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Mathematic Methods HW Solutions 60

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

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Chapter 13 60 4.6 y = 4 hl π 2 v summationdisplay 1 odd n 1 n 2 sin 2 sin nπw l sin nπx l sin nπvt l 4.7 y = 9 hl π 3 v summationdisplay 1 1 n 3 sin 3 sin nπx l sin nπvt l 4.8 y = 4 l π 2 v bracketleftBigg 1 3 sin πx l sin πvt l + π 16 sin 2 πx l sin 2 πvt l - summationdisplay n =3 sin( nπ/ 2) n ( n 2 - 4) sin nπx l sin nπvt l bracketrightBigg 4.9 1. n = 1, ν = v/ (2 l ) 2. n = 2, ν = v/l 3. n = 3, ν = 3 v/ (2 l ), and n = 4, ν = 2 v/l , have nearly equal intensity. 4. n = 2, ν = v/l 5, 6, 7, 8. n = 1, ν = v/ (2 l ) 4.11 The basis functions for a string pinned at x = 0, free at x = l , and with zero initial string velocity, are y = sin ( n + 1 2 ) πx l cos ( n + 1 2 ) πvt l . The solutions for Problems 2, 3, 4, parts (a) and (b) are: (a) y = summationdisplay 0 a n cos ( n + 1 2 ) πx l cos ( n + 1 2 ) πvt l (b) y = summationdisplay 0 b n sin ( n + 1 2 ) πx l cos ( n + 1 2 ) πvt l where the coefficients are: 2(a) a n = 128 h (2 n + 1) 2 π 2 sin 2 (2 n + 1) π 16 cos (2 n + 1) π 8 2(b) b n = 128 h (2 n + 1) 2 π 2 sin 2 (2 n + 1) π 16 sin (2 n + 1) π 8 3(a) a n = 256 h (2 n + 1) 2 π 2 sin 2 (2 n + 1) π 32 cos (2 n + 1) π 16 3(b) b n = 256 h (2 n + 1) 2 π 2 sin 2 (2 n + 1) π 32 sin (2 n + 1) π 16 4(a) a n = 256 h (2 n + 1) 2 π 2 sin 2 (2 n + 1) π 16 sin (2 n + 1) π 8 sin (2 n + 1) π 4 4(b)
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