THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA

Transcription

1 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA PREVIEW When two waves meet in the same medium they combine to form a new wave by the principle of superposition. The result of superposition is called interference. If the two waves are in phase at a point, they will interfere constructively. At a point where the two waves are out of phase they will interfere destructively. Places where maximum constructive and destructive interference are occurring are called antinodes and nodes, respectively. Interference can be caused by point sources, or by a wave passing through a single slit opening or a double slit opening. Interference in a vibrating string, open pipe, or closed pipe produces resonance conditions called harmonics and overtones. QUICK REFERENCE Important Terms antinode a point of constructive interference and maximum reinforcement in a standing wave beat variations in the loudness of sounds due to the slight difference in frequency of interfering waves closed pipe resonator a pipe closed at one end and a sound source at the other, causing the sound to resonate constructive interference addition of two or more waves which are in phase, resulting in a wave of increased amplitude destructive interference addition of two or more waves which are out of phase resulting in a wave of decreased amplitude diffraction the spreading of a wave beyond the edge of a barrier or through an opening harmonics the combination of several simultaneous frequencies that produce a distinct standing wave pattern 188

2 in phase term applied to two or more waves whose crests and troughs arrive at a place at the same time in such a way as to produce constructive interference interference of waves displacements of two or more waves in the same medium at the same time producing either larger or smaller waves node the point of no displacement in a standing wave out of phase term applied to two or more waves for which the crest of one wave arrives at a point at the same time as the trough of a second wave arrives, producing destructive interference overtone a resonance condition (standing wave) which is above the fundamental principle of superposition the displacement due to two or more interfering waves is equal to the sum of the displacement of the individual waves standing wave wave with stationary nodes produced by two identical waves traveling in opposite directions in the same medium at the same time Equations and Symbols λ sinθ = D v f n = n n = 1,2,3,4,...(vibratingstring and open tube) 2L v f n = n n = 1,3,5,...(closed tube) 4L where θ = the angle of spread of a wave after passing through a single slit opening λ = wavelength f n = the n th frequency in a series of resonant frequencies v = speed of the wave L = length D = width of an opening through which a wave will diffract 189

3 DISCUSSION OF SELECTED SECTIONS The Principle of Linear Superposition When two or more waves pass through each other in the same medium at the same time they interfere, as shown in the diagrams on the opening pages of chapter 17 in your textbook. Two waves meeting on the same side of the spring will interfere constructively, producing an antinode, and two waves meeting on opposite sides of the spring will interfere destructively and cancel the waves momentarily, producing a node. Either way, the principle of superposition states that any two waves which meet at a point may be added (or subtracted, if they meet on opposite sides). We say that two waves which interfere constructively are in phase, and two waves that interfere destructively are out of phase. Constructive and Destructive Interference of Sound Waves Have you ever sat in a room or auditorium in a place where it was difficult to hear the speaker? Like other waves, sound waves interfere constructively and destructively as they are produced and reflected around a room. For two sound waves which are in phase (crest on crest), the path difference between the waves is either zero or a whole number of wavelengths apart (Fig. A): Fig. A Waves in phase Fig. B Waves out of phase For two sound waves which are out of phase (crest on trough), the path difference between the waves is a half-number of wavelengths apart (Fig. B). Example 1 in your textbook is a good illustration of whether waves arrive at a particular point in phase or out of phase. 190

4 Transverse Standing Waves The term standing wave is an oxymoron, since waves must move and never stand still. But waves can appear to stand still when two identical waves traveling in opposite directions in the same medium at the same time create a series of nodes and antinodes. Consider a rope tied to a wall. If we send a wave down the rope toward the wall, the wave reflects off the wall on the opposite side from which it was sent, according to the law of reflection. If we continue to send regular waves down the rope and they continue to reflect off the wall, the incident and reflected waves will reinforce each other in some places and cancel each other in other places. The result is a series of antinodes (loops) where constructive interference is occurring, and nodes (points of no displacement between the loops) where destructive interference is occurring, and we call the pattern produced a standing wave. λ L A N A N A The pattern above, called a harmonic or overtone, shows three antinodes (loops), and occupies 2 3 of a wavelength. Example 2 A string is attached to a vibrating machine which has a frequency of 120 Hz. The other end of the string is passed over a pulley of negligible mass and friction and is attached to a weight hanger which holds a mass m = 0.5 kg. L m (a) Determine the tension in the string. (b) The speed of the wave in the string is related to the tension by the equation 191

5 F v = T, where F T is the tension in the string and µ is the linear density of the string. If µ the linear density of this string is 0.05 kg/m, determine the speed of the wave in the string. (c) Determine the wavelength of the wave in the string. (d) Determine the length of the string from the point of attachment on the vibrating machine to the pulley. (e) Would you need to increase or decrease the mass on the hanger to produce a lower number of loops? Explain. Solution F T = mg = 0.5kg 2 10 m/ s = 5 (a) ( )( ) N FT 5 N (b) v = = = 100m / s µ 0.05 kg / m v 100m / s (c) λ = = = 0.83m f 120 Hz (d) L = 3 λ = 3( 0.83m) = 2. 5m (e) A lower number of loops would imply a longer wavelength, which would require a higher speed, which would require a higher tension in the string, which would require increasing the mass on the hanger. Longitudinal Standing Waves An open pipe is one in which both ends of the pipe are open. If we send a sound wave into the pipe and let it reflect off of the air at other end, it will return to the original end of the pipe. If we continually send sound waves down into the pipe to match the rate at which they are being reflected off the closed end, we will set up a resonance condition, that is, a standing wave. as shown in the figure below: A N A A N A N A 1 st Harmonic Fundamental Overtone 2 nd Harmonic 1 st Overtone A standing wave pattern is set up in the open pipe, having a series of nodes and antinodes. Note that there is an antinode at either end of the pipe. When a standing sound wave fits inside the pipe this way, we hear a louder pitch than when the sound does not fit inside the pipe. We say that the sound is resonating inside the pipe. 192

