sinusoidal models.

Conceptually (and graphically), the decreasing exponential function forces a boundary on the model's position values (outputs). The following Interactive Element should provide more understanding than I could write in a few paragraphs.

Interactive Element: Damped Harmonic Model (and the exponential boundaries)

Interact: Move the slider to adjust the damping factor.

Observation: The boundary of the damped harmonic motion is always $ y = \pm A \, e^{-c t} $.

Example $\PageIndex{2}$

Model the equations that fit the two scenarios and use graphing technology to graph the functions: Two mass-spring systems exhibit damped harmonic motion at a quasi-frequency of 0.5 cycles per second. Both have an initial displacement of 10 cm. The first has a damping factor of 0.5, and the second has a damping factor of 0.1.

Solutions

At time $t=0$, the displacement is the maximum of 10 cm, which calls for the cosine function. The cosine function will apply to both models. We are given the quasi-frequency, $f=\frac{\omega}{2\pi}$, of 0.5 cycles per second. Thus,\[\begin{array}{rrcl}
& \dfrac{\omega}{2\pi} & = & 0.5 \\
\\
\implies & \omega & = & \pi \\
\end{array} \nonumber \]The first spring system has a damping factor of $c=0.5$. Following the general model for damped harmonic motion, we have\[s_1(t)=10e^{−0.5t} \cos (\pi t). \nonumber\]

$7.5.2.1 Example.png$

The second spring system has a damping factor of $c=0.1$ and can be modeled as\[s_2(t)=10e^{−0.1t} \cos (\pi t).\nonumber \]

$7.5.2.2 Example.png$

Notice the differing effects of the damping constant. The function's local maximum and minimum values with the damping factor $c=0.5$ decrease much more rapidly than the function with $c=0.1$.

Checkpoint $\PageIndex{2}$

Find and graph a function of the form $s(t)=A \, e^{−ct} \cos\left( \omega t \right)$ that models the information given.

$A=20, \, c=0.05,$ and the quasi-period is $4$
$A=2, \, c=1.5,$ and the quasi-frequency is $3$

Answers

$s(t)=20e^{−0.05t} \cos \left(\frac{\pi}{2}t\right).$
$Graph of f(t) = 20(e^(-.05t))cos(pi/2 * t), which begins with a high amplitude and slowly decreases.$
$s(t)=2e^{−1.5t} \cos \left(6\pi t\right).$
$Graph of f(t) = 2(e^(-1.5t))cos(6pi * t), which begins with a small amplitude and quickly decreases to almost a straight line.$

Example $\PageIndex{3}$

Find and graph a function of the form $s(t) = A \, e^{−ct} \sin\left( \omega t \right)$ that models the information given.

$A=7$, $c=10$, and the quasi-period is $p = \frac{\pi}{6}$
$A=0.3$, $c=0.2$, and the quasi-frequency is $f=20$

Solutions

Calculate the value of $\omega$ and substitute the known values into the model.

As the quasi-period is $\frac{2\pi}{\omega}$, we have\[\begin{array}{rrcl}
& \dfrac{\pi}{6} & = & \dfrac{2\pi}{\omega} \\
\\
\implies & \omega \pi & = & 6(2 \pi) \\
\\
\implies & \omega & = & 12 \\
\end{array} \nonumber \]The damping factor is given as 10 and $ A $ is 7. Thus, the model is $s(t)=7e^{−10t} \sin (12t)$.
$Graph of f(t) = 7(e^(-10t))sin(12t), which spikes close to t=0 and quickly becomes almost a straight line.$
As the quasi-frequency is $\frac{\omega}{2\pi}$, we have\[\begin{array}{rrcl}
& 20 & = & \dfrac{\omega}{2\pi} \\
\\
\implies & 40\pi & = & \omega \\
\end{array} \nonumber \]The damping factor is given as $0.2$ and $A = 0.3$. The model is $s(t)=0.3e^{−0.2t} \sin (40\pi t)$.
$Graph of f(t) = .3e^(-.2t)sin(40pi*t), which has a small amplitude but quickly decreases to the appearance of a straight line. The frequency is so high that, in this scaling, the function looks like a solid shape. The zoom in cut out of the graph shows the actual sinusoidal image of the function.$

