What is the algebraic relationship between the resonant wavelength, X, and length of an open-closed tube, L, for the n=1, 3, 5, and 7 harmonics. The options below refer to the following expressions The wavelength of the n=1 harmonic is given by The wavelength of the n-3 harmonic is given by The wavelength of the n=5 harmonic is given by The wavelength of the n-7 harmonic is given by 1. A = 2L 2. λ = 2L 7 3. λ=5L 4.λ=3L 5. A=4L 6. λ = 4L 7.λ = L 8. λ = 4 9.λ = 7L 10.A4L 5

Classical Dynamics of Particles and Systems
5th Edition
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
Publisher:Stephen T. Thornton, Jerry B. Marion
Chapter13: Continuous Systems; Waves
Section: Chapter Questions
Problem 13.2P
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What is the algebraic relationship between the resonant wavelength, X, and length of an open-closed tube, L, for the n=1, 3, 5, and 7 harmonics. The options below refer to the
following expressions
The wavelength of the n=1 harmonic is given by
The wavelength of the n=3 harmonic is given by
The wavelength of the n=5 harmonic is given by
The wavelength of the n-7 harmonic is given by
1. λ = 2L
2. λ = 2L
3. λ = 5L
4.3L
5. λ = 4
6. λ = 4L
7.λ = L
8. λ = 4
9.λ = 7L
10. = 4L
>
s
s
Transcribed Image Text:What is the algebraic relationship between the resonant wavelength, X, and length of an open-closed tube, L, for the n=1, 3, 5, and 7 harmonics. The options below refer to the following expressions The wavelength of the n=1 harmonic is given by The wavelength of the n=3 harmonic is given by The wavelength of the n=5 harmonic is given by The wavelength of the n-7 harmonic is given by 1. λ = 2L 2. λ = 2L 3. λ = 5L 4.3L 5. λ = 4 6. λ = 4L 7.λ = L 8. λ = 4 9.λ = 7L 10. = 4L > s s
"Open-Closed" Resonator:
Pipe with one open and one closed end
* Only the odd harmonics can be produced
Harmonic Number of
Number: visible nodes:
n=1
1
XxxxX
n = 3
n = 5
n=7
2
3
4
Harmonic
Frequency:
f = fo
f = 3*fo
f = 5*fo
f = 7*fo
Transcribed Image Text:"Open-Closed" Resonator: Pipe with one open and one closed end * Only the odd harmonics can be produced Harmonic Number of Number: visible nodes: n=1 1 XxxxX n = 3 n = 5 n=7 2 3 4 Harmonic Frequency: f = fo f = 3*fo f = 5*fo f = 7*fo
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