Ch17. The Principle of Linear Superposition and Interference Phenomena The Principle of Linear Superposition 1
THE PRINCIPLE OF LINEAR SUPERPOSITION When two or more waves are present simultaneously at the same place, the resultant disturbance is the sum of the disturbances from the individual waves. 2
Check Your Understanding 1 The drawing shows two pulses traveling toward each other at t = 0 s. Each pulse has a constant speed of 1 cm/s. When t = 2 s, what is the height of the resultant pulse at (a) x = 2 cm, (b) x = 4 cm, and (c) x = 6 cm? (a) 0 cm, (b) -2 cm, (c) +2 cm 3
Constructive and Destructive Interference of Sound Waves Reading content When two waves always meet condensation-to-condensation and rarefaction-to-rarefaction (or crest-to-crest and trough-totrough), they are said to be exactly in phase and to exhibit constructive interference. 4
When two waves always meet condensation-to-rarefaction (or crest-to-trough), they are said to be exactly out of phase and to exhibit destructive interference. In either case, this means that the wave patterns do not shift relative to one another as time passes. Sources that produce waves in this fashion are called coherent sources. 5
Destructive interference is the basis of a useful technique for reducing the loudness of undesirable sounds.,, For two wave sources vibrating in phase, a difference in path lengths that is zero or an integer number (1, 2, 3,...) of wavelengths leads to constructive interference; a difference in path lengths that is a half-integer number (,,, ) of wavelengths leads to destructive interference. 6
Interference effects can also be detected if the two speakers are fixed in position and the listener moves about the room. 7
Example 1. What Does a Listener Hear? Two in-phase loudspeakers, A and B, are separated by 3.20 m. A listener is stationed at point C, which is 2.40 m in front of speaker B. The triangle ABC is a right triangle. Both speakers are playing identical 214-Hz tones, and the speed of sound is 343 m/s. Does the listener hear a loud sound or no sound? The listener hears a loud sound. 8
Conceptual Example 2. Out-of-Phase Speakers To make a speaker operate, two wires must be connected between the speaker and the receiver (amplifier. To ensure that the diaphragms of two speakers vibrate in phase, it is necessary to make these connections in exactly the same way. If the wires for one speaker are not connected just as they are for the other speaker, the two diaphragms will vibrate out of phase. Whenever one diaphragm moves outward, the other will move inward, and vice versa. Suppose that in Figures 17.3 and 17.4 the connections are made so that the speaker diaphragms vibrate out of phase, everything else remaining the same. In each case, what kind of interference would result at the overlap point? In Figure 17.3 a rarefaction from the left now meets a condensation from the right at the overlap point, and destructive interference results. In Figure 17.4, a condensation from the left now meets a condensation from the right at the overlap point, leading to constructive interference. 9
Diffraction 10
The bending of a wave around an obstacle or the edges of an opening is called diffraction. All kinds of waves exhibit diffraction. 11
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Example 3. Designing a Loudspeaker for Wide Dispersion A 1500-Hz sound and a 8500-Hz sound each emerges from a loudspeaker through a circular opening whose diameter is 0.30 m. Assuming that the speed of sound in air is 343 m/s, find the diffraction angle q for each sound. 13
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Beats 15
The number of times per second that the loudness rises and falls is the beat frequency and is the difference between the two sound frequencies. 16
A 10-Hz sound wave and a 12-Hz sound wave, when added together, produce a wave with a beat frequency of 2 Hz. The drawings show the pressure patterns (in blue) of the individual waves and the pressure pattern (in red) that results when the two overlap. The time interval shown is one second. 17
Check Your Understanding 2 A tuning fork of unknown frequency and a tuning fork of frequency of 384 Hz produce 6 beats in 2 seconds. When a small piece of putty is attached to the tuning fork of unknown frequency, the beat frequency decreases. What is the frequency of that tuning fork? 387 H Z 18
Transverse Standing Waves 19
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The antinodes are places where maximum vibration occurs. 21
The nodes are places that do not vibrate at all. The frequencies in this series (f 1, 2f 1, 3f 1, etc.) are called harmonics. Frequencies above the fundamental are overtones. 22
f 1 = v/(2l) Standing waves arise because identical waves travel on the string in opposite directions and combine in accord with the principle of linear superposition. A standing wave is said to be standing because it does not travel in one direction or the other, as do the individual waves that produce it. or f n λ n = v = v f n (2L/ n ) = v 23
Example 4. Playing a Guitar The heaviest string on an electric guitar has a linear density of m/l = 5.28 10 3 kg/m and is stretched with a tension of F = 226 N. This string produces the musical note E when vibrating along its entire length in a standing wave at the fundamental frequency of 164.8 Hz. (a) Find the length L of the string between its two fixed ends. (b) A guitar player wants the string to vibrate at a fundamental frequency of 2 164.8 Hz = 329.6 Hz, as it must if the musical note E is to be sounded one octave higher in pitch. To accomplish this, he presses the string against the proper fret and then plucks the string. Find the distance L between the fret and the bridge of the guitar. 24
(a) (b) This length is exactly half that determined in part (a) because the frequencies have a ratio of 2:1. 25
Conceptual Example 5. The Frets on a Guitar 26
