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Retur Recall from the " Introduction to Waves" lab that it was easy calculate the harmonic number (n) and wavelength (A) of standing waves on a string by counting the number of antinodes (An : 215/11). Thatwas a system with nodes xed at the end points. Today you will be working with a system that has one open end and one closed end. lie. 3 nodeflxed at one end and an antinode xed at the other end). In this system An = 4L/'n. \"Open-Closed\" Resonator: Pipe with one open and one closed end ' Only the odd harmonics can be produced I: 3'fo f 5% f = T'fa Why it is only possible to produce the odd harmonics in a system with one open end and one closed end? Drawing pictures will help you answer this question. A node must exist at one end of the system, and an antinode at the other. Thus, when forming the second resonance, one must divide the wavelength bythree in order for the next antinode to line up with the open end of the system, while keeping a node at the closed end of the system. This pattern continues as one increases the frequency, such that only odd fractions (U3, 1.5, 1H, etc.) of the wavelength can exist. A node must exist at one end of the system, and an antinode at the other. Thus, when forming the second resonance, one must mu ltiplythe wavelength by three in order for the next antinode to line up with the closed end of the system,while keeping a node at the open end of the system. This pattern continues as one increases the frequency, such that only odd multiples (3, 5, 7, etc.) of the wavelength can exist A node must exist at one end of the system, and an antinode at the other. Thus, when forming the second resonance, one must mu ltiplythe wavelength by three in order for the next antinode to line up with the open end of the system, while keeping a node at the closed end of the system. This pattern continues as one increases the frequency, such that only odd multiples {3, 5, 7, etc.) of the wavelength can exist A node must exist at one end of the system, and an antinode at the other. Thus, when forming the second resonance, one must divide the wavelength bythree in order for the next antinode to line up with the closed end of the system, while keeping a node at the open end of the system. This pattern continues as one increases the frequency, such that only odd fractions {1/3, US, 117, etc.) of the wavelength can exist. Return Submit What is the algebraic relationship between the resonant wavelength, A, and length of an open-closed tube, L, for the n=1, 3, 5, and 7 harmonics. The options below refer to the ollowing expressions 1.1 = 2L 2. 1 = 3. 1 = 5L 4. X = 3L 5. 1 = 44 6. 1 = 4L 7.1 = L 8. 1 = 44 9. 1 = 71 10. 1 = 4L The wavelength of the n=1 harmonic is given by The wavelength of the n=3 harmonic is given by expression 9. expression 5. The wavelength of the n=5 harmonic is given by expression 2. The wavelength of the n=7 harmonic is given by expression 6. expression 10. expression 1. 0.2 points expression 4. Which of the following would be a general expression for the resonant wavelength in an open-closed resona O 1 = nL expression 3.0.2 points 59 Which of the following would be a general expression for the resonant wavelength in an openeclosed resonator for any n? A=nL A=4nL _ 4L 5*? 2L A2? :L '31 A2211}; 4 02pomts $9\" You have an open closed resonator with a length of 22 cm. What is the wavelength of the n=5 harmonic (in cm)? Type your a issue r... 0.2 points 59' If you measure this same harmonic (from question 4) to have a frequency of 757.1 Hz. what is the speed of the wave (in mls)? Type your a-'.s'.-'-:e r