in the resonance tube experiment first resonant length is l1

This is a simulation of a standard physics demonstration to measure the speed of sound in air. A vibrating tuning fork is held above a tube - the tube has some water in it, and the level of the water in the tube can be adjusted. This gives a column of air in the tube, between the top of the water and the top of the tube. By setting the water level appropriately, the height of the air column can be such that it gives a resonance condition for the sound wave produced by the tuning fork. In the real experiment, resonance is found by listening - the sound from the tube is loudest at resonance. In the simulation, resonance is shown by the amplitude of the wave in the air column. The larger the amplitude, the closer to resonance. Note that at certain special heights of the air column, no sound is heard - this is because of completely destructive interference.

In addition, there is always a node (for displacement of the air molecules) at the water surface. To a first approximation, resonance occurs when there is an anti-node at the top of the tube. Knowing the frequency of the tuning fork, the height of the air column, and the appropriate equation for standing waves in a tube like this, the speed of sound in air can be determined experimentally. What do you get for the speed of sound in air in this simulation?

• CBSE Class 11
• CBSE Class 11 Physics Practical
• To Find The Speed Of Sound In Air At Room Temperature Using A Resonance Tube By Two Resonance Positions

To Find the Speed of Sound in Air at Room Temperature Using a Resonance Tube by Two Resonance Positions

To find the speed of sound in air at room temperature using a resonance tube by two resonance positions.

Apparatus/Materials Required

• Resonance tube
• Thermometer
• Set Squares
• Water in a beaker
• Two-timing forks of known frequency

Let l 1 and l 2 be the length of the air column for the first and second resonance respectively with a tuning fork of frequency f .

The speed is given by the formula

Substituting, we get

• By making base horizontal with the help of levelling screws, set the resonance tube vertical.
• Fix the reservoir R in the uppermost position.
• Loosen the pinch cock P and fill the reservoir and metallic tube completely with water by a beaker.
• Tighten the pinch cock, lower the reservoir and fix it in the lowest position.
• Take a tuning fork of higher frequency

Observation

The temperature of the air in the air column:

(i) in the beginning ____ °C

(ii) at the end _____°C

The mean temperature is calculated as follows:

Frequency of first tuning fork = f 1

Frequency of second tuning fork = f 2

Calculation

From the first tuning fork,

From the second tuning fork,

The mean velocity at room temperature is given as follows:

At room temperature, the velocity of sound in air is _____ m/s.

1. What is the working principle of the resonance tube?

It works on the principle of resonance of the air column with a tuning fork.

2. What types of waves are produced in the air column?

Longitudinal stationary waves are produced in the air column.

3. Do you find the velocity of sound in air column or in the water column?

The velocity of sound is found in the air column above the water column.

4. What are the possible errors in the result?

The two possible errors in the result are:

(i) The enclosed air in the air column is denser than the outside air, this may reduce the velocity of air.

(ii) The humidity above the enclosed water column may increase the velocity of sound.

5. Will the result be affected if we take other liquids than water?

No, it will not be affected.

Sound Visualisation

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• Rotational Dynamics
• Periodic Motion
• Fluid Statics
• First Law of Thermodynamics
• Second Law of Thermodynamics
• Wave Motion
• Mechanical Waves
• Waves in Pipes and Strings
• Acoustic Phenomena
• Nature and Propagation of Light
• Interference
• Diffraction
• Polarization
• Electrical Circuits
• Thermoelectric Effects
• Magnetic Field
• Magnetic Properties of Material
• Electromagnetic Induction
• Alternating Currents
• Semiconductor Devices
• Quantization of Energy
• Radioactivity and Nuclear Reaction

1 Physics -- Waves in Pipes and Strings

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Resonance tube experiment

The experimental setup of resonance tube experiment to determine the velocity of sound in air is shown in figure.

Initially, to find the velocity of sound in air the water level in the tube is changed by changing the place of water reservoir. A tuning fork of frequency ‘f’ is set into vibration by striking it against the rubber pad and held it horizontal over the open end. The length of air column in the tube is increased by changing the position of reservoir in doing so, a condition is reached at which the sound heard is maximum, this condition is known as first resonance.

If l1 be the first resonating length and l be the wave length of sound when,

If we go on increasing the length of air column in the tube the intensity of the sound heard goes on decreasing first become minimum and then again goes on increasing at a certain length. The intensity of sound becomes maximum again, these condition is called second resonating length. If l 2  be the second resonating length,

If U be the velocity of sound at room temperature. Then,

Where f is the frequency of sound which is equal to frequency of turning fork, Velocity of sound at NTP is given by

where, P = pressure at NTP f = aqueous tension r = cubical expansively of air = 1/273

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Discussions

More notes on waves in pipes and strings.

