Page 349 - Schaum's Outline of Theory and Problems of Advanced Calculus

340 FOURIER SERIES [CHAP. 13

difference in tensions is Fðsin sin Þ. This is the force producing the acceleration that accounts for
the vibratory motion. ( tan tan ) @y
Now Ffsin sin g¼ F p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Fftan tan g¼ F ðx þ x; tÞ
2
2
1 þ tan 1 þ tan @x
@y
ðx; tÞ , where the squared terms in the denominator are neglected because the vibrations are small.
@x
Next, equate the two forms of the force, i.e.,
2
@y @y w @ y
F ðx þ x; tÞ ðx; tÞ ¼ x
@x @x g @t 2
r ffiffiffiffiffiffi
Fg
, the resulting equation is
w
divide by x, and then let x ! 0. After letting ¼
2
2
@ y 2 @ y
¼
@t 2 @x 2
This homogeneous second partial derivative equation is the classical equation for the vibrating
string. Associated boundary conditions are
yð0; tÞ¼ 0; yðL; tÞ¼ 0; t > 0
The initial conditions are
@y
2
yðx; 0Þ¼ mðLx x Þ; ðx; 0Þ¼ 0; 0 < x < L
@t
The method of solution is to separate variables, i.e., assume
yðx; tÞ¼ GðxÞHðtÞ
Then upon substituting
2
00 00
GðxÞ H ðtÞ¼ G ðxÞ HðtÞ
Separating variables yields
G 00 H 00 2
¼ k; ¼ k; where k is an arbitrary constant
G H
Since the solution must be periodic, trial solutions are
p ffiffiffiffiffiffiffi p ffiffiffiffiffiffiffi
GðxÞ¼ c 1 sin k x þ c 2 cos k x;< 0
p ffiffiffiffiffiffiffi p ffiffiffiffiffiffiffi
HðtÞ¼ c 3 sin k t þ c 4 cos k t
Therefore
p ffiffiffiffiffiffiffi p ffiffiffiffiffiffiffi p ffiffiffiffiffiffiffi p ffiffiffiffiffiffiffi
y ¼ GH ¼½c 1 sin k x þ c 2 cos k xŠ½c 3 sin k t þ c 4 cos k tŠ
The initial condition y ¼ 0at x ¼ 0 for all t leads to the evaluation c 2 ¼ 0.
Thus
p ffiffiffiffiffiffiffi p ffiffiffiffiffiffiffi p ffiffiffiffiffiffiffi
y ¼½c 1 sin k xŠ½c 3 sin k t þ c 4 cos k tŠ
p ffiffiffiffiffiffiffi p ffiffiffiffiffiffiffi
Now impose the boundary condition y ¼ 0at x ¼ L, thus 0 ¼½c 1 sin k LŠ½c 3 sin k t þ
ffiffiffiffiffiffiffi
c 4 cos k tŠ:
p
c 1 6¼ 0as that would imply y ¼ 0 and a trivial solution. The next simplest solution results from the
n h n ih n n i
ffiffiffiffiffiffiffi
p
choice k ¼ , since y ¼ c 1 sin x c 3 sin t þ c 4 cos t and the first factor is zero when
L L l L
x ¼ L.