Use a computer as a computational aid. It was shown in Section 13.2 that the equation of a vibrating string is I a²ua²u р ах2 at² where T is the constant magnitude of the tension in the string and p is its mass per unit length. Suppose a string of length 120 centimeters is secured to the x-axis at its ends and is released from rest from the initial displacement (0.01x, 0.60-60 100 f(x) = 0 ≤ x ≤ 60 60 < x≤ 120. Use the difference equation U₁+1 = ²U₁+1 +2(1-²)uj + ²u₁-1.jUjj-1 to approximate the solution of the boundary-value problem when h= 20, k = 5√ p/T and where p = 0.0225 g/cm, T = 1.7 x 107 dynes. Use m = 50. (Give the approximations obtained for t= km. Round your answers to four decimal places.) u(20, km) 0.1589 x u(40, km) 0.2785 X X x X u(60, km) 0.3228 u(80, km) 0.2761 u(100, km) 0.1564 Book
Use a computer as a computational aid. It was shown in Section 13.2 that the equation of a vibrating string is I a²ua²u р ах2 at² where T is the constant magnitude of the tension in the string and p is its mass per unit length. Suppose a string of length 120 centimeters is secured to the x-axis at its ends and is released from rest from the initial displacement (0.01x, 0.60-60 100 f(x) = 0 ≤ x ≤ 60 60 < x≤ 120. Use the difference equation U₁+1 = ²U₁+1 +2(1-²)uj + ²u₁-1.jUjj-1 to approximate the solution of the boundary-value problem when h= 20, k = 5√ p/T and where p = 0.0225 g/cm, T = 1.7 x 107 dynes. Use m = 50. (Give the approximations obtained for t= km. Round your answers to four decimal places.) u(20, km) 0.1589 x u(40, km) 0.2785 X X x X u(60, km) 0.3228 u(80, km) 0.2761 u(100, km) 0.1564 Book
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Transcribed Image Text:Use a computer as a computational aid.
It was shown in Section 13.2 that the equation of a vibrating string is
Ta²u
a²u
р ах2 at²'
where T is the constant magnitude of the tension in the string and p is its mass per unit length. Suppose a string of length 120 centimeters is secured to the x-axis at its ends and is released from rest
from the initial displacement
f(x) =
=
0.01x,
0.60
X - 60
100
0 ≤ x ≤ 60
60 < x≤ 120.
Use the difference equation U¡¡+1 = √²U¡+1‚j + 2(1 − λ²)u¡j + λ²u¡-1,j — U¡j-1 to approximate the solution of the boundary-value problem when h = 20, k = 5√ p/T and where p = 0.0225 g/cm,
T = 1.7 x 107 dynes. Use m = 50. (Give the approximations obtained for t = km. Round your answers to four decimal places.)
u(20, km)≈ 0.1589
X
X
u(40, km)≈ 0.2785
u(60, km)≈ 0.3228
u(80, km)≈ 0.2761
u(100, km)≈ 0.1564
eBook
X
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