1
Physics 2210. Homework assignment 7.
T&M 3.1.
(a)
ν
0
=
ω
2
π
=
1
2
π
k
m
=
1
2
π
10
0.1
=
1.59
Hz.
And
τ
0
=1/
ν
0
=0.628 sec.
(b) E
total
=1/2kA
2
=0.5x10x0.03
2
=0.0045 J.
(c) Maximum speed occurs when all the energy turns into kinetic energy.
E
total
=1/2mv
2
max
or
v
max
=
2
E
total
m
=
0.009
0.1
=
0.3
m/s.
T&M 3.7
At the equilibrium, the buoyancy force is balanced by the
weight of the body. Therefore, the weight of the body is the
weight of the displaced water at the equilibrium: mg=
ρ
0
Vg,
where V is the volume of the displaced water at the
equilibrium and m is the mass of the body.
Let positive x direction be upward, and the origin of x at the
equilibrium (i.e., the dotted line). Assume that the body is
perturbed upward by the amount of x (>0) (i.e., the difference
between the centers of mass before and after it is perturbed.
Now the buoyancy is F
b
=
ρ
0
g(VAx) and is positive because it
points upward. The net force is
F
net
=mg+ F
b
=
ρ
0
gV+
ρ
0
g(VAx)=
ρ
0
gAx, and is negative (pointing down).
Therefore, we have:
F
net
=
m˙
˙
x
or
m˙
˙
x
+
ρ
0
gAx
=
0
.
where m is the mass of the body and is
ρ
0
V.
We can write the equation as
˙
˙
x
+
gA
V
x
=
0
.
Therefore,
ω
=
gA
V
and
τ
=
2
π
ω
=
2
π
V
gA
.
For the numbers given here,
τ
=
2
π
0.8
×
10
6
10
⋅
10
4
=
0.178
sec.
T&M 3.9.
Let x
0
be the equilibrium position of the block without the
constant force F. Here x
0
is measured from the ceiling (i.e.,
the ceiling is the origin of x coordinates, see the figure on the
right). It is clear that at the equilibrium, the spring is already
stretched by h=mg/k, compared to when the spring is free.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 '09
 ZHONG
 Energy, Force, Kinetic Energy, Work, equilibrium position, tangential force, e2 βτ, dt dt sinθ

Click to edit the document details