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16.5 Laplace Transform Techniques 591

∞

n
−(L−x)s/c −(L+x)s/c n −2sL/c
= e − e (−1) e
n=0
∞

−(L−x)s/c −(L+x)s/c n −2nsL/c
= e − e (−1) e .
n=0
Here we used ξ = e −2sL/c in the geometric series, using the fact that this exponential is always
less than 1 for x > 0. Finally, we can write
∞
cK
1
Y(x,s) = e −(L−x)s/c (−1) n e −2snL/c
E s 2
n=0
∞
cK
1
− e −(L+x)s/c (−1) n e −2nsL/c
E s 2
n=0
∞
cK
1
= (−1) n e ((−(2n+1)L−x)/c)s
E s 2
n=0
∞
cK
1
− e ((−(2n+1)L+x)/c)s .
E s 2
n=0
Now recall that, for any nonzero number α

1
−1 −αs
L e = (t − α)H(t − α)
s 2
in which H is the Heaviside function. Assuming that we can take the transform of the geometric
series term by term, the solution is
cK
(2n + 1)L − x (2n + 1)L − x
∞
y(x,t) = t − H t −
E c c
n=0
cK
(2n + 1)L + x (2n + 1)L + x
∞
− t − H t − .
E c c
n=0
Look at this solution for a particular choice of the constants. Suppose c = 2 and K = E =
L = 1. Now the solution is
∞

2n + 1 − x 2n + 1 − x
y(x,t) =2 (−1) n t − H t −
2 2
n=0
∞

2n + 1 + x 2n + 1 + x
− 2 (−1) n t − H t − .
2 2
n=0
Graphs of this solution on 0≤ x ≤1 are shown in Figure 16.9. The lower graph is for time t =1.7.
Moving up, the next graph is t = 2.4, then 3.4, then 1.3 and 4.7 (indistinguishable in the scale of
the graphs), then 5.2, and the highest graph is 0.9.
It is interesting to look at the motion of the right end of the bar. The analysis carried out to
find Y(x,s) applies, except now put x = L to obtain

cK 1 sinh(sL/c) cK 1
Y(L,s) = = tanh(sL/c).
2
E s cosh(sL/c) E s 2

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October 14, 2010 15:23 THM/NEIL Page-591 27410_16_ch16_p563-610