General formulae for calculating wave velocity in different media

Swarthmore Physics Demonstrations

Resonance tube.

• Purpose: Demonstrate resonance and measure speed of sound.
• Two different resonance tubes are shown above.

For Smaller Tube

• Strike tuning fork on rubber pad; place fork near larger opening of tube; slide inner tube in and out until resonance is heard. Use tube length and frequency to determine speed of sound.

For larger Tube

• Use speaker and function generator to find resonant frequency of tube. Once resonance is found, use vernier microphone to locate nodes and anti-nodes.Use tube length and frequency to determine speed of sound.
• Located in L02,

Physics of Hearing

Sound interference and resonance: standing waves in air columns, learning objectives.

By the end of this section, you will be able to:

• Define antinode, node, fundamental, overtones, and harmonics.
• Identify instances of sound interference in everyday situations.
• Describe how sound interference occurring inside open and closed tubes changes the characteristics of the sound, and how this applies to sounds produced by musical instruments.
• Calculate the length of a tube using sound wave measurements.

Figure 1. Some types of headphones use the phenomena of constructive and destructive interference to cancel out outside noises. (credit: JVC America, Flickr)

Interference is the hallmark of waves, all of which exhibit constructive and destructive interference exactly analogous to that seen for water waves. In fact, one way to prove something “is a wave” is to observe interference effects. So, sound being a wave, we expect it to exhibit interference; we have already mentioned a few such effects, such as the beats from two similar notes played simultaneously.

Figure 2 shows a clever use of sound interference to cancel noise. Larger-scale applications of active noise reduction by destructive interference are contemplated for entire passenger compartments in commercial aircraft. To obtain destructive interference, a fast electronic analysis is performed, and a second sound is introduced with its maxima and minima exactly reversed from the incoming noise. Sound waves in fluids are pressure waves and consistent with Pascal’s principle; pressures from two different sources add and subtract like simple numbers; that is, positive and negative gauge pressures add to a much smaller pressure, producing a lower-intensity sound. Although completely destructive interference is possible only under the simplest conditions, it is possible to reduce noise levels by 30 dB or more using this technique.

Figure 2. Headphones designed to cancel noise with destructive interference create a sound wave exactly opposite to the incoming sound. These headphones can be more effective than the simple passive attenuation used in most ear protection. Such headphones were used on the record-setting, around the world nonstop flight of the Voyager aircraft to protect the pilots’ hearing from engine noise.

Where else can we observe sound interference? All sound resonances, such as in musical instruments, are due to constructive and destructive interference. Only the resonant frequencies interfere constructively to form standing waves, while others interfere destructively and are absent. From the toot made by blowing over a bottle, to the characteristic flavor of a violin’s sounding box, to the recognizability of a great singer’s voice, resonance and standing waves play a vital role.

Interference

Interference is such a fundamental aspect of waves that observing interference is proof that something is a wave. The wave nature of light was established by experiments showing interference. Similarly, when electrons scattered from crystals exhibited interference, their wave nature was confirmed to be exactly as predicted by symmetry with certain wave characteristics of light.

Suppose we hold a tuning fork near the end of a tube that is closed at the other end, as shown in Figure 3, Figure 4, Figure 5, and Figure 6. If the tuning fork has just the right frequency, the air column in the tube resonates loudly, but at most frequencies it vibrates very little. This observation just means that the air column has only certain natural frequencies. The figures show how a resonance at the lowest of these natural frequencies is formed. A disturbance travels down the tube at the speed of sound and bounces off the closed end. If the tube is just the right length, the reflected sound arrives back at the tuning fork exactly half a cycle later, and it interferes constructively with the continuing sound produced by the tuning fork. The incoming and reflected sounds form a standing wave in the tube as shown.

The standing wave formed in the tube has its maximum air displacement (an  antinode ) at the open end, where motion is unconstrained, and no displacement (a node ) at the closed end, where air movement is halted. The distance from a node to an antinode is one-fourth of a wavelength, and this equals the length of the tube; thus, λ  = 4 L . This same resonance can be produced by a vibration introduced at or near the closed end of the tube, as shown in Figure 7. It is best to consider this a natural vibration of the air column independently of how it is induced.

Figure 7. The same standing wave is created in the tube by a vibration introduced near its closed end.

Figure 8. Another resonance for a tube closed at one end. This has maximum air displacements at the open end, and none at the closed end. The wavelength is shorter, with three-fourths λ′ equaling the length of the tube, so that [latex]\lambda^{\prime}=\frac{4L}{3}\\[/latex]. This higher-frequency vibration is the first overtone.

