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# Resonance with Sound Waves

Difficulty Level: At Grade Created by: CK-12

## Objectives

The student will:

• Understand the conditions for resonance.
• Solve problems with strings and pipes using the condition for resonance.

## Vocabulary

• beat frequency
• natural frequency: The frequency at which a system vibrates normally when given energy without outside interference.
• resonance: Timing force to be the same as natural frequency.
• sympathetic vibrations

### Introduction

Many systems have a tendency to vibrate. When the forced vibration frequency is the same as the natural frequency, the amplitude of vibration can increase tremendously. A well-known example of this situation is pushing a person on a swing, Figure below.  We know from study of simple pendulums that without being pushed, the person in the swing rocks back and forth with a frequency that depends on gravity and the length of the chain.

\begin{align*}f = \frac{1}{2 \pi} \sqrt{\frac{g}{L}}\end{align*}

This is one example of a natural frequency – the frequency at which a system vibrates normally when given energy without outside interference.

Pushing on the person in the swing will affect the amplitude of the swinging.  This is called forced vibration – when a periodic force from one object (the person pushing) affects the vibration of another object (the person swinging).  To get the most effect, the person pushing will start just at the very back of the swing. In other words, the frequency of how often they push is exactly the same as the frequency of the swing. Suppose they do not push at the right time, but instead push at some other frequency.  That would mean that sometimes they are pushing forward when the swing is still going backward.  In that case, the swing would slow down – i.e. the amplitude of the swing will be reduced.

Timing the pushes to be the same as the natural frequency is called resonance.  For this reason, the natural frequency is also known as the resonant frequency.  If the pushes are timed just right, then even if each individual push is small, the vibration will get larger with each push.

A classic example of an unfortunate consequence of a forced vibration at resonant frequency is what happened to the Tacoma Narrows Bridge, in 1940. See the link below.

In Figure below, the bridge is beginning to resonate, in part, due to the frequency of vibration of the wind gusts. In Figure below, we see that the bridge is no longer able to respond elastically to the tremendous amplitude of vibration from the forced vibration of wind energy (at its resonant frequency), and it is torn apart.

Modern bridges are built to avoid this effect, but through history there are a number of documented situations where a forced vibration at resonance had dire results. The Broughton Suspension Bridge (1831) and the Angers Bridge (1850) are two examples of bridges believed to have collapsed due to the effect of soldiers marching at a regular pace that caused resonance. The Albert Bridge in West London, England has been nicknamed The Trembling Lady because it has been set into resonance so often by marching soldiers. Though soldiers no longer march across the bridge, there still remains a sign of concern as shown in Figure below.

### Sympathetic Vibrations

There is a typical classroom physics demonstration where one tuning fork is set into motion and an identical tuning fork, if placed closed enough, will also be set vibrating, though with smaller amplitude. The same effect occurs when tuning a guitar.  One string is plucked and another, whose length is shortened by holding it down some distance from the neck of the guitar, will also be set into vibration. When this condition is met, both strings are vibrating with the same frequency. We call this phenomenon sympathetic vibration.

In Figure above, a set of pendulums are fixed to a horizontal bar that can be easily jostled. Pendulums \begin{align*}A\end{align*} and \begin{align*}E\end{align*} have the same length. If one of them is set swinging, the horizontal bar will be forced into moving with a period equal to that of the pendulum, which, in turn, will cause the other pendulum of the same length to begin swinging. Any pendulum that is close in length to pendulums \begin{align*}A\end{align*} and \begin{align*}E\end{align*}, for example, pendulum \begin{align*}D\end{align*}, will also begin to swing. Pendulum \begin{align*}D\end{align*} will swing with smaller amplitude than pendulums \begin{align*}A\end{align*} and \begin{align*}E\end{align*} since its resonant frequency is not quite the same as pendulums \begin{align*}A\end{align*} and \begin{align*}E\end{align*}. Pendulums with lengths dramatically different from pendulums \begin{align*}A\end{align*} and \begin{align*}E\end{align*} will hardly move at all.