A comparison of Examples $ \PageIndex{ 3a } $ and $ \PageIndex{ 3b } $ illustrates how we choose between the sine or cosine functions to model sinusoidal criteria. We see that the cosine function is at the maximum displacement when $t=0$, and the sine function is at the equilibrium point when $t=0.$ For example, consider the equation\[y=20e^{−0.05t} \cos\left(\dfrac{\pi}{2}t\right)\nonumber \]from Checkpoint $ \PageIndex{ 2 } $. We can see from the graph that when $t=0$, $y=20$, which is the initial amplitude. Check this by setting $t=0$ in the cosine equation:\[ \begin{array}{rcl}
y & = & 20e^{−0.05(0)} \cos\left(\dfrac{\pi}{2} \cdot 0\right) \\
\\
& = & 20(1)(1) \\
\\
& = & 20 \\
\end{array}\nonumber \]Using the sine function yields\[\begin{array}{rcl}
y & = & 20e^{−0.05(0)} \sin\left(\dfrac{\pi}{2} \cdot 0\right) \\
\\
& = & 20(1)(0) \\
\\
& = & 0 \\
\end{array}\nonumber \]Thus, cosine is the correct function.

Checkpoint $\PageIndex{3}$

Write the equation for damped harmonic motion given $A=10$, $c=0.5$, and the quasi-period is $p=2$, where the initial value of the function is $ s(0) = -10 $.

Answer: $s(t)=-10e^{−0.5t} \cos (\pi t)$

Example $\PageIndex{4}$

A spring measuring 10 inches in natural length is compressed by 5 inches and released. It oscillates once every 3 seconds, and its amplitude decreases by 30% every second. Find an equation that models the position of the spring $t$ seconds after being released.

Solution

The amplitude begins at 5 inches and deceases 30% each second. We will make an "amplitude function" as an intermediate step. We can write the amplitude portion of the function as\[A(t) = 5(1−0.30)^t. \nonumber\]We put $(1−0.30)^t$ into the form $e^{ct}$ as follows:\[\begin{array}{rrclcl}
& \left( 1 - 0.3 \right)^t & = & e^{ct} & & \\
\implies & \left( 0.7 \right)^t & = & \left( e^c \right)^t & \quad & \left( \text{Laws of Exponents} \right) \\
\implies & 0.7 & = & e^c & & \\
\implies & \ln\left( 0.7 \right) & = & c & \quad & \left( \text{solving the exponential equation} \right) \\
\implies & c & \approx & −0.357 & & \\
\end{array} \nonumber \]Let’s address the quasi-period. The spring completes a cycle every 3 seconds. This is the quasi-period, and we can use the formula to find $\omega$.\[\begin{array}{rrcl}
& 3 & = & \dfrac{2 \pi }{\omega} \\
\\
\implies & \omega & = & \dfrac{2 \pi }{3} \\
\end{array} \nonumber \]The natural length of 10 inches is the midline. We will use the cosine function since the spring starts at maximum displacement. This portion of the equation is represented as\[y = \cos \left(\dfrac{2 \pi }{3}t\right)+10. \nonumber\]Finally, we put both functions together. Our model for the position of the spring at $t$ seconds is given as\[y \approx 5e^{−0.357t} \cos \left(\dfrac{2 \pi }{3}t\right)+10. \nonumber\]The approximation sign is needed because the coefficient of $ t $ is an approximation.

$Graph of the function y = -5e^(-.35t)cos(2pi/3 t) + 10 from 0 to 24. It starts out as waves with a high amplitude and decreases to almost a straight line very quickly.$

Exercise $\PageIndex{4}$

A mass suspended from a spring is pulled 3 cm below its resting position. The mass is released at time $t=0$ and allowed to oscillate. After $\frac{1}{3}$ second, it is observed that the mass returns to its lowest position. Find a function to model this motion relative to its initial resting position.

Answer: $s(t)=-5 \cos \left(6 \pi t\right)$

Example $ \PageIndex{ 5 } $

A guitar string is plucked and vibrates in a damped harmonic motion. The string is pulled and displaced 2 cm from its resting position. After 3 seconds, the displacement measures 1 cm. Find the damping constant.

