Standing+Waves

=Standing Waves= include page="foo"

In physics, a standing wave is produced under certain circumstances and has many identifiable properties. Standing waves and resonance are both essential parts of the study of sound. They are fundamental aspects of physics in general and have many practical applications.

To begin, a wave is **the motion of a disturbance**. There are two basic kinds of waves. When particle motion is **parallel** to wave motion, a **longitudinal wave** can be observed. When particle motion is **perpendicular** to wave motion, a **transverse wave** can be observed**.** The images below depict each of the two types and their differences. http://jasonwupilly.com/wp-content/uploads/2010/12/waves.jpg

When specific conditions are met, **standing waves** are produced. This commonly observed phenomenon happens when a wave appears to, as the name suggests, stand still. At the right frequency, a wave will seem to vibrate up and down in place rather than propagating or traveling. Here we will observe the special characteristics of standing waves and the mathematical descriptions that define them.

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http://web.njit.edu/phys_lab/Laboratory%20Manual/laboratory231/labW/standing_wave1.gif

Nodes and Antinodes
There is a specific frequency (called the natural frequency) at which each wave will resonate. At this frequency, the wave will no longer appear to travel from side to side. Study the figure above. The blue and green waves travel from one side to the other. However, when they interfere, they produce a standing wave (in red) that stays in place. At spots where the a peak of one traveling wave interferes with the trough of the other (same magnitude, opposite signs), the net displacement sums to zero. This is called a **node**. The standing wave displays no motion at this point. Specific parts of the red wave oscillate dramatically. These spots are located directly between two adjacent nodes and are called **antinodes**. Note the presence of these terms in the diagram below:



http://www.nhn.ou.edu/~kieran/reuhome/vizqm/figs/waven.gif

This picture shows one full wavelength (an entire period). Before we move on to the next section, observe that:



This merely demonstrates that the distance between adjacent nodes is half the full wavelength.

Another useful relationship to consider involves the frequency of the wave. The so-called "universal wave equation" states that:



One Dimensional Systems
In a one dimensional system, there are a few different cases in which we can examine the relationship between wavelength and the length of a string or tube in which the wave is formed. Here we will describe and examine three possibilities.

String Fixed At Both Ends
Consider a string fixed at both ends. This means that the two ends are nodes. In order to establish a pattern, we will examine the first three most basic kinds of waves that can be produced in this system. The simplest way (called the first harmonic, as mentioned earlier) to make a wave in this scenario is shown below. The wave needs to start and end with a node, so the least complicated configuration contains only one antinode between them. Here you will learn the general approach to deriving these equations.



@http://www.ipod.org.uk/reality/reality_standing.gif

A full wavelength goes from one peak of a wave to another. The figure above is, observably, **half a wavelength**. Meaning that:



This, however, is still not enough to determine the relationship between wavelength and the length of the string for any harmonic/overtone in the system. The process of establishing a pattern and deriving an equation should be studied carefully and students should verify that they can create a general formulas on their own.

The next possible standing wave (second harmonic/first overtone) is:



@http://www.ipod.org.uk/reality/reality_standing.gif

This configuration represents one full wavelength. If the wave continued beyond this point, it would repeat the same pattern. When n, the harmonic number, equals 2, we may say that

While the first version of the equation is simplified, the second will allow us to recognize a trend more easily.

The third harmonic/second overtone is as follows:



@http://www.ipod.org.uk/reality/reality_standing.gif

This configuration exceeds a full wavelength. Knowing that the distance from node to adjacent node is 0.5λ (Equation 1), we can determine how much beyond 1λ the wave reaches. When the harmonic number is three,



Now it is time to look back at the last three equations and try to find a general equation that describes this phenomenon. As you can see, the wavelength increases by 0.5λ at each harmonic. Therefore we are able to say that:



This information is useful when designing a simple one string guitar. Waves and sound are a huge part of the construction of musical instruments.
 * Note that n represents the harmonic number and n = 1, 2, 3...**

Tube Closed at One End
Consider an air column that is closed at one end (meaning the wave has a node on this side) and open on the other (ends with an antinode). Since the adjacent parts are a node and an antinode, we do not have to have anything between them to form the first wave. This resembles the structure of a pan flute. Visualize the simplest scenario:



@http://www.electronicsteacher.com/alternating-current/transmission-lines/02375.png

If the distance from node to adjacent node is half a wavelength and antinodes are exactly halfway between two nodes, then the distance from node to adjacent antinode is one-fourth of a wavelength.



This already differs from the pattern seen above, meaning that the equation will not be the same. Let's see what happens when we increment the harmonic number by one:



@http://www.electronicsteacher.com/alternating-current/transmission-lines/02375.png

The new equation is:



For the third harmonic:



@http://www.electronicsteacher.com/alternating-current/transmission-lines/02375.png



At this point, we are able to make certain generalizations about the mathematical relationship between length and wavelength. The numerators in the three equations are consecutive odd numbers. There are two ways to state the formula:





In the first version, the numerator only takes odd numbers. Note that, in the alternative form, the expression in the numerator will always yield an odd number. This is a useful application of a simple algebraic trick.

Tube Open at Both Ends
We have examined what happens when a wave must start an end with a node. We have derived the equation relating length and wavelength for waves that start with a node and end in an antinode. The last logical combination would be a wave starting and ending with an antinode. Finding this equation is usually an exercise for students to try on their own in order to test their understanding. It is advisable that students do the problem themselves and use the explanation below to check their work. Imagine an air column with two open ends. Here is the first harmonic:



@http://www.ipodphysics.com/resources/21803.gif

Just as the distance from adjacent node to node is half a wavelength, as is the distance from adjacent antinode to antinode.



