University of Alabama in Huntsville
Department of Physics
Spring 2010
PH 113: General Physics w/ Calculus III
General Information
Time:
Venue:
Instructor:
Phone:
Email:
Office:
Office hours:
Lecture
MWF 3:00 3:55 PM
MSC 113
Lior Burko
(256) 824-2934
burko
a period, except that the period is shifted by an amount of kx, and therefore the integral is
not changed because of it.
Therefore,
dk
dt
We wont show this here, but
dP
dt
1
2
v 2 ym cos2 (kx t)
2
1
2
=
v 2 ym
4
=
=
dK
dt
dE
dt
so that
1
2
= v 2 ym
2
Exer
Fy = sin sin
(tan tan )
y
y
=
x+dx
x
x
2
y
dx
x2
x
in the innitesimal limit.
Let := M/L be linear mass density, The y component of the force on the string element
dx also equals
Fy = may
2y
= ( dm) 2
t
= dx 1 +
=
dx + O(2 )
( dx)
dy
dx
2
2y
t2
2
y
We now let x x vt to have a wave traveling to the right. I.e.,
f (x, t) = ym sin[k (x vt)] = ym sin(kx kvt) = ym sin(kx t)
where := kv , v =
harmonic motion.
2
.
Let t t +
f (x, t +
Then,
2
Recall that f =
k
=
2
k
and is the angular frequency. At any x, t
PH113 Lecture Notes
Lior M. Burko, University of Alabama in Huntsville
c Lior M. Burko for all original materials
(sources for nonoriginal materials available upon request)
1
Waves
Physics includes only local interactions. The modern viewpoint excludes ac
UAHuntsville
PH 113
Spring, 2010
Homework Problem Set No. 9
Submission deadline: 5/3/2010
Problem 52:
Problem 53:
Problem 54:
Problem 55:
Problem 56:
Problem 57:
Show that the relativistic momentum p and the Lorentz factor are
not independent. Specificall
UAHuntsville
PH 113
Spring 2010
Homework Problem Set No. 7
Submission deadline: 4/7/2010
Problem 41:
Problem 42:
Problem 43:
Problem 44:
Problem 45:
Problem 46:
UAHuntsville
PH 113
Spring 2010
Homework Problem Set No. 6
Submission deadline: 3/31/2009
Problem 35:
Problem 36:
Problem 37:
Plot quantitatively the intensity coming out of a diffraction grating for two cases: (a) the
diffraction envelope is hardly chang
UAHuntsville
PH 113
Spring 2010
Homework Problem Set No. 5
Submission deadline: 3/22/2009
Problem 28:
Solve the equation = tan for the first 5 non-trivial roots, to at least 4 significant
figues.
Problem 29:
1. How many bright fringes appear between the f
UAHuntsville
PH 113
Spring 2010
Homework Problem Set No. 4
Submission deadline: 3/3/2010
Problem 21:
Problem 22:
Problem 23:
Imagine that some electrons are moving at a speed of 0.001c in a magnetic field that
keeps them in circular orbit 20cm in radius.
UAHuntsville
PH 113
Spring 2010
Homework Problem Set No. 3
Submission deadline: 2/10/2010
Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
UAHuntsville
PH 113
Spring 2010
Homework Problem Set No. 2
Submission deadline: 2/1/2010
Problem 7
In lecture we derived the power transmitted by the kinetic energy of a taut string for the
transverse case.
(a) Find the power transmitted by the (elastic)
UAHuntsville
PH 113
Spring 2010
Homework Problem Set No. 1
Submission deadline: Recitation of the week of 1/18/10
Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Find the average of the function
Over a full period of the wave.
