Chapter 17. Linear Superposition and Interference

Transcription

1 Chapter 17 Linear Superposition and Interference

2 Linear Superposition If two waves are traveling through the same medium, the resultant wave is found by adding the displacement of the individual waves point by point. There are three (3) types of Interference

3 Interference Constructive- Waves in phase with each other will form 1 wave with an amplitude that is the sum of the 2 starting waves. 1.

4 Interference Destructive- Waves out of phase with each other will form 1 wave with an amplitude that is the difference of the 2 starting waves.

5 Interference Totally Destructive - Waves out of phase with each other, yet have the same wavelength, will cancel each other. Note: Interference does not physically destroy a wave. It only changes the characteristics at a given point.

6 The Superposition Principle When two or more waves (blue and green) exist in the same medium, each wave moves as though the other were absent. The resultant displacement of these waves at any point is the algebraic sum (yellow) wave of the two displacements. Constructive Interference Destructive Interference

7 Formation of a Standing Wave Incident and reflected waves traveling in opposite directions produce nodes, N, and antinodes, A. The distance between alternate nodes or anti-nodes is one wavelength.

8 Vibrating Air Columns Just as for a vibrating string, there are characteristic wavelengths and frequencies for longitudinal sound waves. Boundary conditions apply for pipes: The open end of a pipe must be a displacement antinode, A. A Open pipe A The closed end of a pipe must be a displacement node, N. N Closed pipe A

9 Velocity and Wave Frequency The period T is the time to move a distance of one wavelength. Therefore, the wave speed is: 1 v but T so v f T f The frequency f is in s -1 or hertz (Hz). The velocity of any wave is the product of the frequency and the wavelength: v f f v

10 Open Pipes OPEN PIPES- have an antinode on BOTH ends of the tube. You will get your FIRST sound when the length of the pipe equals one-half of a wavelength.

11 Open Pipes - Harmonics Since harmonics are MULTIPLES of the fundamental, the second harmonic of an open pipe will be ONE WAVELENGTH. First Harmonic Second Harmonic The picture above is the SECOND harmonic or the FIRST OVERTONE. This also works for string instruments. Treat them as open pipes.

12 Open pipes - Harmonics Another half of a wavelength would ALSO produce an antinode on BOTH ends. In fact, no matter how many halves you add you will always have an antinode on the ends. First Harmonic Second Harmonic Third Harmonic The picture above is the THIRD harmonic or the SECOND OVERTONE. CONCLUSION: Sounds in OPEN pipes are produced at ALL HARMONICS! THEREFORE, we can assume that we will get sounds at 1/2, 1, 1 1/2, 2, 2 1/2, 3, You get sounds at even or odd intervals or at ALL HARMONICS.

13 Possible Waves for Open Pipe Fundamental, n = 1 2 nd Harmonic, or 1st Overtone, n = 2 3 rd Harmonic, or 2nd Overtone, n = 3 4 th Harmonic, or 3rd Overtone, n = 4 L 2L 1 2L 2 2L 3 2L 4 All harmonics are possible for open pipes: n 2L n n 1, 2, 3, 4...

14 Open Pipe. Wave velocity, v= Lf, but now ½ λ = L or 2L =λ L Fundamental, n = 1 1st Overtone, n = 2 2nd Overtone, n = 3 3rd Overtone, n = 4 v 2Lf 2v f 2L 3v f 2L 4v f 2L All harmonics are possible nv for open pipes: fn n1, 2, 3, L

15 Closed Pipes Have an antinode at one end and a node at the other. Each sound you hear will occur when an antinode appears at the top of the pipe. You get your first sound or encounter your first antinode when the length of the actual pipe is equal to a quarter of a wavelength. (L = ¼ λ) This FIRST SOUND is called the FUNDAMENTAL FREQUENCY or the FIRST HARMONIC.

16 Closed Pipes - Harmonics Since the first harmonic was at ¼ λ, the second harmonic should be at the next ¼ λ, or ½ λ. This is the location of a node, so no sound is produced.

17 Closed Pipes - Harmonics In a closed pipe you have an ANTINODE at the 3rd harmonic position, or the next ¼ λ, or ¾ λ, therefore SOUND is produced. First Harmonic Second Harmonic Third Harmonic CONCLUSION: Sounds in CLOSED pipes are produced ONLY at ODD HARMONICS! THEREFORE, we can assume that we will get sounds at 1/4, 3/4, 5/4, 7/4, 9/4 λ.

18 Possible Waves for Closed Pipe. Fundamental, n = 1 3 rd Harmonic, or 1 st Overtone n = 3 L 4L 1 1 4L th Harmonic, or 2 nd Overtone n = 5 7 th Harmonic, or 3 rd Overtone n = L 5 4L 7 Only the odd harmonics are allowed: n 4L n n 1, 3, 5, 7...

19 Possible Waves for Closed Pipe v f L Fundamental, n = v f L 3 rd Harmonic, n = v f L 5 th Harmonic, n = v f L 7 th Harmonic, n = 7 Only the odd harmonics are allowed: L 1, 3, 5, n nv f n L

20 Beats The phenomenon of beats occurs when two tuning forks, or musical instruments, are sounded simultaneously. You hear a beat the Constructive interference reaches your ear.

21 Beats The number of times a beat occurs per second is the beat frequency and is the difference between the two sound frequencies. Two tuning forks have slightly different frequencies of 440 Hz and 438 Hz. You will hear 2 beats/sec.

22 Sound Waves Sound Waves are a common type of standing wave as they are caused by RESONANCE. Resonance when a FORCED vibration matches an object s natural frequency thus producing vibration, sound, or even damage. One example of this involves shattering a wine glass by hitting a musical note that is on the same frequency as the natural frequency of the glass. (Natural frequency depends on the size, shape, and composition of the object in question.) Because the frequencies resonate, or are in sync with one another, maximum energy transfer is possible.

23 Example Problem What length of closed pipe is needed to resonate with a fundamental frequency of 256 Hz? What is the second overtone? Assume that the velocity of sound is 340 m/s. N Closed pipe L =? A f n nv 4L n 1, 3, 5, v v f1 ; L 4L 4f m/s 4(256 Hz) L = m, or 33.2 cm The second overtone occurs when n = 5: f 5 = 5f 1 = 5(256 Hz) 2nd Overtone = 1280 Hz

24 Example The speed of sound waves in air is found to be 340 m/s. Determine the fundamental frequency (1st harmonic) of an open-end air column which has a length of 67.5 cm. v 2Lf 340 m/s 2( m ) f Hz f

25 Example The windpipe of a typical whooping crane is about m long. What is the lowest resonant frequency of this pipe assuming it is a pipe closed at one end? Assume the speed of sound at 37 C is m/s. v f 4lf m/s Hz 4(1.525 m ) f

26 Example The G string on a guitar has a fundamental frequency of 196 Hz and a length of 0.62 m. This string is pressed against the proper fret to produce the note C, whose fundamental frequency is 262 Hz. What is the distance L between the fret and the end of the string at the bridge of the guitar? (Assume the speed of the wave in the string does not change.) Assuming that fretting the string does NOT change the tension, the speed of waves on the string will be the same in both cases. The speed of the waves is given v = λf = 2Lf. Applied to the first case (unfretted), this relation gives v = 2L 1 f 1. Applied to the second case, it gives v = 2L 2 f 2. Setting the above equations equal to each other and rearranging yields L 2 = L 1 (f 1 /f 2 ) = (0.62 m)(196 Hz)/(262 Hz) = 0.46m

27 Assignment Ch 17, Pages , #3, 21, 27, 37, 41, 43, 57