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Solved Problems

∞
v X 1 2nπd nπξ nπx nπct L
u(x, t) = 2nπd + L sin sin sin sin for d = .
2
π c n 2 L L L L 2n
n=1

c) The kinetic energy of the string is

Z L
1 2
E = ρ (u (x, t)) dx,
t
2 0
where ρ is the density of the string per unit length.
Flat Hammer. The n th harmonic is

4Lv nπd nπξ nπx nπct
u = sin sin sin sin .
n
2 2
n π c L L L L
The kinetic energy of the n th harmonic is

ρ Z L ∂u n 2 4Lv 2 nπd nπξ nπct
E = dx = sin 2 sin 2 cos 2 .
n
2 2
2 0 ∂t n π L L L
This will be maximized if

nπξ
sin 2 = 1,
L
nπξ π(2m − 1)
= , m = 1, . . . , n,
L 2
(2m − 1)L
ξ = , m = 1, . . . , n
2n

2
We note that the kinetic energies of the n th harmonic decay as 1/n .
Curved Hammer. We assume that d 6= L . The n th harmonic is
2n
2
8dL v nπd nπξ nπx nπct
u = cos sin sin sin .
n
2
2
2 2
nπ c(L − 4d n ) L L L L
The kinetic energy of the n th harmonic is
Z L 2 2 3 2
ρ ∂u n 16d L v nπd nπξ nπct
E = dx= cos 2 sin 2 cos 2 .
n
2 2 2
2
2
2 0 ∂t π (L − 4d n ) L L L
This will be maximized if

nπξ
sin 2 = 1,
L
(2m − 1)L
ξ = , m = 1, . . . , n
2n

4
We note that the kinetic energies of the n th harmonic decay as 1/n .

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