The frets on the neck of a guitar. They allow the player to produce a complete sequence of musical notes using a single string. Starting with the fret at the top of the neck, each successive fret indicates where the player should press to get the next note in the sequence. Musicians call the sequence the chromatic scale, and every thirteenth note in it corresponds to one octave, or a doubling of the sound frequency. The spacing between the frets is greatest at the top of the neck and decreases with each additional fret further on down. The spacing eventually becomes smaller than the width of a finger, limiting the number of frets that can be used. Why does the spacing between the frets decrease going down the neck? D 1 is greater than D 2, and the frets near the top of the neck have more space between them than those further down. 27
Check Your Understanding 3 A standing wave that corresponds to the fourth harmonic is set up on a string that is fixed at both ends. (a) How many loops are in this standing wave? (b) How many nodes (excluding the nodes at the ends of the string) does this standing wave have? (c) Is there a node or an antinode at the midpoint of the string? (d) If the frequency of this standing wave is 440 Hz, what is the frequency of the lowest-frequency standing wave that could be set up on this string? (a) 4, (b) 3, (c) node, (d) 110 H Z 28
Longitudinal Standing Waves Standing wave patterns can also be formed from longitudinal waves. 29
f n = v/λ n 30
Example 6. Playing a Flute When all the holes are closed on one type of flute, the lowest note it can sound is a middle C, whose fundamental frequency is 261.6 Hz. (a) The air temperature is 293 K, and the speed of sound is 343 m/s. Assuming the flute is a cylindrical tube open at both ends, determine the distance L, the distance from the mouthpiece to the end of the tube. (This distance is only approximate, since the antinode does not occur exactly at the mouthpiece.) (b) A flautist can alter the length of the flute by adjusting the extent to which the head joint is inserted into the main stem of the instrument. If the air temperature rises to 305 K, to what length must the flute be adjusted to play a middle C? 31
(a) (b) v 305 K = 1.02(v 293 K ) = 1.02(343 m/s) = 3.50 10 2 m/s Thus, to play in tune at the higher temperature, a flautist must lengthen the flute by 0.013 m. 32
Standing waves can also exist in a tube with only one end open. Here the standing waves have a displacement antinode at the open end and a displacement node at the closed end, where the air molecules are not free to move. 33
Check Your Understanding 4 A cylindrical bottle, partially filled with water, is open at the top. When you blow across the top of the bottle a standing wave is set up inside it. Is there a node or an antinode (a) at the top of the bottle and (b) at the surface of the water? (c) If the standing wave is vibrating at its fundamental frequency, what is the distance between the top of the bottle and the surface of the water? Express your answer in terms of the wavelength l of the standing wave. (d) If you take a sip, is the fundamental frequency of the standing wave raised, lowered, or does it remain the same? (a) antinode, (b) node, (c) 1 λ 4, (d) lowered 34
Complex Sound Waves 35
The sound wave corresponding to a note produced by a musical instrument or a singer is called a complex sound wave because it consists of a mixture of the fundamental and harmonic frequencies. 36
Concepts & Calculations Example 7. Diffraction in Two Different Media A sound wave with a frequency of 15 khz emerges through a circular opening that has a diameter of 0.20 m. Find the diffraction angle when the θ sound travels in air at a speed of 343 m/s and in water at a speed of 1482 m/s. 37
Concepts & Calculations Example 8. Standing Waves of Sound The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again. Two tubes of gas are identical and are open at only one end. One tube contains neon (Ne) and the other krypton (Kr). Both are monatomic gases, have the same temperature, and may be assumed to be ideal gases. The fundamental frequency of the tube containing neon is 481Hz. What is the fundamental frequency of the tube containing krypton? 38
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Problem 1 REASONING AND SOLUTION In a time of t = 1 s, the pulse on the left has moved to the right a distance of 1 cm, while the pulse on the right has moved to the left a distance of 1 cm. Adding the shapes of these two pulses when t = 1 s reveals that the height of the resultant pulse is a. 2 cm at x = 3 cm b. 1 cm at x = 4 cm. 40
Problem 18 REASONING The beat frequency is the difference between two sound frequencies. Therefore, the original frequency of the guitar string (before it was tightened) was either 3 Hz lower than that of the tuning fork (440.0 Hz - 3 Hz = 337 Hz) or 3 Hz higher (440.0 Hz + 3 Hz = 443 Hz) 443 Hz 440.0 Hz 437 Hz } } 3-Hz beat frequency 3-Hz beat frequency To determine which of these frequencies is the correct one (437 or 443 Hz), we will use the information that the beat frequency decreases when the guitar string is tightened 41
SOLUTION When the guitar string is tightened, its frequency of vibration (either 437 or 443 Hz) increases. As the drawing below shows, when the 437-Hz frequency increases, it becomes closer to 440.0 Hz, so the beat frequency decreases. When the 443-Hz frequency increases, it becomes farther from 440.0 Hz, so the beat frequency increases. Since the problem states that the beat frequency decreases, the original frequency of the guitar string was 437 H Z. 440.0 Hz 443 Hz 437 Hz } } beat frequency increases beat frequency decreases Tuning fork Original string Tightened string 42
Problem 32 (2f The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then 1 ). The distance between the frets is insert it again. REASONING AND SOLUTION We are given f j = 12 2 a. The length of the unfretted string is L 0 = v/(2f 0 ) and the length of the string when it is pushed against fret 1 is L 1 = v/ f j 1 L 0 L 1 = v v 2 f 2 f 0 1 v f = 1 2 f0 f 0 1 = v 2 f 1 1 12 0 2 1 = L0 1 12 = (0.628m)(0.0561) = 0. 0352m 2 43
b. The frequencies corresponding to the sixth and seventh frets are ( ) 6 12 f and. The distance 6 = 2 f ( ) 7 12 0 f between fret 6 and fret 7 is 7 = 2 f0 L 6 L 7 = v f v f = ( ) 6 ( ) 7 12 12 2 f0 2 2 0 2 6 2 7 2 f v 1 1 = 7 2 f0 2 v ( ) ( ) 6 12 12 2 v = 1 1 L0 = (0.628m)(0.0397) = 0. 0249m ( 2) 6 ( 2) 7 12 12 44
Problem 50 t = 1 s t = 2 s t = 3 s t = 4 s 0 2 4 6 8 10 45