Page 22 - Rotating Machinery Pratical Solutions to Unbalance and Misalignment

Introduction to Vibrations

Equations 1.9 and 1.11 are common forms for describing
simple harmonic motion. This is best illustrated by an example.

Example 1.1
A spring is stretched 1 foot with a force of 2 pounds. If a 6-
pound weight is displaced 4 feet from the position of rest, (a) what
is the maximum velocity, (b) what is the maximum acceleration,
and (c) what is the velocity and acceleration when the distance is
2 feet?

F

M

–x o o x +x o

Figure 1-11. Mass Spring System for Example 1.1

Step 1
First, the spring constant must be determined. Using Hooke’s
Law,
F = –kx or –2 lbs. = (–k)(1)ft. Therefore k = 2 lbs./ft.
2
The maximum potential energy is 1/2 k x , or
0
2
1/2k x 2 = 1/2(2 lbs./ft.)(16 ft. ) = 16 ft.-lbs.
0
Step 2
Since the highest velocity occurs when × = 0, refer to Figure
1-3, using Equation (1.9) and substituting the provided data from
the example yields:

6lbs v max = 2 lbs./ft. 16 ft 2
2
32.2ft/sec .

Note: Mass is the weight divided by the acceleration of gravity
2
(32.2 ft/sec. ). In this example, the mass is 6/32.2 or 0.1863 lbm.