Given that maximum air displacements are possible at the open end and none at the closed end, there are other, shorter wavelengths that can resonate in the tube, such as the one shown in Figure 8. Here the standing wave has three-fourths of its wavelength in the tube, or [latex]L=\frac{3}{4}\lambda^{\prime}\\[/latex], so that [latex]\lambda^{\prime}=\frac{4L}{3}\\[/latex]. Continuing this process reveals a whole series of shorter-wavelength and higher-frequency sounds that resonate in the tube. We use specific terms for the resonances in any system. The lowest resonant frequency is called the fundamental , while all higher resonant frequencies are called overtones . All resonant frequencies are integral multiples of the fundamental, and they are collectively called harmonics . The fundamental is the first harmonic, the first overtone is the second harmonic, and so on. Figure 9 shows the fundamental and the first three overtones (the first four harmonics) in a tube closed at one end.

Figure 9. The fundamental and three lowest overtones for a tube closed at one end. All have maximum air displacements at the open end and none at the closed end.

The fundamental and overtones can be present simultaneously in a variety of combinations. For example, middle C on a trumpet has a sound distinctively different from middle C on a clarinet, both instruments being modified versions of a tube closed at one end. The fundamental frequency is the same (and usually the most intense), but the overtones and their mix of intensities are different and subject to shading by the musician. This mix is what gives various musical instruments (and human voices) their distinctive characteristics, whether they have air columns, strings, sounding boxes, or drumheads. In fact, much of our speech is determined by shaping the cavity formed by the throat and mouth and positioning the tongue to adjust the fundamental and combination of overtones. Simple resonant cavities can be made to resonate with the sound of the vowels, for example. (See Figure 10.) In boys, at puberty, the larynx grows and the shape of the resonant cavity changes giving rise to the difference in predominant frequencies in speech between men and women.

Figure 10. The throat and mouth form an air column closed at one end that resonates in response to vibrations in the voice box. The spectrum of overtones and their intensities vary with mouth shaping and tongue position to form different sounds. The voice box can be replaced with a mechanical vibrator, and understandable speech is still possible. Variations in basic shapes make different voices recognizable.

Now let us look for a pattern in the resonant frequencies for a simple tube that is closed at one end. The fundamental has λ  = 4 L , and frequency is related to wavelength and the speed of sound as given by  v w  =  fλ.

Solving for f in this equation gives

[latex]f=\frac{v_{\text{w}}}{\lambda}=\frac{v_{\text{w}}}{4L}\\[/latex],

where v w is the speed of sound in air. Similarly, the first overtone has [latex]\lambda^{\prime}=\frac{4L}{3}\\[/latex] (see Figure 9), so that

[latex]f^{\prime}=3\frac{v_{\text{w}}}{4L}=3f\\[/latex].

Because f ′ = 3 f , we call the first overtone the third harmonic. Continuing this process, we see a pattern that can be generalized in a single expression. The resonant frequencies of a tube closed at one end are

[latex]f_n=n\frac{v_{\text{w}}}{4L},n=1,3,5\\[/latex],

where f 1 is the fundamental, f 3 is the first overtone, and so on. It is interesting that the resonant frequencies depend on the speed of sound and, hence, on temperature. This dependence poses a noticeable problem for organs in old unheated cathedrals, and it is also the reason why musicians commonly bring their wind instruments to room temperature before playing them.

Example 1. Find the Length of a Tube with a 128 Hz Fundamental

• What length should a tube closed at one end have on a day when the air temperature, is 22.0ºC, if its fundamental frequency is to be 128 Hz (C below middle C)?
• What is the frequency of its fourth overtone?

The length L can be found from the relationship in [latex]f_n=n\frac{v_{\text{w}}}{4L}\\[/latex], but we will first need to find the speed of sound v w .

Solution for Part 1

Identify knowns:

• the fundamental frequency is 128 Hz
• the air temperature is 22.0ºC

Use [latex]f_n=n\frac{v_{\text{w}}}{4L}\\[/latex] to find the fundamental frequency ( n  = 1):

[latex]f_1=\frac{v_{\text{w}}}{4L}\\[/latex]

Solve this equation for length: [latex]L=\frac{v_{\text{w}}}{4f_1}\\[/latex].

Find the speed of sound using [latex]v_{\text{w}}=\left(331\text{ m/s}\right)\sqrt{\frac{T}{273\text{ K}}}\\[/latex].

[latex]v_{\text{w}}=\left(331\text{ m/s}\right)\sqrt{\frac{295\text{ K}}{273\text{ K}}}=344\text{ m/s}\\[/latex]

Enter the values of the speed of sound and frequency into the expression for L .