Solution: The displacement factor represents the amplitude and is determined by the coefficient $A \, e^{−ct}$ in the model for damped harmonic motion. The damping constant $ c $ is included in the term $e^{−ct}$. It is known that after 3 seconds, maximum displacement measures one-half of its original value. Therefore, we have the equation\[A \, e^{−c(t+3)}=\dfrac{1}{2} A \, e^{−ct}.\nonumber \]Using Algebra and the Laws of Exponents, we solve for $c$.\[\begin{array}{rrclcl}
& A \, e^{−c(t+3)} & = & \dfrac{1}{2}A \, e^{−ct} & & \\
\\
\implies & e^{−c(t+3)} & = & \dfrac{1}{2} e^{−ct} & \quad & \left( \text{divide both sides by }A \right) \\
\\
\implies & e^{-ct-3c} & = & \dfrac{1}{2} e^{−ct} & \quad & \left( \text{distribute} \right) \\
\\
\implies & e^{−ct} e^{−3c} & = & \dfrac{1}{2}e^{−ct} & \quad & \left( \text{Laws of Exponents} \right) \\
\\
\implies & e^{−3c} & = & \dfrac{1}{2} & \quad & \left(\text{divide both sides by }e^{−ct}\right) \\
\\
\implies & e^{3c} & = & 2 & \quad & \left(\text{take the reciprocal of both sides} \right) \\
\\
\implies & 3c & = & \ln\left( 2 \right) & \quad & \left( \text{Laws of Logarithms} \right) \\
\\
\implies & c & = & \dfrac{1}{3} \ln\left( 2 \right) & \quad & \left( \text{divide both sides by }3 \right) \\
\\
\end{array}\nonumber \]Thus, the damping constant is $c = \frac{\ln(2)}{3}$.

Footnotes

¹ Within both simple and damped harmonic motion, you also encounter free and driven (or forced) motion. These topics are reserved for Differential Equations; however, you can think of free motion as motion without extra external forces being applied. Driven (or forced) motion occurs when the object gains or loses kinetic energy due to an extra external force being applied.

For example, if we attach a mass to a spring, allow it to stretch to equilibrium, pull down slightly once it has reached equilibrium, and then let go, the motion of the mass will be free, damped motion. It starts oscillating, and we do not add any other forces to its movement; however, the spring constantly tries to pull (or push) the mass back to equilibrium. As time goes on, the amplitude of the oscillations decreases as the mass settles to the equilibrium position.

On the other hand, if we attach a mass to a spring, allow it to stretch to equilibrium, pull down slightly once it has reached equilibrium, let go, and then give it an extra push each time it reaches its lowest point (kind of like pushing a child on a swing), the motion is driven, damped motion. It's still damped because we are trying to be realistic by assuming that, when left alone, the amplitude of the oscillations should get smaller; however, it is driven because we add energy to the mass by pushing it each time it reaches its lowest point.

² From Algebra, $ F $ is directly proportional to $ x $ if there exists a constant, $ k $, such that $ F = k x $. $ k $ is called the constant of proportionality.

Homework

Vocabulary Check

The force associated with an object's tendency to return to equilibrium is called the ___ force.
Of the two types of harmonic motion discussed in this section, ___ harmonic motion is more realistic as it considers friction.
While an object undergoing simple harmonic motion has a period and frequency, an object undergoing damped harmonic motion has a ___ and ___.
Another name for simple harmonic motion is ___ harmonic motion.

Concept Check

Explain the effect of a damping factor on the graphs of harmonic motion functions.

True or False? For Problems 6 - 8, determine if the statement is true or false. If true, cite the definition or theorem stated in the text supporting your claim. If false, explain why it is false and, if possible, correct the statement.

A particle undergoing harmonic motion will theoretically oscillate forever.
A particle undergoing damped harmonic motion will theoretically oscillate forever.
The frequency of the particle undergoing damped harmonic motion described by\[ s(t) = 2 e^{-0.01t} \sin\left( \dfrac{2\pi}{3} t \right) \nonumber \]is $ f = \frac{\omega}{2 \pi} = \frac{2\pi/3}{2\pi} = \frac{1}{3} $.

Basic Skills

For Problems 9 - 13, find the amplitude, period, and frequency of the given function.

The displacement, $h(t)$, in centimeters of a mass suspended by a spring is modeled by the function\[h(t) = 14 \sin\left(120\pi t\right), \nonumber \]where $t$ is measured in seconds.
The displacement, $h(t)$, in centimeters of a mass suspended by a spring is modeled by the function\[h(t)=8\sin(6 \pi t),\nonumber \]where $t$ is measured in seconds.
The displacement, $h(t)$, in centimeters of a mass suspended by a spring is modeled by the function\[h(t)=11\sin(12 \pi t),\nonumber \]where $t$ is measured in seconds.
The displacement, $h(t)$, in centimeters of a mass suspended by a spring is modeled by the function\[h(t)=4\cos(2 \pi t),\nonumber \]where $t$ is measured in seconds.
The displacement, $h(t)$, in centimeters, of a mass suspended by a spring is modeled by the function\[h(t)=−5\cos(60 \pi t), \nonumber \]where $t$ is measured in seconds.

