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


DisciplinaFísico-química I6.504 materiais97.838 seguidores
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n 2 m 2 ) 1 1 2 
= v : r r - + -
a 2 b 2 
( 2 . 4 7 ) 
A c c o r d i n g t o P r o b l e m 2 - 1 5 , E q u a t i o n 2 . 4 6 c a n b e w r i t t e n a s 
T n m ( t ) = G n m c o s ( w n m t + < P n m ) 
( 2 . 4 8 ) 
T h e c o m p l e t e s o l u t i o n t o E q u a t i o n 2 . 2 8 i s 
0 0 0 0 
u ( x , y , t ) = L L u n m ( x , y , t ) 
n = l m = l 
0 0 0 0 
' &quot; &quot; &quot; ' ' &quot; &quot; &quot; ' . n : r r x m : r r y 
= L - J L - J A n m c o s ( w n m t + ¢ n m ) s m - - s i n - -
n = l m = l a b 
( 2 . 4 9 ) 
A s i n t h e o n e - d i m e n s i o n a l c a s e o f a v i b r a t i n g s t r i n g , w e s e e t h a t t h e g e n e r a l 
v i b r a t i o n a l m o t i o n o f a r e c t a n g u l a r d r u m c a n b e e x p r e s s e d a 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 , u n m ( x , y , t ) . S o m e o f t h e s e m o d e s a r e s h o w n i n F i g u r e 2 . 6 . N o t e t h a t i n t h i s 
t w o - d i m e n s i o n a l p r o b l e m w e o b t a i n n o d a l l i n e s . I n t w o - d i m e n s i o n a l p r o b l e m s , t h e 
n o d e s a r e l i n e s , a s c o m p a r e d w i t h p o i n t s i n o n e - d i m e n s i o n a l p r o b l e m s . F i g u r e 2 . 6 
s h o w s t h e n o r m a l m o d e s f o r a c a s e i n w h i c h a = f . b . T h e c a s e i n w h i c h a = b i s a n 
u i i 
u 2 I 
u 3 I 
F I G U R E 2 . 6 
T h e f i r s t f e w n o r m a l m o d e s o f a r e c t a n g u l a r m e m b r a n e w i t h s h a d e d a n d c l e a r s e c t i o n s h a v i n g 
o p p o s i t e s i n u s o i d a l d i s p l a c e m e n t s a s i n d i c a t e d . 
5 3 
54 
F I CURE 2.7 
The normal modes of a square membrane, illustrating the occurrence of degeneracy in this 
system. The normal modes u 12 and u21 have different orientations but the same frequency, given 
by Equation 2.50. The same is true for the normal modes u13 and u31 . 
interesting one. The frequencies of the normal modes are given by Equation 2.47. 
When a = b in Equation 2.4 7, we have 
VJT ( 2 2)112 w =- n +m 
nm a (2.50) 
We see from Equation 2.50 that w12 = w21 = 5
112 fa in this case; yet the normal modes 
u12 (x, y, t) and u21 (x, y, t) are not the same, as seen from Figure 2.7. This is an 
example of a degeneracy, and we say that the frequency w12 = w21 is doubly degenerate 
or two-fold degenerate. Note that the phenomenon of degeneracy arises because of the 
symmetry introduced when a= b. This phenomenon can be seen easily by comparing 
the modes u 12 and u21 in Figure 2.7. Equation 2.50 shows that there will be at least 
a twofold degeneracy when m =/=- n because m 2 + n 2 = n 2 + m 2 \u2022 We will see that the 
concept of degeneracy arises in quantum mechanics also. 
This chapter has presented a discussion of the wave equation and its solutions. In 
Chapter 3, we will use the mathematical methods developed here, and so we recommend 
doing many of the problems at the end of this chapter before going on. Several problems 
involve physical systems and serve as refreshers or introductions to classical mechanics. 
Problems 
2-1. Find the general solutions to the following differential equations. 
d2y dy 
b. d2y+6dy=0 
dy 
a. --4- +3y =0 c. - +3y = 0 dx 2 dx dx 2 dx dx 
d. 
d2y dy d2y dy 
-+2--y=O e. --3-+2y=0 
dx2 dx dx 2 dx 
P r o b l e m s 
2 - 2 . S o l v e t h e f o l l o w i n g d i f f e r e n t i a l e q u a t i o n s : 
d 2 y 
a . 4 
d x
2
- Y = 0 
d y 
y ( O ) = 2 ; - ( a t x = 0 ) = 4 
d x 
d
2
y d y 
b . - - 5 - + 6 y = 0 
d x
2 
d x 
d y 
y ( O ) = - 1 ; d x ( a t x = 0 ) = 0 
d y 
c . - - 2 y = 0 
d x 
y ( O ) = 2 
2 - 3 . P r o v e t h a t x ( t ) = c o s i l l t o s c i l l a t e s w i t h a f r e q u e n c y v = i l l / 2 r r . P r o v e t h a t 
x ( t ) = A c o s i l l t + B s i n i l l f o s c i l l a t e s w i t h t h e s a m e f r e q u e n c y , i l l j 2 n . 
