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Unformatted text preview: he spring is proportional to the extension
is sometimes known as Hooke’s law and k is sometimes called the spring constant
or Hooke’s constant. It is also related to the Young’s modulus of the elastic spring.
Comment: If we were to draw the tension in the stretched spring as a function of
the extension, the graph would look something like one of these. For small l the
graph is linear with
✻ H (hard) T L (linear) S (soft) ✚ ✚ ✚
✚
✚k
slope ✲
l constant slope k , but if certain springs are stretched too far, they weaken as in S .
On the other hand some spring like structures can actually stiﬀen i.e. become more
resistant. They are shown as hard (H).
In this course, for the most part, we will assume T = kl for all l.
Comment: Most mechanical systems encounter frictional losses due to imperfect
materials, bad joints, air friction, etc. The energy lost from the spring goes into
heat or sound. We will model the frictional loss on the spring as being proportional
to the instantaneous velocity of the mass at the end,
Ffriction = −b dx
.
dt Newtons’s 2nd Law
d2 x
dt2 = sum of forces = m mg − k (x + l) − b dx
+ F ( t)
dt 2 x
where m is constant, d 2 is acceleration mg is gravity k (x + l) is tension b dx is
dt
dt
friction and F (t) is some external applied force. Since kl = mg , d2 x
b dx
k
1
+
+ x = F ( t)
dt2
m dt
m
m
3 (6.4) is the equation that describes the motion of the spring, or to be more precise, the
motion of the mass at its end.
The motion of the spring under the combined action of tension restoring forces,
friction forces and applied forces is one of the most useful paradigms in all of
physics, chemistry and mechanics. It allows us to gain a picture of behaviors of
bridges, proteins and dielectric media in very simple terms.
Let us write (6.4) as a system of two ﬁrst order ode’s by writing
dx
=u
dt (6.5a) du
b
k
1
= − u − x + F ( t)
dt
m
m
m (6.5b) Before we learn to solve (6.4), (6.5) analytically, let us use the program JOde for
6.5a,b where we take m = 1, call k by a and take F (t) = 0.
Namely rewrite (6.5) as
dx
=u
dt (6.6a) du
= −bu − ax
dt (6.6b) Solve for the two cases a = 1, b = 0 and a = 1, b = 0.2.
Recall pg. 5 in Lectures 710 and equation (4.2) for the charge q (t) in an LRC
1
circuit. Compare with (6.4). Note m plays the role of L; b of R and k of C . The
force F (t) in (6.4) is analogous to the applied voltage V (t) in (4.2). 4 The property of superposition
Linear homogeneous ode’s such as (6.2) have a very important property, the property of superposition, which says that any linear combination (A, B are constants)
y (t) = Ay1 (t) + By2 (t)
of two solutions y1 (t), y2 (t) of (6.2) is also a solution. Why? Let us check by
substituting y (t) = Ay1 (t) + By2 (t) into =
=
= d2 y
dy
+ p( t )
+ q ( t) y
2
dt
dt
d2
d
(Ay1 (t) + By2 (t)) + p(t) (Ay1 (t) + By2 (t)) + q (t)(Ay1 (t) + By2 (t))
2
dt
dt
2
...
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This note was uploaded on 01/22/2014 for the course MATH 254 taught by Professor Indik during the Fall '08 term at Arizona.
 Fall '08
 INDIK

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