Page 873 - Advanced_Engineering_Mathematics o'neil

Answers to Selected Problems 853

Section 16.6 Characteristics and d’Alembert’s Solution
1 1 x+t
2
2
2
2
1. y(x,t) = [(x − t) + (x + t) ]+ −ξ dξ = x + t − xt
2 2 x−t
1 49 1 49
3
2
2
3. y(x,t) = [cos(π(x − 7t)) + cos(π(x + 7t))]+ t − x t − t = cos(πx)cos(7πt) + t − x t − t 3
2 3 2 3
1
x
5. y(x,t) = [e x−14t + e x+14t ]+ xt = e cosh(14t) + xt
2
1 1 1
2
7. y(x,t) = x + (e −x+4t − e −x−4t ) + xt + t 3
8 2 6
1 1
4
2
2
9. y(x,t) = (sin(2(x + 8t)) − sin(2(x − 8t))) + xt + x + 64t − x
32 12
1 9
11. y(x,t) = [cosh(x − 3t) + cosh(x + 3t)]+ t + xt 5
2 10
1
13. (sin(2(x − ct)) + sin(2(x + ct)))
2
1
15. (cos(x − ct) + cos(x + ct))
2
Section 16.7 Vibrations in a Circular Membrane
1. We find that (approximately),
1
2 xJ 0 (2.405x)dx 0.1057
a 1 = 0 ≈ 2 = 0.78442,
[J 1 (2.405)] 2 0.2695
a 2 ≈ 0.06869,a 3 ≈ 0.05311,a 4 ≈ 0.01736,a 5 ≈ 0.01698
The fifth partial sum gives the approximation
z(r,t) ≈ 0.78442J 0 (2.405r)cos(2.405t) + 0.05311J 0 (5.520r)cos(5.520t)
+ 0.06869J 0 (8.654r)cos(8.654t) + 0.01736J 0 (11.792r)cos(11.792t)
+ 0.01698J 0 (14.931r)cos(14.931t).
3. We find the approximation
z(r,t) ≈ 1.2534J 0 (2.405r)cos(2.405t) − 0.80469J 0 (5.520r)cos(5.520t)
− 0.11615J 0 (8.654r)cos(8.654t) − 0.09814J 0 (11.792r)cos(11.792t)
− 0.03740J 0 (14.931r)cos(14.931t)

Section 16.8 Vibrations in a Circular Membrane II
1. The solution is
∞
j 0k

z(r,θ,t) = α k J 0 r cos( j 0k t)
2
k=1
∞
j 2k
+ cos(2θ) β k J 2 r cos( j 2k t)
2
k=1
∞ ∞

j pq
+ sin(pθ) δ pq J p r sin( j pq t)
2
p=1 q=1
where
1
2
2
α k = ξ(1 − ξ )J 0 ( j 0k ξ)dξ ,
[J 1 ( j 0k )] 2 0
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October 14, 2010 17:50 THM/NEIL Page-853 27410_25_Ans_p801-866