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# [Donald A. McQuarrie, John D. Simon] Physical Chem(BookZZ.org)

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in Figure 2.2. Note that it oscillates cosinusoidally in time,
with an amplitude A. The wavelength of the oscillation is 2n I w and the frequency v
is given by (see Problem 2-3)
FIGURE 2.2
(J)
v=-
2n
A plot of x(t) =A cos wt, the solution to the problem in Example 2--4. The amplitude is A, the
wavelength is 2n I w, and the frequency is w 12n.
2-4. The General Solution to the Wave Equation Is a Superposition
of Normal Modes
Let us assess where we are now. We have obtained Equations 2.8 and 2.9 by applying
the method of separation of variables to the wave equation. We have already shown
that if the separation constant K is zero, then only a trivial solution results. Now let's
assume that K is positive. To this end, write K as {3 2 , where f3 is real. This assures that
2 - 4 . T h e G e n e r a l S o l u t i o n t o t h e W a v e E q u a t i o n I s a S u p e r p o s i t i o n o f N o r m a l M o d e s
K i s p o s i t i v e b e c a u s e i t i s t h e s q u a r e o f a r e a l n u m b e r . I n t h e c a s e K = { 3
2
, t h e g e n e r a l
s o l u t i o n t o E q u a t i o n 2 . 8 i s
X ( x ) = c
1
e f 3 x + c
2
e - f 3 x
W e c a n e a s i l y s h o w t h a t t h e o n l y w a y t o s a t i s f y t h e b o u n d a r y c o n d i t i o n s ( E q u a t i o n 2 . 1 4 )
i s f o r c
1
= c
2
= 0 , a n d s o o n c e a g a i n w e f i n d o n l y a t r i v i a l s o l u t i o n .
L e t ' s h o p e t h a t a s s u m i n g K t o b e n e g a t i v e g i v e s u s s o m e t h i n g i n t e r e s t i n g . I f w e
s e t K = - { 3
2
, t h e n K i s n e g a t i v e i f f 3 i s r e a l . I n t h i s c a s e E q u a t i o n 2 . 8 i s
d 2 X ( x ) + f 3 2 X ( x ) = 0
d x
2
R e f e r r i n g t o E x a m p l e 2 - 4 , w e s e e t h a t t h e g e n e r a l s o l u t i o n c a n b e w r i t t e n a s
X ( x ) = A c o s f 3 x + B s i n f 3 x
T h e b o u n d a r y c o n d i t i o n t h a t X ( 0 ) = 0 i m p l i e s t h a t A = 0 . T h e c o n d i t i o n a t t h e b o u n d -
a r y x = l s a y s t h a t
X ( l ) = B s i n f 3 l = 0
( 2 . 1 8 )
E q u a t i o n 2 . 1 8 c a n b e s a t i s f i e d i n t w o w a y s . O n e i s t h a t B = 0 , b u t t h i s a l o n g w i t h t h e
f a c t t h a t A = 0 y i e l d s a t r i v i a l s o l u t i o n . T h e o t h e r w a y i s t o r e q u i r e t h a t s i n f 3 l = 0 .
B e c a u s e s i n e = 0 w h e n e = 0 , J T , 2 n , 3 n , . . . , E q u a t i o n 2 . 1 8 i m p l i e s t h a t
f 3 l = n J T
n = 1 , 2 , 3 , . . .
( 2 . 1 9 )
w h e r e w e h a v e o m i t t e d t h e n = 0 c a s e b e c a u s e i t l e a d s t o f 3 = 0 , a n d a t r i v i a l s o l u t i o n .
E q u a t i o n 2 . 1 9 d e t e r m i n e s t h e p a r a m e t e r f 3 a n d h e n c e t h e s e p a r a t i o n c o n s t a n t K = - { 3
2
\u2022
S o f a r t h e n , w e h a v e t h a t
n n x
X ( x ) = B s i n - l -
( 2 . 2 0 )
R e m e m b e r t h a t w e h a v e E q u a t i o n 2 . 9 t o s o l v e a l s o . E q u a t i o n 2 . 9 c a n b e w r i t t e n a s
d 2 T ( t ) + f 3 2 v 2 T ( t ) = 0
d t
2
( 2 . 2 1 )
w h e r e E q u a t i o n 2 . 1 9 s a y s t h a t f 3 = m r / l . R e f e r r i n g t o t h e r e s u l t o b t a i n e d i n E x a m p l e
2 - 4 a g a i n , t h e g e n e r a l s o l u t i o n t o E q u a t i o n 2 . 2 1 i s
T ( t ) = D c o s w n t + E s i n w n t
( 2 . 2 2 )
4 7
48 Chapter 2 I The Classical Wave Equation
where w n = f3 v = mr vI l. We have no conditions to specify D and E, so the amplitude
u(x, t) is (cf. Equation 2.3)
u(x, t) = X(x)T(t)
= ( B sin n7x) (Dcoswnt + E sinwnt)
nnx
= (Fcosw t+Gsinw t)sin--
n n l n = 1, 2, ...