[latex]L=\frac{v_{\text{w}}}{4f_1}=\frac{344\text{ m/s}}{4(128\text{ Hz})}=0.672\text{ m}\\[/latex]

Discussion on Part 1

Many wind instruments are modified tubes that have finger holes, valves, and other devices for changing the length of the resonating air column and hence, the frequency of the note played. Horns producing very low frequencies, such as tubas, require tubes so long that they are coiled into loops.

Solution for Part 2

• the first overtone has n  = 3
• the second overtone has n  = 5
• the third overtone has n  = 7
• the fourth overtone has n  = 9

Enter the value for the fourth overtone into [latex]f_n=n\frac{v_{\text{w}}}{4L}\\[/latex] :

[latex]f_9=9\frac{v_{\text{w}}}{4L}=9f_1=1.15\text{ kHz}\\[/latex]

Discussion on Part 2

Whether this overtone occurs in a simple tube or a musical instrument depends on how it is stimulated to vibrate and the details of its shape. The trombone, for example, does not produce its fundamental frequency and only makes overtones.

Another type of tube is one that is open at both ends. Examples are some organ pipes, flutes, and oboes. The resonances of tubes open at both ends can be analyzed in a very similar fashion to those for tubes closed at one end. The air columns in tubes open at both ends have maximum air displacements at both ends, as illustrated in Figure 11. Standing waves form as shown.

Figure 11. The resonant frequencies of a tube open at both ends are shown, including the fundamental and the first three overtones. In all cases the maximum air displacements occur at both ends of the tube, giving it different natural frequencies than a tube closed at one end.

Based on the fact that a tube open at both ends has maximum air displacements at both ends, and using Figure 11 as a guide, we can see that the resonant frequencies of a tube open at both ends are:

[latex]f_n=n\frac{v_{\text{w}}}{2L},n=1,2,3,\dots,\\[/latex]

where f 1 is the fundamental, f 2 is the first overtone, f 3 is the second overtone, and so on. Note that a tube open at both ends has a fundamental frequency twice what it would have if closed at one end. It also has a different spectrum of overtones than a tube closed at one end. So if you had two tubes with the same fundamental frequency but one was open at both ends and the other was closed at one end, they would sound different when played because they have different overtones. Middle C, for example, would sound richer played on an open tube, because it has even multiples of the fundamental as well as odd. A closed tube has only odd multiples.

Real-World Applications: Resonance in Everyday Systems

Resonance occurs in many different systems, including strings, air columns, and atoms. Resonance is the driven or forced oscillation of a system at its natural frequency. At resonance, energy is transferred rapidly to the oscillating system, and the amplitude of its oscillations grows until the system can no longer be described by Hooke’s law. An example of this is the distorted sound intentionally produced in certain types of rock music.

Figure 12. String instruments such as violins and guitars use resonance in their sounding boxes to amplify and enrich the sound created by their vibrating strings. The bridge and supports couple the string vibrations to the sounding boxes and air within. (credits: guitar, Feliciano Guimares, Fotopedia; violin, Steve Snodgrass, Flickr)

Figure 13. Resonance has been used in musical instruments since prehistoric times. This marimba uses gourds as resonance chambers to amplify its sound. (credit: APC Events, Flickr)

Wind instruments use resonance in air columns to amplify tones made by lips or vibrating reeds. Other instruments also use air resonance in clever ways to amplify sound. Figure 12 shows a violin and a guitar, both of which have sounding boxes but with different shapes, resulting in different overtone structures. The vibrating string creates a sound that resonates in the sounding box, greatly amplifying the sound and creating overtones that give the instrument its characteristic flavor. The more complex the shape of the sounding box, the greater its ability to resonate over a wide range of frequencies. The marimba, like the one shown in Figure 13 uses pots or gourds below the wooden slats to amplify their tones. The resonance of the pot can be adjusted by adding water.

We have emphasized sound applications in our discussions of resonance and standing waves, but these ideas apply to any system that has wave characteristics. Vibrating strings, for example, are actually resonating and have fundamentals and overtones similar to those for air columns. More subtle are the resonances in atoms due to the wave character of their electrons. Their orbitals can be viewed as standing waves, which have a fundamental (ground state) and overtones (excited states). It is fascinating that wave characteristics apply to such a wide range of physical systems.

Check Your Understanding

Describe how noise-canceling headphones differ from standard headphones used to block outside sounds.

Regular headphones only block sound waves with a physical barrier. Noise-canceling headphones use destructive interference to reduce the loudness of outside sounds.

How is it possible to use a standing wave’s node and antinode to determine the length of a closed-end tube?

When the tube resonates at its natural frequency, the wave’s node is located at the closed end of the tube, and the antinode is located at the open end. The length of the tube is equal to one-fourth of the wavelength of this wave. Thus, if we know the wavelength of the wave, we can determine the length of the tube.