Resonance is a very common phenomenon, especially with sound.  The length of any instrument is related to what note it plays.  If you blow into the top of a bottle, for example, the note will vary depending on the height of air in the bottle.  This plays an important role in human voice generation. The length of the human vocal tube is between 17 cm and 18 cm.  The typical frequencies of human speech are in the range of 100 Hz to 5000 Hz.

By using the muscles in their throat, singers change the note they sing.  A dramatic example of this is breaking glass with the human voice. By singing at exactly the resonant frequency of a delicate wine glass, the glass will resonate with the note and shatter.

The resonance of sound is also a mechanical analogue to how a radio set receives a signal.  The Figure below shows one of the earliest radio designs, called a crystal radio because the element which detected the radio waves was a crystalline mineral such as galena.

An old crystal radio set.

In modern times, the air is filled with all manner of radio waves. In order to listen to your favorite radio station, you must tune your radio to resonate with only the frequency of the radio station.  When you hear the tuning number of a radio station, such as “101.3 FM”, that is the measure in megahertz,  \begin{align*}101.3 ~ \text{MHz} = 1.013 \times 10^8 ~ \text{Hz}\end{align*}.  The coiled wire (called an inductor) and the capacitor in Figure above act together to tune in a specific radio station. Effectively, the capacitor and inductor act analogously to a pendulum of a specific length that will only respond to vibrations of another pendulum of the same length. So when tuned, only a specific radio frequency will cause resonance in the radio antenna.

### Strings Fixed at Both Ends

A case of natural frequency that you can observe plainly is when you pluck a string or stretched rubber band.  Normally, the string will vibrate at a single widest point in the middle.  This is called the fundamental or first harmonic resonance of the string. This is the same as the natural frequency of a simple pendulum or mass on a spring.  Because it vibrates all along its length, though, the string also lets us see further patterns of resonance.

By vibrating the end of the string rather than just plucking it, we can force vibration at frequencies other than the first harmonic.  When the string is set into vibration, energy will travel down the string and reflect back toward the end where the waves are being generated.  This steady pattern of vibration is called a standing wave. The points where the reflecting waves interfere destructively with the “generated’ waves are called nodes. The points where the reflecting waves interfere constructively with the generated waves are called anti-nodes.

Figure below shows a string fixed at both ends vibrating in its fundamental mode. There are two nodes shown and one antinode. The dashed segment represents the reflected wave.

If you compare the wave shape of the first harmonic to the wave shape of Figure above, it will be apparent that the first harmonic contains one-half of a wavelength, \begin{align*}\lambda\end{align*}. Therefore \begin{align*}L\end{align*}, the length of the unstretched string, is equal to one-half the wavelength, which is \begin{align*}\frac{1}{2} \lambda_1 = L \rightarrow \lambda_1 =2L.\end{align*}

The second harmonic contains an entire wavelength \begin{align*}\frac{2}{2} \lambda_2 = L \rightarrow \lambda_2 = L\end{align*} as shown in Figure below.

And the third harmonic contains one and one-half wavelengths \begin{align*}\frac{3}{2} \lambda_3 = L \rightarrow \lambda_3 = \frac{2}{3} L\end{align*}.

If the pattern continues then the fourth harmonic will have a wavelength of \begin{align*}\frac{4}{2} \lambda_4 = L \rightarrow \lambda_4 = \frac{1}{2}L\end{align*}. Looking at the expressions for the length of the string in terms of the wavelength, a simple pattern emerges: \begin{align*}L = \frac{1}{2} \lambda_1, \frac{2}{2} \lambda_2, \frac{3}{2}\lambda_3, \frac{4}{2}\lambda_4 \ldots\end{align*}. We can express the condition for standing waves (and of resonance, as well) as \begin{align*}L = \frac{n}{2} \lambda_n \end{align*} or \begin{align*}\lambda_n = \frac{2}{n}L\end{align*}, where \begin{align*}L\end{align*} is the length of string and \begin{align*}n = 1,2,3 \ldots\end{align*}.

### Check Your Understanding

1. How many nodes and anti-nodes are shown in Figure above?

Answer: There are three nodes and two anti-nodes.