2 - 4 . S o l v e t h e f o l l o w i n g d i f f e r e n t i a l e q u a t i o n s : 
d
2
x d x 
a . -
2 
+ i l l
2
x ( t ) = 0 x ( O ) = 0 ; - ( a t t = 0 ) = v
0 
d t d t 
d
2
x 
b . -
2 
+ i l l
2
X ( t ) = 0 
d t 
d x 
x ( O ) = A ; d t ( a t t = 0 ) = v
0 
P r o v e i n b o t h c a s e s t h a t x ( t ) o s c i l l a t e s w i t h f r e q u e n c y i l l j 2 n . 
2 - 5 . T h e g e n e r a l s o l u t i o n t o t h e d i f f e r e n t i a l e q u a t i o n 
d
2
x 
-
2 
+ i l l
2
X ( t ) = 0 
d t 
i s 
x ( t ) = c
1 
c o s i l l t + c
2 
s i n i l l t 
F o r c o n v e n i e n c e , w e o f t e n w r i t e t h i s s o l u t i o n i n t h e e q u i v a l e n t f o r m s 
x ( t ) = A s i n ( M + ¢ ) 
o r 
x ( t ) = B C O S ( i l l t + 1 / J ) 
S h o w t h a t a l l t h r e e o f t h e s e e x p r e s s i o n s f o r x ( t ) a r e e q u i v a l e n t . D e r i v e e q u a t i o n s f o r A 
a n d ¢ i n t e r m s o f c 
1 
a n d c
2
, a n d f o r B a n d 1 / 1 i n t e r m s o f c 
1 
a n d c
2
\u2022 S h o w t h a t a l l t h r e e f o r m s 
o f x ( t ) o s c i l l a t e w i t h f r e q u e n c y i l l / 2 r r . H i n t : U s e t h e t r i g o n o m e t r i c i d e n t i t i e s 
s i n ( a + f J ) = s i n a c o s f J + c o s a s i n f J 
a n d 
c o s ( a + { J ) = c o s a c o s f J - s i n a s i n f J 
2 - 6 . I n a l l t h e d i f f e r e n t i a l e q u a t i o n s w e h a v e d i s c u s s e d s o f a r , t h e v a l u e s o f t h e e x p o n e n t s a t h a t 
w e h a v e f o u n d h a v e b e e n e i t h e r r e a l o r p u r e l y i m a g i n a r y . L e t u s c o n s i d e r a c a s e i n w h i c h 
a t u r n s o u t t o b e c o m p l e x . C o n s i d e r t h e e q u a t i o n 
d
2
y d y 
- + 2 - + l O y = O 
d x
2 
d x 
5 5 
56 Chapter 2 I The Classical Wave Equation 
If we substitute y(x) = e&quot;x into this equation, we find that a 2 + 2a + 10 = 0 or that 
a = -1 ± 3i. The general solution is 
y(x) = c!e(-1+3i)x + c2e(-l-3i)x 
= c!e-xe3ix + c2e-xe-3ix 
Show that y(x) can be written in the equivalent form 
Thus we see that complex values of the a's lead to trigonometric solutions modulated by 
an exponential factor. Solve the following equations. 
dy 
y(O) = 1; dx (at x = 0) = -3 
2-7. This problem develops the idea of a classical harmonic oscillator. Consider a mass m 
attached to a spring as shown in Figure 2.8. Suppose there is no gravitational force acting 
on m so that the only force is from the spring. Let the relaxed or undistorted length of the 
spring be xo- Hooke's law says that the force acting on the mass m is f = -k(x- x0), 
where k is a constant characteristic of the spring and is called the force constant of the 
spring. Note that the minus sign indicates the direction of the force: to the left if x > x0 
(extended) and to the right if x < x0 (compressed). The momentum of the mass is 
dx d(x- x0) p=m-=m 
dt dt 
Newton's second law says that the rate of change of momentum is equal to a force 
Replacing f(x) by Hooke's law, show that 
d2(x- x0) m = -k(x- x) dt2 0 
m 
X FIGURE 2.8 
A body of mass m connected to a wall by a spring. 
P r o b l e m s 
U p o n l e t t i n g ~ = x - x
0 
b e t h e d i s p l a c e m e n t o f t h e s p r i n g f r o m i t s