PhET Explorations: Sound

This simulation lets you see sound waves. Adjust the frequency or volume and you can see and hear how the wave changes. Move the listener around and hear what she hears.

Click to download the simulation. Run using Java.

Section Summary

• Sound interference and resonance have the same properties as defined for all waves.
• In air columns, the lowest-frequency resonance is called the fundamental, whereas all higher resonant frequencies are called overtones. Collectively, they are called harmonics.
• The resonant frequencies of a tube closed at one end are: [latex]{f}_{n}=n\frac{{v}_{w}}{4L}\text{, }n=1, 3, 5\dots\\[/latex],  f 1  is the fundamental and L  is the length of the tube.
• The resonant frequencies of a tube open at both ends are: [latex]{f}_{n}=n\frac{{v}_{w}}{2L}\text{, }n=1, 2, 3\dots\\[/latex]

Conceptual Questions

• How does an unamplified guitar produce sounds so much more intense than those of a plucked string held taut by a simple stick?
• You are given two wind instruments of identical length. One is open at both ends, whereas the other is closed at one end. Which is able to produce the lowest frequency?
• What is the difference between an overtone and a harmonic? Are all harmonics overtones? Are all overtones harmonics?

Problems & Exercises

1. A “showy” custom-built car has two brass horns that are supposed to produce the same frequency but actually emit 263.8 and 264.5 Hz. What beat frequency is produced?

2. What beat frequencies will be present: (a) If the musical notes A and C are played together (frequencies of 220 and 264 Hz)? (b) If D and F are played together (frequencies of 297 and 352 Hz)? (c) If all four are played together?

3. What beat frequencies result if a piano hammer hits three strings that emit frequencies of 127.8, 128.1, and 128.3 Hz?

4. A piano tuner hears a beat every 2.00 s when listening to a 264.0-Hz tuning fork and a single piano string. What are the two possible frequencies of the string?

5. (a) What is the fundamental frequency of a 0.672-m-long tube, open at both ends, on a day when the speed of sound is 344 m/s? (b) What is the frequency of its second harmonic?

6. If a wind instrument, such as a tuba, has a fundamental frequency of 32.0 Hz, what are its first three overtones? It is closed at one end. (The overtones of a real tuba are more complex than this example, because it is a tapered tube.)

7. What are the first three overtones of a bassoon that has a fundamental frequency of 90.0 Hz? It is open at both ends. (The overtones of a real bassoon are more complex than this example, because its double reed makes it act more like a tube closed at one end.)

8. How long must a flute be in order to have a fundamental frequency of 262 Hz (this frequency corresponds to middle C on the evenly tempered chromatic scale) on a day when air temperature is 20.0ºC? It is open at both ends.

9. What length should an oboe have to produce a fundamental frequency of 110 Hz on a day when the speed of sound is 343 m/s? It is open at both ends.

10. What is the length of a tube that has a fundamental frequency of 176 Hz and a first overtone of 352 Hz if the speed of sound is 343 m/s?

11. (a) Find the length of an organ pipe closed at one end that produces a fundamental frequency of 256 Hz when air temperature is 18.0ºC. (b) What is its fundamental frequency at 25.0ºC?

12. By what fraction will the frequencies produced by a wind instrument change when air temperature goes from 10.0ºC to 30.0ºC? That is, find the ratio of the frequencies at those temperatures.

13. The ear canal resonates like a tube closed at one end. (See Figure 5 in Hearing .) If ear canals range in length from 1.80 to 2.60 cm in an average population, what is the range of fundamental resonant frequencies? Take air temperature to be 37.0ºC, which is the same as body temperature. How does this result correlate with the intensity versus frequency graph of the human ear (Figure 14)?

Figure 14. The shaded region represents frequencies and intensity levels found in normal conversational speech. The 0-phon line represents the normal hearing threshold, while those at 40 and 60 represent thresholds for people with 40- and 60-phon hearing losses, respectively.

14. Calculate the first overtone in an ear canal, which resonates like a 2.40-cm-long tube closed at one end, by taking air temperature to be 37.0ºC. Is the ear particularly sensitive to such a frequency? (The resonances of the ear canal are complicated by its nonuniform shape, which we shall ignore.)

15. A crude approximation of voice production is to consider the breathing passages and mouth to be a resonating tube closed at one end. (See Figure 10.) (a) What is the fundamental frequency if the tube is 0.240-m long, by taking air temperature to be 37.0ºC? (b) What would this frequency become if the person replaced the air with helium? Assume the same temperature dependence for helium as for air.

16. (a) Students in a physics lab are asked to find the length of an air column in a tube closed at one end that has a fundamental frequency of 256 Hz. They hold the tube vertically and fill it with water to the top, then lower the water while a 256-Hz tuning fork is rung and listen for the first resonance. What is the air temperature if the resonance occurs for a length of 0.336 m? (b) At what length will they observe the second resonance (first overtone)?

