Section 9
Mechanical Vibrations
 Repetitive motion relative to an
equilibrium position.
 Evident in virtually all machines.
 Vibrational loads can exceed static loads
by orders of magnitude.
Undamped, Free Vibration
One-degree of motion
stiffness, k
mass, m
FBD
of non-equilibrium forces
a
Fs = kx
Fi
SF = 0:
-kx - ma = 0
Motion Equation
 Equilibrium:
kx + ma = 0
2
 Rewrite as
 Define:
d x
m
dt
k
m
p
2
 kx  0
(circular frequency)
2
 Then:
d x
dt
2
 p x0
2
Motion Equation
 Second order, differential equation has the
solution
x = xm sin(pt+f)
 where:
xm 
f  tan
 v0

 p
1
2

2
  x 0

 px 0 


v
 0 
Motion Curve
V o lts
x = xm sin(pt+f)
Displacement
xm
T im e
t
 Period, t
 Frequency, f
t 
2
p
f 
p
2

1
k
2
m
Differentiating the Solution:
 Position
x = xm sin(pt+f)
 Velocity
v = xm p cos(pt+f)
 Acceleration
a = -xm p2 sin(pt+f)
Problem 9-1:
A 100 lb electronics case is supported by the spring
suspension as shown. The case is able to move along two
linear guides. The spring has a constant of 75 lb/in. The
case is displaced 0.5 in then released. Determine the
resulting frequency of vibrations. Also determine the
maximum velocity and acceleration of the case. Neglect
the effects of friction.
Problem 9-10:
The bent link shown has negligible mass and
supports a 4 kg collar at its end. Determine the
frequency of vibration if the collar is
displaced a small amount and released.
4 kg
100 mm
250 mm
300 N/m
Equivalent Springs
 Springs in parallel
kequiv = k1 + k2 + ...
k1
 Springs in series
1
k equiv

1
k1

1
k2
 ...
k1
k2
k2
Problem 9-3:
A 40 lb machine base is supported by the double spring
suspension as shown. The machine base is able to move
along two linear guides. Both springs have a constant of
15 lb/in. The machine base is displaced 1.25 in then
released. Determine the resulting frequency of
vibrations. Also determine the maximum velocity and
acceleration of the base.
Neglect the effects of friction
machine base
Undamped, Forced Vibration
One-degree of motion
F0sin wt
stiffness, k
mass, m
FBD
of non-equilibrium forces
a
Fi
Fs = kx
SF = 0:
-kx - ma = F0 sin wt
F0sin wt
Motion Equation
kx + ma = F0sin wt
 Equilibrium:
2
 Rewrite as
 Define:
 Then:
m
d x
dt
k
 p
2
 kx  F0 sin w t
(circular frequency)
m
2
d x
dt
2
 p x  F0 sin w t
2
Motion Equation
 Second order, differential equation has the
solution
x  x m sin  pt  f  
Free vibration
F0 / k
1  (w / p )
2
sin w t
Forced vibration
Motion Curve
Displacement
Forced
Full solution
Time
Free
 In time, the free vibration will dampen out.
 Forced vibration term is called the steady
state solution.
Steady State Solution
 The steady state vibration is:
x ss 
F0 / k
1  (w / p )
2
sin w t
 Differentiating the solution:
v ss 
a ss 
w F0 / k
1  (w / p )
w
2
cos w t
2
F0 / k
1  (w / p )
2
sin w t
Problem 9-13
The electric motor has a mass of 50 kg and is
supported by four springs, each having a stiffness of
100 N/m. The motor turns a 7 kg disk, which is
mounted eccentrically, 20 mm from the disks center.
Determine the speed of the motor at which resonance
occurs. Assume the motor only vibrates in the
vertical direction
Problem 9-14
Determine the amplitude of the steady-state
vibration of the motor described in the
previous problem, when it is running at
1200 rpm.
Frequency Response Plot
Displacement Excitation
 Some machines have a periodic
displacement of the support.
d0sin wt
stiffness, k
mass, m
 Simply replace F0 with kd0
Problem 9-24:
The 18 lb instrument shown is used for on site
measurements, and is carried in a truck. It is centered
uniformly on a platform, which is isolated from the
truck by four springs, each having a stiffness of 13
lb/in. Determine the frequency of the vibrations of
the truck body, which will cause resonance to occur.
The platform is only able to vibrate in the vertical
direction
Problem 9-25:
Determine the amplitude of the steady-state
vibration of the instrument described in the
previous problem, when the truck floor is
vibrating at 7 Hz with an amplitude of 2 in.
Viscous Damping
 Many cases, damping is attributed to
resistance created by a substance, such as
oil, air or water.
 This type of viscous force is proportional to
the speed of the rigid body.
Damping
Force
c
1
Relative Velocity
Damped, Forced Vibration
One-degree of motion
stiffness, k
mass, m
F0sin wt
damping, c
FBD
of non-equilibrium forces
a
Fs = cv
Fs = kx
Fi
SF = 0:
-kx - cv - ma = F0 sin wt
F0sin wt
Motion Equation
kx +cv+ ma = F0sin wt
 Equilibrium:
2
 Rewrite as
 Define:
 Then:
k
m
d x
dt
p
m
2
dt
2
 kx  F0 sin w t
dt
(circular frequency)
2
d x
c
dx
 p
dx
dt
 p x  F0 sin w t
2
Motion Equation
 Second order, differential equation has the
solution

x  xm e
(c / 2 m )t
 sin(
pd t  f ) 
Free vibration
F0 / k
1  (w / p )   2 ( c / c
2 2
)( w / p ) 
sin( w t  f ' )
2
c
Forced vibration
Damping Ratio
 Critical damping ratio
cc  2 m
k
 2 mp
m
– if c > cc system does not oscillate
 Damped natural frequency
pd 
2
 c
 c 

  p 1  
m  2m 
 cc
k




2
Motion Curve
Displacement
Forced
Full solution
Time
Free
 Again, the free vibration will dampen out.
 Forced vibration term is called the steady
state solution.
Steady State Solution
 The steady state vibration is:
x ss 
F0 / k
1  (w / p )   2 ( c / c
2 2
)( w / p ) 
c
sin( w t  f ' )
2
 Differentiating the solution:
v ss 
w F0 / k
1  (w / p )   2 ( c / c
2 2
cos w t
)( w / p ) 
c
2
 w F0 / k
2
a ss 
1  (w / p )   2 ( c / c
2 2
)( w / p ) 
c
sin w t
2
Problem 9-28:
The electric motor has a mass of 40 kg and is
supported by four springs, each having a stiffness of
200 N/m. The motor turns a 4 kg disk, which is
mounted eccentrically, 60 mm from the disks center.
Determine the speed of the motor at which resonance
occurs. The damping factor c/cc = 0.15. Assume the
motor only vibrates in the vertical direction
Problem 9-29:
Determine the amplitude of the steady-state
vibration of the motor described in the
previous problem, when it is running at
100 rpm.
Frequency Response