2. If the length of the unstretched string is 20 cm, what is the wavelength for the 10th harmonic?

Answer: \begin{align*}\lambda_n = \frac{2}{n} L \rightarrow \lambda_{10} = \frac{2}{10}L = \frac{1}{5}(20 ~ \text{cm}) = 4 ~ \text{cm}\end{align*}

### Strings Fixed at One End and Opened at One End

A string fixed at one only end displays a different standing wave pattern. In this case, the unbounded end of a string of length \begin{align*}L\end{align*} is an antinode. The fundamental mode (the first harmonic) for the length \begin{align*}L\end{align*} of string contains only one-fourth of a wavelength as shown in Figure below. Therefore L, the length of the unstretched string, is equal to one-quarter the wavelength, which is \begin{align*}\frac{1}{4} \lambda_1 = L \rightarrow \lambda_1 = 4L\end{align*}.

The second harmonic contains three-quarters of a wavelength \begin{align*}\frac{3}{4} \lambda_2 = L \rightarrow \lambda_2 = \frac{4}{3}L\end{align*} as shown in Figure below.

The third harmonic contains five-fourths of a wavelength \begin{align*}\frac{5}{4} \lambda_2 = L \rightarrow \lambda_2 = \frac{4}{5}L\end{align*} as shown in Figure below.

If the pattern continues, then the fourth harmonic will have a wavelength of \begin{align*}\frac{7}{4}\lambda_4 = L \rightarrow \lambda_4 = \frac{4}{7}L\end{align*}. Looking at the expressions for the length of the string in terms of the wavelength, a simple pattern emerges \begin{align*}\frac{1}{4}\lambda_1, \frac{3}{4}\lambda_2, \frac{5}{4}\lambda_3, \frac{7}{4}\lambda_4 \ldots\end{align*}. We can express the condition for resonance as \begin{align*}L = \frac{n}{4}\lambda_n\end{align*} or \begin{align*}\lambda_n = \frac{4}{n}L\end{align*}, where \begin{align*}L\end{align*} is the length of string and \begin{align*}n = 1,3,5 \ldots\end{align*}.

As long as the tension in the string remains fixed, the velocity of the wave along the string remains constant. Does it seem reasonable that a sagging string will not support the same wave velocity as a taut string? Since \begin{align*}v = \lambda f\end{align*} product \begin{align*}\lambda f\end{align*} is constant as long as the wave velocity remains constant. Therefore, for a string vibrating in many different modes, we have \begin{align*}v = \lambda_1f_1 = \lambda_2f_2 = \lambda_3f_3 \ldots\end{align*}.

#### Illustrative Example 1

1a. If the frequency of the first harmonic for a string fixed at both ends is \begin{align*}f_1\end{align*}, determine the frequency for successive harmonics in terms of \begin{align*}f_1\end{align*}.

Answer: We know that \begin{align*}\lambda_n = \frac{2}{n}L\end{align*} and \begin{align*}v = \lambda f\end{align*}. Combining, we have \begin{align*}v = \left ( \frac{2}{n} L \right ) f \rightarrow f_n = \frac{n}{2L}v\end{align*}. But \begin{align*}v\end{align*} can be expressed \begin{align*}v=\lambda_1f_1=2Lf_1\end{align*} and substituted into \begin{align*}f_n = \frac{n}{2L}v\end{align*}, giving

\begin{align*}f_n = \frac{n}{2L}(2Lf_1)=nf_1 \rightarrow f_n=nf_1\end{align*}

1b. If the first harmonic has frequency of 261 Hz, what frequencies do the second and third harmonics have?

Since

\begin{align*}f_n &= nf_1 \rightarrow 2(261 ~ \text{Hz}) = 522 ~ \text{Hz} \\ f_n &= nf_1 \rightarrow 3(261 ~ \text{Hz}) = 783 ~ \text{Hz}\end{align*}

All whole number multiples of the first harmonic (the fundamental) are called harmonics. String instruments, as well as non-string instruments, can actually vibrate with many different frequencies simultaneously (called modes). For example, a string may vibrate with frequencies 261 Hz, 522 Hz and 783 Hz simultaneously.

One of attributes of the “quality” or “timbre” of musical instruments depends upon the combination of the various overtones produced by the instrument.