The next wave will add another node and antionode:



http://www.ipodphysics.com/resources/21803.gif

This pattern might start to seem familiar:

Finally, the third harmonic:



http://www.ipodphysics.com/resources/21803.gif



It is clear that this case resembles the first we studied. Therefore, the general equation is the same:




 * where n represents the harmonic number and n = 1, 2, 3...**

Note: Do not be intimidated by the number of equations presented above. Only general equations are important. Instead of memorizing, make sure you are able to derive these equations using the process shown above.

Other General Equations and Relationships
Let's consider the **first harmonic** of a wave formed by a string fixed at both ends or a pipe open at both ends. This means that the wave is created in the most basic way possible at the fundamental frequency. In the first harmonic, the length of the string (L) is the distance from node to node (or antinode to antinode) in Equation 1. This can also be determined by studying Equations 3 and 12. Solve for lambda in these two equations and substitute your result into Equation 2. By now solving Equation 2 for the fundamental frequency, we see that:



A pipe closed at one end and open at the other is somewhat different. This system has a node at the closed end and an antinode at the open end. A wave can be formed from this alone. Solve Equation 7 for lambda and plug your result into Equation 2. Solve for fundamental frequency and find that:



Standing waves are often studied in relation to **harmonic series**. The harmonics of a wave are integer multiples of the fundamental frequency at which the wave becomes standing. It is important for students to realize that the frequencies of the first, second, third, etc. harmonics are positive integer multiples of the fundamental frequency. These are some of the first equations presented in your textbook.

Sample Problems
1) A stretched string fixed at each end has a mass of 40.0 g and a length of 8.00 m. The tension in the string is 49.0 N. (a) Find the wavelength of the string. (b) Determine the positions of the nodes and antinodes for the third harmonic.

2) Consider a pipe open on both ends that is 3.45 meters long. Determine the wavelengths for the first, third, and sixth harmonics.

3) Using the same pipe and harmonics, find the frequencies. Assume that the speed of sound in air is 343 m/s.

4) Consider the string in Sample Problem 1. Solve for the wavelength and frequency of the 2nd harmonic. Again, assume the speed of sound in air is 343 m/s.

5) Using the string in Sample Problem 4, solve for the frequency of the second harmonic. Again, the speed of sound in air should be set to 343 m/s.

6) A monkey is swinging from vines in a jungle. It loses its grip and falls, smacking its head on a conveniently placed copper tube, producing a clear tone. Sadly, the monkey didn't quite survive to hear it. The tube is closed on one end and has a length of 4.6 meters. Find the wavelength of its fundamental tone.

Solutions
1) A stretched string fixed at each end has a mass of 40.0 g and a length of 8.00 m. The tension in the string is 49.0 N. (a) Find the wavelength of the string. (b) Determine the positions of the nodes and antinodes for the third harmonic.

(a) When starting a problem like this, **begin by asking yourself: what kind of one dimensional system is this?** Once that has been determined, chose relevant equations based on what information you know and what is missing. This is a one dimensional system with a string with both ends fixed. We must find the wavelength. First solve the appropriate equation for wavelength **with variables only**:





Now we can plug in values. **Always verify that the length provided is in meters, otherwise a conversion is necessary**. We want the third harmonic, meaning that n = 3. The problem also provides L, the length of the string.





(b) Next, we have to use our knowledge of the construction of the third harmonic:

http://www.ipod.org.uk/reality/reality_standing.gif

The nodes and antinodes break the wave into six equal sections. The length of each of these sections is 8.00/6 meters. Therefore, the positions of the nodes and antinodes are multiples of this number. **Nodes are at 0 m, 2.67 m, 5.33 m, and 8.00 m** from the left. **Antinodes are at 1.33 m, 4.00 m, 6.67 m** from the left.

2) Consider a pipe open on both ends that is 3.45 meters long. Determine the wavelengths and frequencies for the first, third, and sixth harmonics. Assume the speed of sound in the air is 343 m/s.

It is important to first determine which equations we will be employing to solve this problem. As we are dealing with a pipe open at both ends, it is important to use Equation 15: Set the equation equal to lambda since we are attempting to find the wavelength first: After this, now plug in the values for n that we established in the problem: 1, 3, and 6. You should obtain wavelengths of:

1st harmonic- **6.9 meters**

3rd harmonic- **2.3 meters**

6th harmonic- **1.15 meters**

3) Using the same pipe and harmonics, find the frequencies. Assume that the speed of sound in air is 343 m/s.

Now, we'll use Equation 2 to solve for the frequency of the harmonics:

If you plug in the numbers we obtained for the wavelengths and use the speed of sound to be 343 m/s, you should arrive at these results:

1st harmonic- **49.71 Hz**

3rd harmonic- **149.13 Hz**

6th harmonic- **298.26 Hz**

4) Consider the string in Sample Problem 1. Solve for the wavelength of the second harmonic.

We will use the same equations used in Sample Problem 1 to find the wavelength of the harmonic:

We can use this to find that the wavelength of the second harmonic should be equal to **8 meters**.

5) Using the string in Sample Problem 4, solve for the frequency of the second harmonic. Again, the speed of sound in air should be set to 343 m/s.

We can now use Equation 2 to find the frequency of the 2nd harmonic.

You should arrive at a frequency of **42.875 Hz**.

6) A monkey is swinging from vines in a jungle. It loses its grip and falls, smacking its head on a conveniently placed copper tube, producing a clear tone. Sadly, the monkey didn't quite survive to hear it. The tube is closed on one end and has a length of 4.6 meters. Find the wavelength of its fundamental tone.

Since we are dealing with a tube with one end open, we must use Equation 11:

Since the fundamental is just the first harmonic, we can easily salve for lambda.

You should yield an answer of about **6.13 meters**.