17. What frequencies will a 1.80-m-long tube produce in the audible range at 20.0ºC if: (a) The tube is closed at one end? (b) It is open at both ends?

antinode:  point of maximum displacement

node:  point of zero displacement

fundamental:  the lowest-frequency resonance

overtones:  all resonant frequencies higher than the fundamental

harmonics:  the term used to refer collectively to the fundamental and its overtones

Selected Solutions to Problems & Exercises

3. 0.3 Hz, 0.2 Hz, 0.5 Hz

5. (a) 256 Hz; (b) 512 Hz

7. 180 Hz, 270 Hz, 360 Hz

11. (a) 0.334 m; (b) 259 Hz

13. 3.39 to 4.90 kHz

15. (a) 367 Hz; (b) 1.07 kHz

17. (a) f n = n (47.6 Hz), n  = 1, 3, 5,…, 419; (b) f n   = n (95.3 Hz), n  = 1, 2, 3,…, 210

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• PhET Interactive Simulations . Provided by : University of Colorado Boulder . Located at : http://phet.colorado.edu . License : CC BY: Attribution

16.6 Standing Waves and Resonance

Learning objectives.

By the end of this section, you will be able to:

• Describe standing waves and explain how they are produced
• Describe the modes of a standing wave on a string
• Provide examples of standing waves beyond the waves on a string

Throughout this chapter, we have been studying traveling waves, or waves that transport energy from one place to another. Under certain conditions, waves can bounce back and forth through a particular region, effectively becoming stationary. These are called standing wave s .

Another related effect is known as resonance . In Oscillations , we defined resonance as a phenomenon in which a small-amplitude driving force could produce large-amplitude motion. Think of a child on a swing, which can be modeled as a physical pendulum. Relatively small-amplitude pushes by a parent can produce large-amplitude swings. Sometimes this resonance is good—for example, when producing music with a stringed instrument. At other times, the effects can be devastating, such as the collapse of a building during an earthquake. In the case of standing waves, the relatively large amplitude standing waves are produced by the superposition of smaller amplitude component waves.

Standing Waves

Sometimes waves do not seem to move; rather, they just vibrate in place. You can see unmoving waves on the surface of a glass of milk in a refrigerator, for example. Vibrations from the refrigerator motor create waves on the milk that oscillate up and down but do not seem to move across the surface. Figure 16.25 shows an experiment you can try at home. Take a bowl of milk and place it on a common box fan. Vibrations from the fan will produce circular standing waves in the milk. The waves are visible in the photo due to the reflection from a lamp. These waves are formed by the superposition of two or more traveling waves, such as illustrated in Figure 16.26 for two identical waves moving in opposite directions. The waves move through each other with their disturbances adding as they go by. If the two waves have the same amplitude and wavelength, then they alternate between constructive and destructive interference. The resultant looks like a wave standing in place and, thus, is called a standing wave.

Consider two identical waves that move in opposite directions. The first wave has a wave function of y 1 ( x , t ) = A sin ( k x − ω t ) y 1 ( x , t ) = A sin ( k x − ω t ) and the second wave has a wave function y 2 ( x , t ) = A sin ( k x + ω t ) y 2 ( x , t ) = A sin ( k x + ω t ) . The waves interfere and form a resultant wave

This can be simplified using the trigonometric identity

where α = k x α = k x and β = ω t β = ω t , giving us

which simplifies to

Notice that the resultant wave is a sine wave that is a function only of position, multiplied by a cosine function that is a function only of time. Graphs of y ( x , t ) as a function of x for various times are shown in Figure 16.26 . The red wave moves in the negative x -direction, the blue wave moves in the positive x -direction, and the black wave is the sum of the two waves. As the red and blue waves move through each other, they move in and out of constructive interference and destructive interference.

Initially, at time t = 0 , t = 0 , the two waves are in phase, and the result is a wave that is twice the amplitude of the individual waves. The waves are also in phase at the time t = T 2 . t = T 2 . In fact, the waves are in phase at any integer multiple of half of a period:

At other times, the two waves are 180 ° ( π radians ) 180 ° ( π radians ) out of phase, and the resulting wave is equal to zero. This happens at

Notice that some x -positions of the resultant wave are always zero no matter what the phase relationship is. These positions are called node s . Where do the nodes occur? Consider the solution to the sum of the two waves

Finding the positions where the sine function equals zero provides the positions of the nodes.

There are also positions where y oscillates between y = ± A y = ± A . These are the antinode s . We can find them by considering which values of x result in sin ( k x ) = ± 1 sin ( k x ) = ± 1 .

