Response to Periodic

Force
 Response Under a General Periodic
Force
• The equation of motion can be expressed as
a0 
mx  cx  kx  F (t )    a j cos jt   b j sin jt

(4.8)
2 j 1 j 1

• The steady-state solution of the equation is derived as:

a0  (a j / k )
x p (t )   cos( jt   j )
2k j 1 (1  j 2 r 2 ) 2  (2jr ) 2
(b j / k )
 sin( jt   j )


j 1 (1  j r )  (2jr )
2 2 2 2

2 © 2011 Mechanical Vibrations Fifth Edition in SI Units

 Response Under a Periodic Force

 Using Fourier coefficients:

2 N
a0   Fi ( 4 .9 )
N i 1
2 N 2 jti
a j   Fi cos , j  1,2,... (4.10)
N i 1 
2 N 2 jti
b j   Fi sin , j  1,2,... (4.11)
N i 1 

3 © 2011 Mechanical Vibrations Fifth Edition in SI Units

 Response Under a Periodic Force

 An irregular forcing function:

 Once the Fourier coefficients a0, aj, and bj are known, the steady-state response of the
system can be found using Eq.(4.13) with

 2 
r   

 n
4 © 2011 Mechanical Vibrations Fifth Edition in SI Units
 Response Under a Periodic Force

Example
Steady-State Vibration of a Hydraulic Valve

Find the steady-state response of the valve in the figure below if the pressure
fluctuations in the chamber are found to be periodic. The valves of pressure
measured at 0.01 second intervals in one cycle are given below.

5 © 2011 Mechanical Vibrations Fifth Edition in SI Units

 Response Under a Periodic Force

Example
Steady-State Vibration of a Hydraulic Valve
Solution
Since the pressure fluctuations on the valve are periodic, the Fourier analysis of
the given data of pressures in a cycle gives:
p(t )  34083.3  26996.0 cos 52.36t  8307.7 sin 52.36t
1416.7 cos104.72t  3608.3 sin 104.72t
5833.3 cos157.08t  2333.3 sin 157.08t  ... N/m 2 (E.1)
Other quantities needed for the computation are

2 2 
   52.36 rad/s , n  100 rad/s , r  0.5236
6  0.12
© 2011 Mechanical Vibrations Fifth Edition in SI Units n
 Response Under a Periodic Force

Example
Steady-State Vibration of a Hydraulic Valve
Solution
  0.2
We have also
A  0.000625 m 2
 2r  1  2  0.2  0.5236 
1  tan 1    tan    16.1
1 r   1  0.5236 
2 2

 4r  1  4  0.2  0.5236 

2  tan 1    tan  2 
 77.01
 1  4r   1  4  0.5236 
2

 6r  1  6  0.2  0.5236 

3  tan 1    tan  2 
 23.18
 1  9r   1  9  0.5236 
2

7 © 2011 Mechanical Vibrations Fifth Edition in SI Units

 Response Under a Periodic Force

Example
Steady-State Vibration of a Hydraulic Valve
Solution
The steady-state response of the valve can be expressed as

34083.3 A (26996.0 A / k ) (8309.7 A / k )

x p (t )   cos(52.36t  1 )  sin(52.36t  1 )
k (1  r )  (2r )
2 2 2
(1  r )  (2r )
2 2 2

(1416.7 A / k ) (3608.3 A / k )
 cos(104.72t  2 )  sin(104.72t  2 )
(1  4r 2 ) 2  (4r ) 2 (1  4r 2 ) 2  (4r ) 2
(5833.3 A / k ) (2333.3 A / k )
 cos(157.08t  3 )  sin(157.08t  3 )
(1  9r 2 ) 2  (6r ) 2 (1  9r 2 ) 2  (6r ) 2

8 © 2011 Mechanical Vibrations Fifth Edition in SI Units