Physics 5300, Theoretical Mechanics Spring 2015
Assignment 8 solutions
The problems numbers below are from Classical Mechanics, John R. Taylor, University
Science Books (2005).
Problem 1
Solution:
Taylor 16.2
Each mass is m, and the spacing is b. The force on the mass is
T sin θ1 − T sin θ2 ≈ T (tan θ1 − tan θ2 ) = T [
yn+1 − yn yn − yn−1
d2 y
−
] ≈ Tb 2
b
b
dx
(1)
The motion is given by
mÿ ≈ T b
ÿ ≈
Thus taking a limit where
Tb
m
d2 y
dx2
T b d2 y
m dx2
(4)
Taylor 16.5
Solution:
d2
f (x − ct) = f 00 (x − ct)
dx2
d
f (x − ct) = f 0 (x − ct),
dx
d
f (x − ct) = −cf 0 (x − ct),
dt
Thus
Problem 3
(3)
= 1, we get the wave equation
ÿ = y 00
Problem 2
(2)
d2
f (x − ct) = (−c)2 f 00 (x − ct)
dx2
2
d2 f
2d f
=
c
dx2
dt2
Taylor 14.2
1
(5)
(6)
(7)
Solution: (a)The radius is r = 5 × 10−15 m, so the area is πr2 = 25π10−30 m2 ≈
8 × 10−29 m2 . We thus have σ = .8 barn.
For an atom, r = 10−9 m, so σ = π × 10−18 m2 . This is σ = 3 × 1010 barns.
Problem 4
Taylor 14.3
Solution: The mass of each atom is m. The total mass is M . Thus the number of
atoms is M/m. The cross section area we take to be A. The depth is L. Thus the volume
is V = AL. The density is ρ. Thus the mass is M = ρV = ρAL. Thus
N
M
ρAL
ρL
.07 × 50
= 2 × 1024
=
=
=
=
A
mA
mA
m
1.67 × 10−24
Problem 5
Solution:
(8)
Taylor 14.14
(a) We have
Thus
dσ = 2db
(9)
dΩ = 2dθ
(10)
dσ
|db
=
|
dΩ
dθ
(11)
θ
2
(12)
(b) We have
b = R cos
db
R
θ
= − sin
dθ
2
2
dσ
db
R
θ
= | | = sin
dΩ
dθ
2
2
(c)
Z
π
σ=
dθ
−π
Problem 6
θ
R
| sin | = 2R
2
2
Taylor 14.24
2
(13)
(14)
(15)
Solution:
(a) We have
cos θcm = cos(2θlab ) = 2cos2 θlab − 1
θcm = 2θlab ,
(16)
d cos θcm
= −2 cos θlab
d cos(θlab
(17)
d cos θcm
| = 42 cos θlab
d cos(θlab
(18)
Thus
|
and the result follows from eq 14.45:
(
d cos θcm dσ
dσ
)lab = |
|( )cm
dΩ
d cos(θlab dΩ
(19)
R2
dσ
)cm =
dΩ
4
(20)
(b)
(
Thus
dσ
dσ
R2
)lab = 4 cos θlab ( )cm = 4 cos θlab
= R2 | cos θlab |
dΩ
dΩ
4
ranges between 0 and π, so θlab ranges between 0 and π2 . Thus
(
Note that θcm
Z
2
Z
dΩR cos θlab = 2π
(21)
1
d cos θlab R2 | cos θlab | = πR2
0
3
(22)