Page 71 - Singiresu S. Rao-Mechanical Vibrations in SI Units, Global Edition-Pearson (2017)

68 Chapter 1 Fundamentals oF Vibration
3. Although the two springs appear to be connected to the rigid bar in parallel, the formula of
parallel springs (Eq. 1.12) cannot be used because the displacements of the two springs are not
the same.
■
1.7.5 In some applications, a restoring force or moment due to gravity is developed when a
spring Constant mass undergoes a displacement. In such cases, an equivalent spring constant can be associ-
associated with ated with the restoring force or moment of gravity. The following example illustrates the
procedure.
the restoring
Force due to
Gravity

spring Constant associated with restoring Force due to Gravity
example 1.10
Figure 1.34 shows a simple pendulum of length l with a bob of mass m. Considering an angular dis-
placement u of the pendulum, determine the equivalent spring constant associated with the restoring
force (or moment).

Solution: When the pendulum undergoes an angular displacement u, the mass m moves by a dis-
tance l sin u along the horizontal (x) direction. The restoring moment or torque (T) created by the
weight of the mass (mg) about the pivot point O is given by
T = mg1l sin u2 (E.1)
For small angular displacements u, sin u can be approximated as sin u u (see Appendix A) and
Eq. (E.1) becomes
T = mglu (E.2)

x
O

u

l

m

y mg
l sin u

FiGure 1.34 Simple pendulum.