Lecture 12
0.1
Shock response
One of the typical vibration engineering design tasks involves protecting
a device from a shock impulse, such as when one drops a cell phone
on the ﬂoor. A more extreme example is that of a torpedo hitting a
ship whose ﬁberoptics for communications should still survive  video
of shock vibration testing for that case was shown in class.
The basic physics that plays in shock response is that the maximum
force that the device body will experience is proportional to its max
imum decceleration. Consider the standard springmass system with
mass
M
and stiﬀness
k
. Assume we drop an object of mass
m
o
from
height
h
. We set the
y
coordinate to be 0 at the place we dropped
the object from and
h
at the position where it hits the mass
M
of the
massspring system. By integrating
m
o
¨
y
=
m
o
g
we get
˙
y
(
t
) =
gt
and
y
(
t
) =
gt
2
/
2
.
Thus, time of impact is
t
imp
=
p
(2
h/g
)
.
Plugging this into the velocity equation, we get the velocity at impact
is going to be
v
imp
=
v
0
=
p
2
hg.
Now we consider the continuing motion of the mass
m
0
and the testing
equipment of mass
M
. Thus the joint mass is
m
=
m
0
+
M
. The
diﬀerence in the total linear momentum of the system after the collision
can be attributed to the impulse of all the forces acting on it during
the collision. This leads to
m
o
˙
x
(0 +
δt
) +
M
˙
x
(0 +
δt
)

m
o
v
imp
=
Z
δt
0
F
s
dt,
1
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h
y
0
k
m
x
Figure 1: Testing the forces on an object falling from height
h
.
where
F
s
is the force in the spring. Now, since the impulse lasts for a
very short time, we can take
δt
to zero and obtain
m
˙
x
(0
+
) =
m
o
v
imp
,
where, as before 0
+
denotes the time right after the impact. Thus, the
initial velocity of the massspring system is
v
0
=
m
o
m
v
imp
.
We take the initial position to be
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 Winter '08
 Mezic,I
 Laplace

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