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for the coefficients. We can make (8.11) match (8.9) by choosing h(x) = ϕ(x) and
a n = A n . This tells us that

l
Z
2 nπ
A n = ϕ(x) sin x dx.
l l
0
πnc
Similarly, we can make (8.11) match (8.10) by choosing h(x) = ψ(x) and a n = B n .
l
This tells us that
l
Z
πnc 2 nπ
B n = ψ(x) sin x dx.
l l l
0
So we have a solution:

∞
X πnc πnc πn
u(x, t) = A n cos t + B n sin t sin x (8.12)
l l l
n=1
with
l
Z
2 πn
A n = ϕ(x) sin x dx, (8.13)
l l
0
and
l
Z
2 πn
B n = ψ(x) sin x dx. (8.14)
πnc l
0
While the sum (8.12) can be very complicated, each term, called a “mode”, is quite
simple. For each fixed t, the mode

πnc πnc πn
A n cos t + B n sin t sin x (8.15)
l l l
πn
is just a constant times sin x .
l
nπ
As x runs from 0 to l, the argument of sin x runs from 0 to πn, which is
l
n half–periods of sin . Here are graphs, at fixed t, of the first three modes, called the
fundamental tone, the first harmonic and the second harmonic.
πnc
For each fixed x, the mode (8.15) is just a constant times cos t plus a constant
l
πnc πnc πnc
times sin t . As t increases by one second, the argument, t of both cos t
l l
πnc πnc nc
l
and cos t increases by which is cycles (i.e. periods). So the fundamental
l l 2l
c c
oscillates at cps, the first harmonic oscillates at 2 cps, the second harmonic oscillates
2l 2l
c
at 3 cps and so on.
2l

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