What results is a standing wave as shown in Figure 16.27 , which shows snapshots of the resulting wave of two identical waves moving in opposite directions. The resulting wave appears to be a sine wave with nodes at integer multiples of half wavelengths. The antinodes oscillate between y = ± 2 A y = ± 2 A due to the cosine term, cos ( ω t ) cos ( ω t ) , which oscillates between ± 1 ± 1 .

The resultant wave appears to be standing still, with no apparent movement in the x -direction, although it is composed of one wave function moving in the positive, whereas the second wave is moving in the negative x -direction. Figure 16.27 shows various snapshots of the resulting wave. The nodes are marked with red dots while the antinodes are marked with blue dots.

A common example of standing waves are the waves produced by stringed musical instruments. When the string is plucked, pulses travel along the string in opposite directions. The ends of the strings are fixed in place, so nodes appear at the ends of the strings—the boundary conditions of the system, regulating the resonant frequencies in the strings. The resonance produced on a string instrument can be modeled in a physics lab using the apparatus shown in Figure 16.28 .

The lab setup shows a string attached to a string vibrator, which oscillates the string with an adjustable frequency f . The other end of the string passes over a frictionless pulley and is tied to a hanging mass. The magnitude of the tension in the string is equal to the weight of the hanging mass. The string has a constant linear density (mass per length) μ μ and the speed at which a wave travels down the string equals v = F T μ = m g μ v = F T μ = m g μ Equation 16.7 . The symmetrical boundary conditions (a node at each end) dictate the possible frequencies that can excite standing waves. Starting from a frequency of zero and slowly increasing the frequency, the first mode n = 1 n = 1 appears as shown in Figure 16.29 . The first mode, also called the fundamental mode or the first harmonic, shows half of a wavelength has formed, so the wavelength is equal to twice the length between the nodes λ 1 = 2 L λ 1 = 2 L . The fundamental frequency , or first harmonic frequency, that drives this mode is

where the speed of the wave is v = F T μ . v = F T μ . Keeping the tension constant and increasing the frequency leads to the second harmonic or the n = 2 n = 2 mode. This mode is a full wavelength λ 2 = L λ 2 = L and the frequency is twice the fundamental frequency:

The next two modes, or the third and fourth harmonics, have wavelengths of λ 3 = 2 3 L λ 3 = 2 3 L and λ 4 = 2 4 L , λ 4 = 2 4 L , driven by frequencies of f 3 = 3 v 2 L = 3 f 1 f 3 = 3 v 2 L = 3 f 1 and f 4 = 4 v 2 L = 4 f 1 . f 4 = 4 v 2 L = 4 f 1 . All frequencies above the frequency f 1 f 1 are known as the overtone s . The equations for the wavelength and the frequency can be summarized as:

The standing wave patterns that are possible for a string, the first four of which are shown in Figure 16.29 , are known as the normal mode s , with frequencies known as the normal frequencies. In summary, the first frequency to produce a normal mode is called the fundamental frequency (or first harmonic). Any frequencies above the fundamental frequency are overtones. The second frequency of the n = 2 n = 2 normal mode of the string is the first overtone (or second harmonic). The frequency of the n = 3 n = 3 normal mode is the second overtone (or third harmonic) and so on.

The solutions shown as Equation 16.15 and Equation 16.16 are for a string with the boundary condition of a node on each end. When the boundary condition on either side is the same, the system is said to have symmetric boundary conditions. Equation 16.15 and Equation 16.16 are good for any symmetric boundary conditions, that is, nodes at both ends or antinodes at both ends.

Example 16.7

Standing waves on a string.

• The velocity of the wave can be found using v = F T μ . v = F T μ . The tension is provided by the weight of the hanging mass.
• The standing waves will depend on the boundary conditions. There must be a node at each end. The first mode will be one half of a wave. The second can be found by adding a half wavelength. That is the shortest length that will result in a node at the boundaries. For example, adding one quarter of a wavelength will result in an antinode at the boundary and is not a mode which would satisfy the boundary conditions. This is shown in Figure 16.31 .
• Begin with the velocity of a wave on a string. The tension is equal to the weight of the hanging mass. The linear mass density and mass of the hanging mass are given: v = F T μ = m g μ = 2 kg ( 9.8 m s ) 0.006 kg m = 57.15 m/s . v = F T μ = m g μ = 2 kg ( 9.8 m s ) 0.006 kg m = 57.15 m/s .
• The frequencies of the first three modes are found by using f = v w λ . f = v w λ . f 1 = v w λ 1 = 57.15 m/s 4.00 m = 14.29 Hz f 2 = v w λ 2 = 57.15 m/s 2.00 m = 28.58 Hz f 3 = v w λ 3 = 57.15 m/s 1.333 m = 42.87 Hz f 1 = v w λ 1 = 57.15 m/s 4.00 m = 14.29 Hz f 2 = v w λ 2 = 57.15 m/s 2.00 m = 28.58 Hz f 3 = v w λ 3 = 57.15 m/s 1.333 m = 42.87 Hz

Significance

Interactive.