### Check Your Understanding

1. A tuning fork has a frequency of 512 Hz stamped on it. When it is struck, a student claims she can hear higher frequencies from the tuning fork. Is this possible?

Answer: Yes, it is. The tuning fork may be producing harmonics, in which case the student may be hearing frequencies in multiples of 512 Hz, such as 1,024 Hz and 1,536 Hz.

2. A string with a fundamental frequency of 220 Hz vibrates in its third harmonic with a wavelength of 60 cm. What is the wave velocity on the string?

Answer: \begin{align*}v = \lambda f \end{align*} but \begin{align*}f = 3f_1 = 3(220 ~ \text{Hz}) = 660 ~ \text{Hz}\end{align*}, so

\begin{align*}\lambda &= 0.60 ~ \text{m} \\ v &= (660 ~ \text{Hz})(0.60 ~ \text{m}) = 396 ~ \text{m/s}\end{align*}.

### Open and Closed Pipes and Tubes

In our discussions of pipes, the length of the pipe will be assumed to be much greater than the diameter of the pipe.

An open pipe, as the name implies, has both ends open. Though open pipes have antinodes at their ends, the resonant conditions for standing waves in an open pipe are the same as for a string fixed at both ends. Thus for an open pipe we have: for \begin{align*}n = 1,2,3..., L = \frac{n}{2} \lambda_n\end{align*}, or \begin{align*}\lambda_n = \frac{2}{n}L\end{align*}.

There is a simple experiment your instructor may have you do in class that demonstrates resonance in an open tube. Roll two sheets of long paper into two separate tubes and use a small amount of tape to keep them rolled. Have the diameter of one tube just small enough to fit inside the other tube so the inside tube can freely slide back and forth. Hold a struck tuning fork (your instructor will make sure the frequency is adequate) close to the end of the outer tube while the inside tube is moved slowly. When the total length of the tubes is the proper length to establish resonance, you’ll hear a noticeable increase in the volume of the sound. At this moment, there are standing waves present in the tubes.

A closed pipe is closed at only one end. Closed pipes have the same standing wave patterns as a string fixed at one end and unbound at the other end. They therefore have the same resonant conditions as a string fixed at only one end, for \begin{align*}n = 1,3,5 \ldots, L = \frac{n}{4} \lambda_n\end{align*} or \begin{align*}\lambda_n = \frac{4}{n}L\end{align*}.

A closed pipe supporting the first harmonic (the fundamental frequency) will fit one-fourth of the wavelength, the second harmonic will fit three-fourths, and so on, as shown in Figure below. Compare these pictures to those in the figures above for a string fixed at only one end

A standard physics laboratory experiment is to find the velocity of sound by using a tuning fork that vibrates over a closed pipe as shown in Figure below. The water level in a pipe is slowly changed until the first harmonic is heard.

### Interactive Simulation

#### Illustrative Example 2

Resonance is established in a hollow tube similar to that shown in Figure above with a tuning fork of 512 Hz. The distance from the tube opening to the water level is 16.8 cm.

a. What is the velocity of sound according to this experiment?

Answer:  The wave velocity equation is \begin{align*}v = \lambda f\end{align*}. One-fourth of the wave occupies the length of the tube for the first harmonic. So the wavelength of the resonant wave must be four times the length of the hollow tube.  That is,

\begin{align*}\lambda_1 &= 4L = 4 \times 16.8 ~ \text{cm} = 67.2 ~ \text{cm} = 0.672 ~ \text{m} \\ v &= (0.672 ~ \text{m}) ( 512 ~ \text{Hz}) = 344 ~ \text{m/s}\end{align*}

b. The velocity of sound changes with temperature as given by the formula \begin{align*}v = 330 + 0.6T\end{align*}, where \begin{align*}T\end{align*} is the temperature in degrees centigrade. Using the result of part A, determine the temperature at the location the experiment was conducted.

Answer: We simply set the result of part A equal to the given equation:

\begin{align*}344 = 330 + 0.6 T \rightarrow T = 23.3^\circ C\end{align*} (or about \begin{align*}74^\circ F\end{align*}).

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Date Created:
Mar 11, 2013