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6. Find the deflection u(x, t) of the string of length L when \(\pi\) 1, the initial velocity is zero, and the initial deflection in the interval \((0, L)\) is (a) f(x) = kx(1 - x) with L = 1 2 Ans: u(x, t) = \sum_{n=1}^{\infty} \frac{12k(-1)^{n+1}}{n^3 \pi^3} \cos n\pi t \sin nx (b) f(x) = k(\sin(\pi x) - \frac{1}{3}\sin(3\pi x)) with L = 1 Ans: u(x, t) = k \cos \pi t \sin \pi x - (k/3) \cos 3\pi t \sin 3\pi x (c) f(x) = \begin{cases} \frac{\pi}{L}x & 0 \le x \le L/3\\ \frac{\pi}{L}(L - 2x) & L/3 \le x \le 2L/3\\ \frac{\pi}{L}(x - L) & 2L/3 \le x \le L \end{cases} Ans: u(x, t) = \sum_{n=1}^{\infty} B_n \cos n\pi t \sin nx. B_n = \frac{186 \sin n\pi/3}{n^3 \pi^3}(1 - 2 \cos n\pi/3)

          6. Find the deflection u(x, t) of the string of length L when
\(\pi\)
1, the initial velocity is zero, and the initial deflection in the
interval \((0, L)\) is
(a) f(x) = kx(1 - x) with L = 1
2
Ans: u(x, t) = \sum_{n=1}^{\infty} \frac{12k(-1)^{n+1}}{n^3 \pi^3} \cos n\pi t \sin nx
(b) f(x) = k(\sin(\pi x) - \frac{1}{3}\sin(3\pi x)) with L = 1
Ans: u(x, t) = k \cos \pi t \sin \pi x - (k/3) \cos 3\pi t \sin 3\pi x
(c) f(x) = \begin{cases} \frac{\pi}{L}x & 0 \le x \le L/3\\ \frac{\pi}{L}(L - 2x) & L/3 \le x \le 2L/3\\ \frac{\pi}{L}(x - L) & 2L/3 \le x \le L \end{cases}
Ans: u(x, t) = \sum_{n=1}^{\infty} B_n \cos n\pi t \sin nx.
B_n = \frac{186 \sin n\pi/3}{n^3 \pi^3}(1 - 2 \cos n\pi/3)

6. Find the deflection u(x, t) of the string of length L when
π
1, the initial velocity is zero, and the initial deflection in the
interval (0, L) is
(a) f(x) = kx(1 - x) with L = 1
2
Ans: u(x, t) = ∑n=1^∞ (12k(-1)^n+1)/(n^3 π^3) cosnπt sinnx
(b) f(x) = k(sin(πx) - (1)/(3)sin(3πx)) with L = 1
Ans: u(x, t) = k cosπt sinπx - (k/3) cos3πt sin3πx
(c) f(x) = (π)/(L)x 0 ≤x ≤L/3
(π)/(L)(L - 2x) L/3 ≤x ≤2L/3
(π)/(L)(x - L) 2L/3 ≤x ≤L
Ans: u(x, t) = ∑n=1^∞ Bn cosnπt sinnx.
Bn = (186 sinnπ/3)/(n^3 π^3)(1 - 2 cosnπ/3)

Added by Jonathan H.

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