Problem 6
A step wave func
University of Alabama in Huntsville
Department of Physics
Spring 2010
PH 113: General Physics with Calculus III
Homework Assignments
Homework submission deadline is changed to lecture on posted day
Homework #
HW1
HW2
HW3
HW4
Due date during
recitation of
UAHuntsville Spring, 2010 PH 113: Homework Solutions No. 9 Solution 52: (a) We use:
Vstop
1240 eV nm hc 1.8 eV hf 400 nm = = = = 1.3 V e e e
(b) The speed v of the electron satisfies
K max
1 21 v2 2 = mv = mc 2 = E photon 2 2 c
We find
v=
2 (Ephoton )
me
UAHuntsville
Spring, 2010
PH 113: Homework Solutions No. 9
Solution 52:
(a) We use:
Vstop
1240 eV nm
hc
- 1.8 eV
-F
hf - F
400 nm
l
=
=
=
= 1.3 V
e
e
e
(b) The speed v of the electron satisfies
K max
1 21
v2
2
= mv = mc 2 = E photon - F
2
2
c
We find
v=
2
proportional to the (second) spatial derivative of the wave function (recall the wave equation!), we may write the condition on the acceleration as / x|x=0 = / x|x=0+ . (The
second spatial derivative being continuous guarantees that the rst derivative is
yIII = A3 eik1 x + A3 eik1 x eit
Notice that A3 = 0 because there are no reected waves from innity. The boundary conditions are the continuity of the wave and its xderivative at the two points of discontinuity.
Specically,
A1 + A1 = A2 + A2
k1 (A1 A1 ) =
x = 0. Therefore the slope at x = 0 is sin(kx) = 0 without restriction. At x = L, the
slope is sin(kL) = 0, so that kL = n f = 2n v as in the two xed endpoints case.
L
These considerations are important, say, for wind instruments, such as organ pipes, ute
shortest travel time. We want to arrange the curved surface such that each ray starting from O, at any point P , will be bent such that it gets to O . We need to shape the surface so that the light travel time along OP + nP O is a constant independent of
Optics
In the study of optics, the study of light and its interactions, we are going to see three main
dierent approximation regimes, according to the following table:
Table 1: Dierent regions of approximation in optics. Here, is the light wavelength, D i
On the screen, these two spectral lines, dierent in wavelength by 1 nm, are 0.258 mm apart.
The plate factor is
1
1
nm
=
,
mm = 3.87
fD
0.258 nm
mm
which means that one mm of distance on the screen spans a range of nearly 4nm.
The Poisson Bright Spot
At t
N
2
j =1
Fig. 8. The case N = 7 and b
N
2
Fj = 2b
j =1
P =
ei(2j 1) =
sin(N )
2 sin
.
sin sin(N )
sin sin(N )
=b
2 sin
sin
EL i(kr0 t) sin sin(N )
e
b
r0
sin
and the intensity is
I
P P
=
EL
b
r0
2
sin
2
diraction o wide slit
sin(N )
sin
2
interfere
Diraction grating:
There are dierent kinds of diraction grating, with many applications in spectroscopy. We
consider here one type of grating, which is a set of N
1 slits or rulings, each of width b,
at distance a from one another, so that the total width
so that they get to a single point on the focal plane, and images from dierent stars get to
dierent such points.) Because of the diraction eect, we are here interested in the question
of whether these images can be resolved. In practice, if we observe a b
these two curves correspond to the x values of the solution to the transcendental equation.
Another useful method to solve this equation is numerically, using, say, the NewtonRaphson
method. The reason why we obtain this diraction pattern is that as we go
we moved the mirror, the nth interference fringe after the mirror is moved is seen where the
2nth interference fringe was before we moved the mirror. Similarly, also the other fringes are
shifted in position. Dark fringes are found when = (m + 1 ) , or =
when cos2 L (n2 n1 ) = 0, or L (n2 n1 ) = (2m +1) , i.e., L(n2 n1 ) = (2m +1) . There
2
2
is no interference when I = 2A2 (i.e., the sum of the intensities of the two individual waves),
i.e., when cos2 L (n2 n1 ) = 1 , or L (n2 n2 ) = + m , i.e., L(n2 n1
Argument from polarization that EM waves are transverse: We revisit now the transversality of electromagnetic waves. Assume there were a longitudinal component to the E eld.
Consider now the light falling on two crossed polaroids. This longitudinal compon