The motion of string deflection u, can be expressed by the one-dimensional wave equation, shown in Eq. 1. 8² Eq. (1) Here, e²=T/p, where 7 is the tension in the string and p is the mass per unit length of the string. A string hold at both its ends is released from an initial deflection f(x) and initial velocity of zero. Since the string is hold at both its ends, the boundary conditions of the string can be expressed as u(0, t) = 0 and u(L, t) = 0. Here, L is the length of the string. By using method of separation of variables, a) show that the general solution for u(x, t) can be expressed as in Eq. (2) 1(x, t) - cos (t) sin(T) Eq. (2) E₂ E. =*/(x) sin (7) dx Eq. (3) b) Therefore, if e² = 1 and L=1, for an initial deflection shown in Figure 2, show that the specific solution for u(x, t) is as follows, (x, t) = (2-√2) cos(nt) sin(xx)-(2+√2) cos(3mt) sin(3x) +(2+ √2) cos(5mt) sin(5mx)-(2- Eq. (4) √2) cos(7mt) sin(7xx)+...] M 3/4 t 1/4 1/2 AM

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Answer 2 b only

QUESTION 2
The motion of string deflection u, can be expressed by the one-dimensional wave equation,
shown in Eq. 1.
8²μ
a²u
Eq. (1)
at²
ax²
Here, e² = T/p, where 7 is the tension in the string and p is the mass per unit length of the
string.
A string hold at both its ends is released from an initial deflection f(x) and initial velocity of
zero. Since the string is hold at both its ends, the boundary conditions of the string can be
expressed as u(0, t) = 0 and u(L, t) = 0. Here, L is the length of the string.
By using method of separation of variables,
a) show that the general solution for u(x, t) can be expressed as in Eq. (2)
1(x, t) = £, cos(
e) sin
Eq. (2)
MIX
E₂ - F(x) sin(
dx
Eq. (3)
b) Therefore, if c² = 1 and L = 1, for an initial deflection shown in Figure 2, show that
the specific solution for u(x, t) is as follows,
z(x, t) = (2-√2) cos(xt) sin(xx)-(2+ √2) cos(3mt) sin(3xx)
Eq. (4)
+(2+ √2) cos(5mt) sin(5mx)--(2-
√2) cos(7mt) sin(7xx) +..]
4/4
ħ
1/4
24
Figure 2-Initial deflection of string
Transcribed Image Text:QUESTION 2 The motion of string deflection u, can be expressed by the one-dimensional wave equation, shown in Eq. 1. 8²μ a²u Eq. (1) at² ax² Here, e² = T/p, where 7 is the tension in the string and p is the mass per unit length of the string. A string hold at both its ends is released from an initial deflection f(x) and initial velocity of zero. Since the string is hold at both its ends, the boundary conditions of the string can be expressed as u(0, t) = 0 and u(L, t) = 0. Here, L is the length of the string. By using method of separation of variables, a) show that the general solution for u(x, t) can be expressed as in Eq. (2) 1(x, t) = £, cos( e) sin Eq. (2) MIX E₂ - F(x) sin( dx Eq. (3) b) Therefore, if c² = 1 and L = 1, for an initial deflection shown in Figure 2, show that the specific solution for u(x, t) is as follows, z(x, t) = (2-√2) cos(xt) sin(xx)-(2+ √2) cos(3mt) sin(3xx) Eq. (4) +(2+ √2) cos(5mt) sin(5mx)--(2- √2) cos(7mt) sin(7xx) +..] 4/4 ħ 1/4 24 Figure 2-Initial deflection of string
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