The manner in which a string is plucked determines the mixture of harmonic amplitudes in the resulting wave. Consider a string exactly 1/2-m long that is fixed at both its ends located at x = 0.0 and x = m. The first five harmonics of this string have wavelengths λ₁ = 1.0 m, λ₂ = 1/2 m, λ = 1/3 m, 14 = 1/4 m, and λ5 = 1/5 m. According to Fourier's theorem, any shape of this string can be formed by a sum of its harmonics, with each harmonic having its own unique amplitude A. We limit the sum to the first five harmonics in the expression m D(x) = m An sin A, (²x) n=1 and D is the displacement of the string at a time t = 0. Imagine plucking this string at its midpoint (Fig. below (a)) or at a point two-thirds from the left end (Fig. below (b)). Use Excel, MATLAB, or other graphing software, to create two plots showing that the above expression can fairly accurately represent the shape in Fig. (a), if A₁ = 1.00, A₂ = 0.00, A3 = -0.11, A4 = 0.00, and A5 = 0.040 and in (b), if A₁ = 0.87, A₂ = -0.22, A3 = -0.00, A4 = 0.054, and A5 = -0.035. ¹ m 211 m (a) (b) Must provide a print out of a spread sheet, code, or input parameters along with the appropriately labeled figures. Full credit will not be given for only the figures.
The manner in which a string is plucked determines the mixture of harmonic amplitudes in the resulting wave. Consider a string exactly 1/2-m long that is fixed at both its ends located at x = 0.0 and x = m. The first five harmonics of this string have wavelengths λ₁ = 1.0 m, λ₂ = 1/2 m, λ = 1/3 m, 14 = 1/4 m, and λ5 = 1/5 m. According to Fourier's theorem, any shape of this string can be formed by a sum of its harmonics, with each harmonic having its own unique amplitude A. We limit the sum to the first five harmonics in the expression m D(x) = m An sin A, (²x) n=1 and D is the displacement of the string at a time t = 0. Imagine plucking this string at its midpoint (Fig. below (a)) or at a point two-thirds from the left end (Fig. below (b)). Use Excel, MATLAB, or other graphing software, to create two plots showing that the above expression can fairly accurately represent the shape in Fig. (a), if A₁ = 1.00, A₂ = 0.00, A3 = -0.11, A4 = 0.00, and A5 = 0.040 and in (b), if A₁ = 0.87, A₂ = -0.22, A3 = -0.00, A4 = 0.054, and A5 = -0.035. ¹ m 211 m (a) (b) Must provide a print out of a spread sheet, code, or input parameters along with the appropriately labeled figures. Full credit will not be given for only the figures.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![The manner in which a string is plucked determines the mixture of harmonic amplitudes in the resulting wave. Consider a string exactly 1/2-m long that is fixed at both its ends located at \( x = 0.0 \) and \( x = \frac{1}{2} \, \text{m} \). The first five harmonics of this string have wavelengths \( \lambda_1 = 1.0 \, \text{m}, \, \lambda_2 = \frac{1}{2} \, \text{m}, \, \lambda_3 = \frac{1}{3} \, \text{m}, \, \lambda_4 = \frac{1}{4} \, \text{m}, \, \text{and} \, \lambda_5 = \frac{1}{5} \, \text{m} \). According to Fourier’s theorem, any shape of this string can be formed by a sum of its harmonics, with each harmonic having its own unique amplitude \( A \). We limit the sum to the first five harmonics in the expression
\[
D(x) = \sum_{n=1}^{5} A_n \sin \left( \frac{2 \pi}{\lambda_n} x \right)
\]
where \( D \) is the displacement of the string at a time \( t = 0 \).
Imagine plucking this string at its midpoint (Fig. below (a)) or at a point two-thirds from the left end (Fig. below (b)). Use Excel, MATLAB, or other graphing software, to create two plots showing that the above expression can fairly accurately represent the shape in Fig. (a), if \( A_1 = 1.00, \, A_2 = 0.00, \, A_3 = -0.11, \, A_4 = 0.00, \, \text{and} \, A_5 = 0.040 \) and in (b), if \( A_1 = 0.87, \, A_2 = -0.22, \, A_3 = -0.00, \, A_4 = 0.054, \, \text{and} \, A_5 = -0.035 \).
*Figures Description:*
- **Fig](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd7b1f94a-aeb1-4e21-a353-b14674dcbf02%2F5e200d5f-4fe3-4219-82e6-6db46ac19acc%2F6hk206r_processed.png&w=3840&q=75)
Transcribed Image Text:The manner in which a string is plucked determines the mixture of harmonic amplitudes in the resulting wave. Consider a string exactly 1/2-m long that is fixed at both its ends located at \( x = 0.0 \) and \( x = \frac{1}{2} \, \text{m} \). The first five harmonics of this string have wavelengths \( \lambda_1 = 1.0 \, \text{m}, \, \lambda_2 = \frac{1}{2} \, \text{m}, \, \lambda_3 = \frac{1}{3} \, \text{m}, \, \lambda_4 = \frac{1}{4} \, \text{m}, \, \text{and} \, \lambda_5 = \frac{1}{5} \, \text{m} \). According to Fourier’s theorem, any shape of this string can be formed by a sum of its harmonics, with each harmonic having its own unique amplitude \( A \). We limit the sum to the first five harmonics in the expression
\[
D(x) = \sum_{n=1}^{5} A_n \sin \left( \frac{2 \pi}{\lambda_n} x \right)
\]
where \( D \) is the displacement of the string at a time \( t = 0 \).
Imagine plucking this string at its midpoint (Fig. below (a)) or at a point two-thirds from the left end (Fig. below (b)). Use Excel, MATLAB, or other graphing software, to create two plots showing that the above expression can fairly accurately represent the shape in Fig. (a), if \( A_1 = 1.00, \, A_2 = 0.00, \, A_3 = -0.11, \, A_4 = 0.00, \, \text{and} \, A_5 = 0.040 \) and in (b), if \( A_1 = 0.87, \, A_2 = -0.22, \, A_3 = -0.00, \, A_4 = 0.054, \, \text{and} \, A_5 = -0.035 \).
*Figures Description:*
- **Fig
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