Applications

For Problems 14 - 17, construct an equation that models the described behavior.

A spring attached to the ceiling is pulled 10 cm down from equilibrium and released. The amplitude decreases by 15% each second. The spring oscillates 18 times each second. Find a function that models the distance, $D$, the end of the spring is from equilibrium in terms of seconds, $t$, since the spring was released.
A spring attached to the ceiling is pulled 7 cm down from equilibrium and released. The amplitude decreases by 11% each second. The spring oscillates 20 times each second. Find a function that models the distance, $D$, the end of the spring is from equilibrium in terms of seconds, $t$, since the spring was released.
A spring attached to the ceiling is pulled 17 cm down from equilibrium and released. After 3 seconds, the amplitude has decreased to 13 cm. The spring oscillates 14 times each second. Find a function that models the distance, $D$, the end of the spring is from equilibrium in terms of seconds, $t$, since the spring was released.
A spring attached to the ceiling is pulled 19 cm down from equilibrium and released. After 4 seconds, the amplitude has decreased to 14 cm. The spring oscillates 13 times each second. Find a function that models the distance, $D$, the end of the spring is from equilibrium in terms of seconds, $t$, since the spring was released.

For Problems 18 - 22, create a function modeling the described behavior. Then, calculate the desired result using a calculator.

A spring attached to a ceiling is pulled down 11 cm from equilibrium and released. After 2 seconds, the amplitude has decreased to 6 cm. The spring oscillates 8 times each second. Find when the spring first comes between $−0.1$ and $0.1$ cm, effectively at rest.
A spring attached to a ceiling is pulled down 21 cm from equilibrium and released. After 6 seconds, the amplitude has decreased to 4 cm. The spring oscillates 20 times each second. Find when the spring first comes between $−0.1$ and $0.1$ cm, effectively at rest.
Two springs are pulled down from the ceiling and released at the same time. The first spring, which oscillates 8 times per second, was initially pulled down 32 cm from equilibrium, and the amplitude decreases by 50% each second. The second spring, oscillating 18 times per second, was initially pulled down 15 cm from equilibrium and after 4 seconds has an amplitude of 2 cm. Which spring comes to rest first, and at what time? Consider “rest” as an amplitude less than $0.1$ cm.
Two springs are pulled down from the ceiling and released at the same time. The first spring, which oscillates 14 times per second, was initially pulled down 2 cm from equilibrium, and the amplitude decreases by 8% each second. The second spring, oscillating 22 times per second, was initially pulled down 10 cm from equilibrium and after 3 seconds has an amplitude of 2 cm. Which spring comes to rest first, and at what time? Consider “rest” as an amplitude less than $0.1$ cm.
A spring attached to a ceiling is pulled down 20 cm. After 3 seconds, wherein it completes 6 full periods, the amplitude is only 15 cm. Find the function modeling the position of the spring $t$ seconds after being released. At what time will the spring come to rest? In this case, use $1$ cm amplitude as rest.

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Interactive Element: Simple Harmonic Motion

Theorem: Simple Harmonic Motion

Example \(\PageIndex{1}\)

Interactive Element: Damped Harmonic Motion

Theorem: Damped Harmonic Motion

Interactive Element: Damped Harmonic Model (and the exponential boundaries)

Example \(\PageIndex{2}\)

Checkpoint \(\PageIndex{2}\)

Example \(\PageIndex{3}\)

Checkpoint \(\PageIndex{3}\)

Example \(\PageIndex{4}\)

Exercise \(\PageIndex{4}\)

Example \( \PageIndex{ 5 } \)

Suggested Prerequisite Topics

Learning Objectives

Simple Harmonic Motion

Interactive Element: Simple Harmonic Motion

Theorem: Simple Harmonic Motion

Example \(\PageIndex{1}\)

Damped Harmonic Motion

Interactive Element: Damped Harmonic Motion

Theorem: Damped Harmonic Motion

Interactive Element: Damped Harmonic Model (and the exponential boundaries)

Example \(\PageIndex{2}\)

Checkpoint \(\PageIndex{2}\)

Example \(\PageIndex{3}\)

Checkpoint \(\PageIndex{3}\)

Example \(\PageIndex{4}\)

Exercise \(\PageIndex{4}\)

Example \( \PageIndex{ 5 } \)

Footnotes

Homework

Vocabulary Check

Concept Check

Basic Skills

Applications