Mathematic Methods HW Solutions 60

Mathematic Methods HW Solutions 60 - Chapter 13 4.6 4.7 4.8...

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Unformatted text preview: Chapter 13 4.6 4.7 4.8 y= y= y= 60 4hl π2 v 9hl π3 v 4l π2 v 4.9 ∞ 1 odd n ∞ 1 nπ nπw nπx nπvt 1 sin sin sin sin 2 n 2 l l l nπ nπx nπvt 1 sin sin sin n3 3 l l ∞ 1 sin(nπ/2) πx πvt π 2πx 2πvt nπx nπvt sin sin + sin sin − sin sin 3 l l 16 l l n(n2 − 4) l l n=3 1. n = 1, ν = v/(2l) 2. n = 2, ν = v/l 3. n = 3, ν = 3v/(2l), and n = 4, ν = 2v/l, have nearly equal intensity. 4. n = 2, ν = v/l 5, 6, 7, 8. n = 1, ν = v/(2l) 4.11 The basis functions for a string pinned at x = 0, free at x = l, and (n+ 1 )πx 1 (n+ 2 )πvt . l 2 cos with zero initial string velocity, are y = sin l The solutions for Problems 2, 3, 4, parts (a) and (b) are: ∞ 1 1 (n + 2 )πx (n + 2 )πvt (a) y = an cos cos l l 0 ∞ 1 (n + 2 )πx (n + 1 )πvt 2 cos l l 0 where the coefficients are: (2n + 1)π (2n + 1)π 128h sin2 cos 2(a) an = 2 π2 (2n + 1) 16 8 128h (2n + 1)π 2 (2n + 1)π 2(b) bn = sin sin (2n + 1)2 π 2 16 8 (2n + 1)π (2n + 1)π 256h sin2 cos 3(a) an = (2n + 1)2 π 2 32 16 256h (2n + 1)π (2n + 1)π 3(b) bn = sin2 sin (2n + 1)2 π 2 32 16 (2n + 1)π (2n + 1)π 256h 2 (2n + 1)π sin sin sin 4(a) an = (2n + 1)2 π 2 16 8 4 (2n + 1)π (2n + 1)π 256h (2n + 1)π sin cos 4(b) bn = sin2 (2n + 1)2 π 2 16 8 4 8 4.12 With bn = 3 3 , odd n, the six solutions on (0, 1) are: nπ 1. Temperature in semi-infinite plate: T = bn e−nπy sin nπx 2. Temperature in finite plate of height H : bn sinh nπ (H − y) sin nπx T= sinh (nπH ) 2 3. 1-dimensional heat flow: u = bn e−(nπα) t sin nπx h ¯ 2 n2 π 2 ¯ 4. Particle in a box: Ψ = bn sin nπx e−iEn t/h , En = 2m 5. Plucked string: y = bn sin nπx cos nπvt bn 6. String with initial velocity: y = sin nπx sin nπvt nπv (b) y= bn sin ...
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This note was uploaded on 02/29/2012 for the course MHF 2312 taught by Professor Dr.chet during the Fall '11 term at UNF.

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