6.3 Solutions
261
The probability density
d p(x)
dx
=
C
ω
√
A 2 − x 2
(3)
C can be found by normalization of distribution
A
−A
d p(x) =
C
ω
A
−A
dx
√
A 2 − x 2
= 1
or
Cπ
ω
= 1 →
C
ω
=
1
π
∴
d p(x)
dx
=
1
π
√
A 2 − x 2
6.20 U =
1
2
kx 2
Using the result of prob. (6.19)
U =
U d p(x) =
A
−A
1
2
kx
2
dx
π
√
A 2 − x 2
Put x = A sin θ, dx = A cos θ dθ
U =
k A 2
2π
π/2
−π/2
sin
2
θ dθ =
1
4
k A
2
Also, K = E − U =
1
2
k A
2
−
1
4
k A
2
=
1
4
k A
2
6.21 K trans + K rot + U = constant
1
2
mv
2
+
1
2
I ω
2
+
1
2
kx
2
= constant
But I =
1
2
m R
2 and ω =
v
R
∴
3
4
m
dx
dt
2
+
1
2
kx
2
= 0 constant
Differentiating
3
2
m
d 2 x
dt 2
dx
dt
+ kx
dx
dt
= 0