Engage the Phet simulation below to play with a 1D or 2D system of coupled mass-spring oscillators. Vary the number of masses, set the initial conditions, and watch the system evolve. See the spectrum of normal modes for arbitrary motion. See longitudinal or transverse modes in the 1D system.

Check Your Understanding 16.7

The equations for the wavelengths and the frequencies of the modes of a wave produced on a string:

were derived by considering a wave on a string where there were symmetric boundary conditions of a node at each end. These modes resulted from two sinusoidal waves with identical characteristics except they were moving in opposite directions, confined to a region L with nodes required at both ends. Will the same equations work if there were symmetric boundary conditions with antinodes at each end? What would the normal modes look like for a medium that was free to oscillate on each end? Don’t worry for now if you cannot imagine such a medium, just consider two sinusoidal wave functions in a region of length L , with antinodes on each end.

The free boundary conditions shown in the last Check Your Understanding may seem hard to visualize. How can there be a system that is free to oscillate on each end? In Figure 16.32 are shown two possible configuration of a metallic rods (shown in red) attached to two supports (shown in blue). In part (a), the rod is supported at the ends, and there are fixed boundary conditions at both ends. Given the proper frequency, the rod can be driven into resonance with a wavelength equal to length of the rod, with nodes at each end. In part (b), the rod is supported at positions one quarter of the length from each end of the rod, and there are free boundary conditions at both ends. Given the proper frequency, this rod can also be driven into resonance with a wavelength equal to the length of the rod, but there are antinodes at each end. If you are having trouble visualizing the wavelength in this figure, remember that the wavelength may be measured between any two nearest identical points and consider Figure 16.33 .

Note that the study of standing waves can become quite complex. In Figure 16.32 (a), the n = 2 n = 2 mode of the standing wave is shown, and it results in a wavelength equal to L . In this configuration, the n = 1 n = 1 mode would also have been possible with a standing wave equal to 2 L . Is it possible to get the n = 1 n = 1 mode for the configuration shown in part (b)? The answer is no. In this configuration, there are additional conditions set beyond the boundary conditions. Since the rod is mounted at a point one quarter of the length from each side, a node must exist there, and this limits the possible modes of standing waves that can be created. We leave it as an exercise for the reader to consider if other modes of standing waves are possible. It should be noted that when a system is driven at a frequency that does not cause the system to resonate, vibrations may still occur, but the amplitude of the vibrations will be much smaller than the amplitude at resonance.

A field of mechanical engineering uses the sound produced by the vibrating parts of complex mechanical systems to troubleshoot problems with the systems. Suppose a part in an automobile is resonating at the frequency of the car’s engine, causing unwanted vibrations in the automobile. This may cause the engine to fail prematurely. The engineers use microphones to record the sound produced by the engine, then use a technique called Fourier analysis to find frequencies of sound produced with large amplitudes and then look at the parts list of the automobile to find a part that would resonate at that frequency. The solution may be as simple as changing the composition of the material used or changing the length of the part in question.

There are other numerous examples of resonance in standing waves in the physical world. The air in a tube, such as found in a musical instrument like a flute, can be forced into resonance and produce a pleasant sound, as we discuss in Sound .

At other times, resonance can cause serious problems. A closer look at earthquakes provides evidence for conditions appropriate for resonance, standing waves, and constructive and destructive interference. A building may vibrate for several seconds with a driving frequency matching that of the natural frequency of vibration of the building—producing a resonance resulting in one building collapsing while neighboring buildings do not. Often, buildings of a certain height are devastated while other taller buildings remain intact. The building height matches the condition for setting up a standing wave for that particular height. The span of the roof is also important. Often it is seen that gymnasiums, supermarkets, and churches suffer damage when individual homes suffer far less damage. The roofs with large surface areas supported only at the edges resonate at the frequencies of the earthquakes, causing them to collapse. As the earthquake waves travel along the surface of Earth and reflect off denser rocks, constructive interference occurs at certain points. Often areas closer to the epicenter are not damaged, while areas farther away are damaged.

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• Authors: William Moebs, Samuel J. Ling, Jeff Sanny
• Publisher/website: OpenStax
• Book title: University Physics Volume 1
• Publication date: Sep 19, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/university-physics-volume-1/pages/1-introduction
• Section URL: https://openstax.org/books/university-physics-volume-1/pages/16-6-standing-waves-and-resonance

© Jul 23, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

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