2.5 Free Vibration with Coulomb Damping
37
F D
k
=
120
80
or
F D =
120
80
k =
120
80
×
P
6
=
P
4
=
1
4
P (P will be the normal reaction).
Therefore, the coefficient of friction is 1/4.
The variation of the displacement is shown in Fig. 2.19.
2.6 Energy Method and Free Torsional Vibration
Using the principle of conservation of energy, natural frequencies of vibrating
systems can be conveniently determined. For a mechanical vibrating system, the
energy is partly kinetic and partly potential. The kinetic energy T E is associated with
the velocity of the mass and the potential energy U is due to the strain energy stored
in the spring. Thus,
T E + U = Total mechanical energy = constant
(2.40)
Therefore, the rate of change of energy is zero. Mathematically
d
dt
(T E + U ) = 0
(2.41)
The equation of motion of the spring–mass system of Fig. 1.4 can be derived from
energy consideration. The kinetic energy of the mass is given by
T E =
1
2
m ˙
x
2
(2.42)
The potential energy of the system is due to strain energy stored in the spring
(Fig. 2.20) and is given by
U =
x
0
kx dx =
1
2
kx
2
(2.43)
Substituting Eqs. (2.42) and (2.43) into Eq. (2.41), we get
d
dt
1
2
m ˙
x
2
+
1
2
kx
2
= 0
(2.44)
or
37
F D
k
=
120
80
or
F D =
120
80
k =
120
80
×
P
6
=
P
4
=
1
4
P (P will be the normal reaction).
Therefore, the coefficient of friction is 1/4.
The variation of the displacement is shown in Fig. 2.19.
2.6 Energy Method and Free Torsional Vibration
Using the principle of conservation of energy, natural frequencies of vibrating
systems can be conveniently determined. For a mechanical vibrating system, the
energy is partly kinetic and partly potential. The kinetic energy T E is associated with
the velocity of the mass and the potential energy U is due to the strain energy stored
in the spring. Thus,
T E + U = Total mechanical energy = constant
(2.40)
Therefore, the rate of change of energy is zero. Mathematically
d
dt
(T E + U ) = 0
(2.41)
The equation of motion of the spring–mass system of Fig. 1.4 can be derived from
energy consideration. The kinetic energy of the mass is given by
T E =
1
2
m ˙
x
2
(2.42)
The potential energy of the system is due to strain energy stored in the spring
(Fig. 2.20) and is given by
U =
x
0
kx dx =
1
2
kx
2
(2.43)
Substituting Eqs. (2.42) and (2.43) into Eq. (2.41), we get
d
dt
1
2
m ˙
x
2
+
1
2
kx
2
= 0
(2.44)
or
