Ph12a Solution Set 4
November 1, 2013
3.28 We recall Eqns. (53), (54), (55), (16), and (17). In steady state, the
absorptive and elastic amplitudes are:
F0
2
M 2 2 + (0 2 )2
2
F0
0 2
Ael =
2
M 2 2 + (0 2 )2
(1)
Aab =
(2)
Setting = 0 as the problem asks, w
Ph12a Solution Set 7
November 22, 2013
6.14 The dispersion relation for a system of coupled pendula is given by eq. 90
of section 2.4:
K
ka
g
+ 4 sin2
l
M
2
2 (k ) =
g
l
The lower cuto frequency is
cuto frequency is given by
p.86 of Crawford).
g
l
corres
Ph12a Solution Set 8
December 6, 2013
9.4 The geometry of this problem is illustrated in Fig. 1. The light entering
the two slits will be reasonably coherent when the path dierence to the
two slits from an edge of the source is small compared to the wavel
1
Physics 12a Fall Term, 2013
Cover Page
Final Examination
This exam covers Chapters 16 and Chapter 9.19.6 in Crawford, as well as the
topics discussed in class, in the handouts, and in the homework.
Due Date: Friday, 13 December 2013, at 5:00 pm SHARP to
1
Physics 12a Fall Term, 2013
Cover Page
Midterm Examination
This exam covers Chapters 13 in Crawford, as well as the topics discussed in
class and in the handouts through 29 October, as well as the homework through
Chapter 3.
Due Date: Wednesday, 6 Novem
Ph12a Solution Set 6
November 19, 2013
5.5 a) The reection coecient is given in Eq. 37 on page 243:
RV =
Z2 Z1
.
Z2 + Z1
So RV = 1 . This gives the amplitude of the reected wave:
3
Aref = Ainc RV = 3.3 V
and the transmitted wave:
Atrans = Ainc (1 + RV ) =
Ph12a Solution Set 5
November 7, 2013
4.8 For a single cylindrical wire of radius r, making a Gaussian pillbox of
cylindrical shape with radius R, r < R < r + D, parallel to the axis of the
wire, and applying Gausss law:
E da = |E |2Ra =
Q
,
o
where E is
Ph12a Solution Set 1
October 17, 2013
1.15 The equations of motion for two coupled oscillators can be put in the form
d2 a
= a11 a a12 b
(1)
dt2
d2 b
= a21 a a22 b
(2)
dt2
Suppose we have two solutions to these equations, one for initial positions
xa1 and
Ph12a Solution Set 2
October 18, 2013
2.12 In a given mode with wavenumber k , the displacement is given as
= cos(t + )(A sin kz + B cos kz ),
where = cs k and cs = T0 /0 . Since the ends of the string are free,
it must have horizontal tangents at both e
Ph12a Solution Set 3
October 25, 2013
2.35 The general motion of the string can be expressed as a linear combination
of normal modes, each with its own phase:
(x, t) =
An sin (n t + n ) sin kn x
n
where kn = n = nk1 , n = kn v , and v = T . We use the co
Solutions for 2013 Midterm Exam
Phys. 12a
H. J. Kimble
(Dated: November 6, 2013)
PROBLEM 1: SIMPLE OSCILLATORS
Write an equation of motion and give the frequency for free oscillation of each of the
systems shown in Figure 1 below.
k1
a. 5 points
m
k2
b.