Mathematic Methods HW Solutions 61

# Mathematic Methods HW Solutions 61 - 16 n odd the six...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 13 61 16 , n odd, the six solutions on (0, π ) are nπ (4 − n2 ) 1. T = bn e−ny sin nx bn sinh n(H − y ) sin nx 2. T = sinh nH 2 3. u = bn e(−nα) t sin nx h ¯ 2 n2 ¯ 4. Ψ = bn sin nx e−iEn t/h , En = 2m 5. y = bn sin nx cos nvt bn 6. y = sin nx sin nvt nv 12(−1)n+1 4.14 Same as 4.12 with bn = , all n, on (0, 1) n3 π 3 4.13 With bn = 5.1 5.2 (a) u ∼ 9.76◦ = (a) 2 J1 (km r ) e−km z sin θ, km = zeros of J1 km J2 (km ) m=1 ∞ 2a J1 (km r/a) e−km z/a sin θ, km = zeros of J1 km J2 (km ) m=1 u(r = 1, z = 1, θ = π/2) ∼ 0.211 = ∞ 200 J0 (km r ) sinh km (10 − z ), km = zeros of J0 (a) u = km J1 (km ) sinh(10km ) m=1 (b) 5.3 (b) u = 5.4 (b) u ∼ 9.76◦ = ∞ ∞ 200 km (H − z ) J0 (km r/a) sinh , k J (k ) sinh(km H/a) a m=1 m 1 m km = zeros of J0 ∞ 2 200 u= J0 (km r/a)e−(km α/a) t , km = zeros of J0 km J1 (km ) m=1 ∞ 5.5 2 200a J1 (km r/a)e−(km α/a) t sin θ, km = zeros of J1 km J2 (km ) m=1 5.6 amn = 5.7 5.8 2π f (r, θ)Jn (kmn r/a) cos nθ r dr dθ 2 πa2 Jn+1 (kmn ) 0 0 a 2π 2 f (r, θ)Jn (kmn r/a) sin nθ r dr dθ bmn = 2 πa2 Jn+1 (kmn ) 0 0 400 nπz 1 nπr u= sin + I0 π nI0 (3nπ/20) 20 20 u = 40 + o dd n ∞ 2 2 120 J0 (km r )e−km α t , where J0 (km ) = 0 k J (k ) m=1 m 1 m 1600 π2 o dd m 6400 5.10 u = 3 π 5.9 a 2 u= n sin(nπx/10) sin(mπy/10) sinh[π (n2 + m2 )1/2 (10 − z )/10] mn sinh[π (n2 + m2 )1/2 ] o dd n nπx mπy pπz −(απ/l)2 (n2 +m2 +p2 )t 1 sin sin sin e nmp l l l o dd n o dd m o dd p −n 5.11 R = r , r , n = 0; R = ln r, const., n = 0 R = r l , r − l− 1 200 r n sin nθ 5.12 u = 50 + π a n o dd n 1 r 4n 400 sin 4nθ 5.13 u = π n 10 o ddn ...
View Full Document

## 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.

Ask a homework question - tutors are online