Homework Set 6 (Section 4.1-4.4)
This week, we had exercises from sections 4.2, 4.4, with some computer homework carried over from 3.6 and
4.4. The questions from the text (to turn in) are 4.2.1, 4.4.1
1. Exercise 4.2.1(a) Using
ρ0 (x)utt = T0 uxx + Q(x, t)ρ0 (x)
compute the sagged equilibrium solution uE (x) if Q = −g.
SOLUTION: Simplifying the PDE, we get:
utt =
T0
uxx − g
ρ0 (x)
Since we’re looking for the equilibrium solution (which does not depend on t), we set utt = 0, and
using the simplification at the top of p. 133, we take c2 = T0 /ρ0 (x). Therefore, we solve:
0 = c2 uxx − g
Solving, we get:
u00 =
g
c2
⇒
g
x + C1
c2
⇒
uE (x) =
uE (0) = 0
⇒
C2 = 0
u0 =
g 2
x + C1 x + C2
2c2
Solving for the coefficients, we have
uE (L) = 0
⇒
so that
uE (x) =
g 2
L + C1 L = 0
2c2
⇒
C1 = −
g
L
2c2
2
gρ0 2
(x2 − Lx) =
(x − Lx)
2c2
2T0
2. Exercise 4.2.1(b) Show that v(x, t) = u(x, t) − uE will solve the simplified PDE: utt = c2 uxx :
SOLUTION: Using the definition of v, we have:
vtt = utt
And
c2 vxx = c2 (uxx − u00E ) = c2 (uxx −
g
) = c2 uxx − g
c2
And since u solved the original expression,
c2 uxx − g = utt = vtt
Therefore,
vtt = c2 vxx
3. Exercise 4.4.1
Consider the vibrating string of uniform density and tension (so these are constants). (a) What are
the natural frequencies? (b) What if one end is fixed and one is free? Sketch a few modes. (c) Show
that the modes for the odd harmonics of part (a) are identical to modes of part (b) if H = L/2. Verify
that the natural frequencies are the same, and briefly explain using symmetry.
SOLUTIONS:
1
For part (a), we had:
u(x, t) =
∞
X
n=1
sin
nπc nπx h
nπc i
An cos
t + Bn sin
t
L
L
L
Therefore, the natural frequencies (in time) are nπc/L for n = 1, 2, 3, · · · .
For part (b), we showed in class that the wave equation with one end fixed and one end free yielded
the following solution (See the class website for the derivation):
u(x, t) =
∞
X
n=1
sin
(2n − 1)πx
2H
(2n − 1)πc
(2n − 1)πc
An cos
t + Bn sin
t
2H
2H
Therefore, the natural frequencies (in time) are
(2n − 1)πc
for n = 1, 2, 3, · · · .
2H
Plotting the first few functions in Maple:
φn (x) = sin
(2n − 1)πx
2H
with H = 1 gave me the following pics (these are φ1 (x) − φ4 (x) respectively).
For part (c), we really just need to note that if H = L/2, then
(2n − 1)πc
(2n − 1)πc
=
2H
L
which are the odd natural frequencies from part (a). What is being shown here?
“A string with a free end is equivalent to a string that is twice as long with two fixed ends.”
To compare modes with part (a), consider the following plots with L = 2 (since H = 1 in our previous
plots):
mπx φm (x) = sin
L
for m = 1, 3, 5, 7:
2