where we have let F = DB and G = E B. Because there is a u (x, t) for each integer n
and because the values ofF and G may depend on n, we should write u(x, t) as
nnx
u (x, t) = (F cosw t + G sinw t) sin--
n n n n n [ n = 1, 2, ... (2.23)
Because each un (x, t) is a solution to the linear differential equation, Equation 2.1, their
sum is also a solution of Equation 2.1 and is, in fact, the general solution. Therefore,
for the general solution we have
n = 1, 2, ... (2.24)
No matter how the string is plucked initially, its shape will evolve according to Equa-
tion 2.24. We can easily verify that Equation 2.24 is a solution to Equation 2.1 by
direct substitution. Problem 2-5 shows that F cos wt + G sin wt can be written in the
equivalent form, A cos(wt + ¢ ), where A and ¢ are constants expressible in terms of
F and G. The quantity A is the amplitude ofthe wave and¢ is called the phase angle.
Using this relation, we can write Equation 2.24 in the form
oo nnx oo
u(x, t) =&quot;A cos (w t +¢)sin-=&quot; u (x, t) L...tn n n / L...tn
n=l n=l
(2.25)
Equation 2.25 has a nice physical interpretation. Each un(x, t) is called a normal
mode, and the time dependence of each normal mode represents harmonic motion of
frequency
V
_ wn _ vn
- -
n 2n 21
(2.26)
where we have used the fact that wn = f3v = nnvll (cf. Equation 2.19). The spatial
dependence of the first few terms in Equation 2.25 is shown in Figure 2.3. The first
term, u 1 (x, t ), called the fundamental mode or first harmonic, represents a sinusoidal
(harmonic) time dependence of frequency v 121 of the motion depicted in Figure 2.3a.
The second harmonic or first overtone, u2 (x, t), vibrates harmonically with frequency
vI l and looks like the motion depicted in Figure 2.3b. Note that the midpoint of this
harmonic is fixed at zero for all t. Such a point is called a node, a concept that arises in
quantum mechanic as well. Notice that u (0) and u (l) are also equal to zero. These terms
are not nodes because their values are fixed by the boundary conditions. Note that the
u u
u
~
0 l 0 0
( a ) ( b )
( c )
F I G U R E 2 . 3
T h e f i r s t t h r e e n o r m a l m o d e s o f a v i b r a t i n g s t r i n g . N o t e t h a t e a c h n o r m a l m o d e i s a s t a n d i n g
w a v e a n d t h e t h e n t h h a r m o n i c h a s n - 1 n o d e s .
s e c o n d h a r m o n i c o s c i l l a t e s w i t h t w i c e t h e f r e q u e n c y o f t h e f i r s t h a r m o n i c . F i g u r e 2 . 3 c
s h o w s t h a t t h e t h i r d h a r m o n i c o r s e c o n d o v e r t o n e h a s t w o n o d e s . I t i s e a s y t o c o n t i n u e
a n d s h o w t h a t t h e n u m b e r o f n o d e s i s e q u a l t o n - 1 ( P r o b l e m 2 - 1 0 ) . T h e w a v e s s h o w n
i n F i g u r e 2 . 3 a r e c a l l e d s t a n d i n g w a v e s b e c a u s e t h e p o s i t i o n s o f t h e n o d e s a r e f i x e d i n
t i m e . B e t w e e n t h e n o d e s , t h e s t r i n g o s c i l l a t e s u p a n d d o w n .
C o n s i d e r a s i m p l e c a s e i n w h i c h u ( x , t ) c o n s i s t s o f o n l y t h e f i r s t t w o h a r m o n i c s
a n d i s o f t h e f o r m ( c f . E q u a t i o n 2 . 2 5 )
r r x 1 ( r r ) 2 r r x
u ( x , t ) = c o s w
1
t s i n -
1
- + l c o s W i + 2 s i n -
1
-
( 2 . 2 7 )
E q u a t i o n 2 . 2 7 i s i l l u s t r a t e d i n F i g u r e 2 . 4 . T h e l e f t s i d e o f F i g u r e 2 . 4 s h o w s t h e t i m e d e -
p e n d e n c e o f e a c h m o d e s e p a r a t e l y . N o t