6 Note that in the first harmonic of the open pipe, ½ of a wavelength just fits inside the length of the pipe. In the 2 nd harmonic, one full wavelength just fits inside the pipe. 1 st Harmonic Fundamental Overtone 3 rd Harmonic 1 st Overtone 5 th Harmonic 2 nd Overtone A closed pipe is one in which one end of the pipe (bottom) is closed and the other end is open (top). If we send a sound wave into the pipe and let it reflect off of the closed end, it will return to the top of the pipe, as shown in the figures above. This results in an antinode at the top open end of the pipe, and a node at the bottom closed end. Again, when a standing sound wave fits inside the pipe this way, we hear a louder pitch (resonance) than when the sound does not fit inside the pipe. Note that in the 1 st harmonic of the closed pipe, ¼ of a wavelength just fits inside the length of the pipe, ¾ wavelength for the 3 rd harmonic, and 5/4 of a wavelength for the 5 th harmonic. The harmonics are named after each quarter-wavelength that fits in the pipe as the sound resonates. If we know the frequency of the sound waves and the length of the pipe, we can find their wavelength and then their speed by v = fλ. The relationships among all these quantities in a resonating pipe are illustrated in the review questions that follow. 193

7 REVIEW QUESTIONS For each of the multiple choice questions below, choose the best answer. Questions 1 2: 1. Two waves approach each other in the same rope at the same time, as shown. When the two waves are exactly between points P and Q, the shape of the rope will be (A) (B) P Q 2. Which of the following diagrams best represents the shape of the rope just after the two waves have completely passed through the region between points P and Q? (A) (B) P P Q (C) P Q Q (C) (D) P Q (D) (E) (E) 3. In general, the higher the frequency of the interfering waves, (A) the higher the wavelength (B) the higher the speed of the waves (C) the greater the number of nodes and antinodes. (D) the smaller the number of nodes and antinodes. (E) the higher the amplitude of the waves. 194

8 8. A N A 1.5 m 1 st Harmonic Fundamental Overtone 6. A standing wave is produced in a vibrating string as shown. If the length of the string is 1.5 m and the frequency of the vibrating motor is 60 Hz, the speed of the wave is (A) 15 m/s (B) 20 m/s (C) 40 m/s (D) 60 m/s (E) 90 m/s The figure above represents the 1 st harmonic for an open pipe. If the length of the pipe is 40 cm, the wavelength of the sound wave resonating inside the pipe is (A) 10 cm (B) 20 cm (C) 40 cm (D) 80 cm (E) 160 cm 7. Which of the following is true of a sound which is resonating in a pipe which is closed at one end? (A) Nodes are formed at both ends of the pipe (B) Antinodes are formed at both ends of the pipe (C) An antinode is formed at the closed end of the pipe and a node is formed at the open end. (D) An antinode is formed at the open end of the pipe and a node is formed at the closed end. (E) A sound wave cannot resonate in a pipe which is closed at only one end. 195

9 Free Response Question Directions: Show all work in working the following question. The question is worth 10 points, and the suggested time for answering the question is about 10 minutes. The parts within a question may not have equal weight. 1. (10 points) A N A N A 2 nd Harmonic 1 st Overtone A sound wave resonates inside an open pipe filled with air at room temperature, as shown above. The length of the pipe is 33 cm. (a) Determine the wavelength of the resonating sound wave. (b) Determine the frequency of the tuning fork. (c) Determine the next higher frequency that will resonate in a pipe of this length. 1 st Harmonic Fundamental Overtone (d) If the open pipe is replaced with a pipe which is closed at one end, what would have to be the length of the closed pipe for the original tuning fork to resonate at its fundamental frequency? 196

10 ANSWERS AND EXPLANATIONS TO CHAPTER 17 REVIEW QUESTIONS Multiple Choice 1. B The two waves are in phase (crest on crest), and so they will constructively interfere and produce a larger wave between points P and Q. 2. A After passing through the region between points P and Q, the waves will simply have switched places, and since they are identical, the rope appears as it did before the waves met. 3. C More frequency results in more waves interfering with each other, producing more nodes and antinodes. 4. D A wave diffracts when it must go around an obstacle or through an opening. 5. B The longer wavelength passing through the same opening will have a greater angle of spread. 6. D The wavelength is the length of two loops and is 1 m. The speed of the wave is v = f λ= (60 Hz)(1 m) = 60 m/s. 7. D The wave is reflected off the closed end, creating a node at the closed end and an antinode at the open end. 8. D The wave pattern in the 40-cm pipe represents half of a wavelength. Thus, the wavelength is 2(40 cm) = 80 cm. 197

11 Free Response Question Solution (a) 2 points One wavelength just fits inside the pipe for the 2 nd harmonic, so L = λ = 33 cm. (b) 3 points v 343m / s f = = = 1029 Hz λ 0.33m (c) 3 points The next higher frequency would correspond to two wavelengths in the pipe. Thus, L = 2λ, and λ = ½ L = 16.5 cm. The speed of sound is 343 m/s, so the next higher frequency is A N A N A v 343m / s f = = = 2058 Hz λ 0.165m This is simply twice the original frequency. (d) 2 points The first resonance condition in a closed pipe requires the length of the pipe to be onequarter the wavelength, or 1 1 L = λ = ( 33cm) = 8. cm 198

12 Chapter 17 The Principle of Linear Siperposition and Interference Phenomena 199