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Transcript
Physics 218 LAB: INVESTIGATING SPRINGS
Name
Section Number
Team Number
Introduction
The purpose of this lab is study the well-known force exerted by a spring. The force,
as given by Hooke’s Law, is a function of the amount the spring is stretched or compressed
and therefore the constant force formulae and resulting trivial potential energy functions do
not apply. The motion of an object suspended from an ideal spring, in the absence of any
friction, is called Simple Harmonic Motion. This type of motion occurs in a wide variety
of physical systems and is itself worth studying. You can also obtain the potential energy
function for the spring force and test the Law of Conservation of Energy. Everything seems
to work out exactly as expected. However, when the force of the spring is investigated as
a function of the mass of the object suspended from it, there is a puzzling result.
Determining the Spring Constant
When a mass rn is suspended from a spring the extension of the spring at the equi
librium position can easily be measured. From the free body diagram
ya
and Newton’s Second Law, with the mass at rest,
you
have
Fspring
=
mg.
Assuming that the force of the spring is given by Hooke’s Law, which states that the
magnitude of the force is a constant times the amount the spring is stretched or compressed,
we have
Fsprjng
=
kxeq
1
=
mg.
Here Xeq is the extension of the spring at equilibrium, k is called the spring constant and
from the above we have
k=.
Xeq
Procedure
Suspend a spring so that it hangs vertically. Place two masses on the hanger, one of
50 grams, the other of 100 grams. Slowly lower the hanger until the force of the spring
cancels the force of gravity. Measure the new length of the spring and the amount by
which the spring has been stretched is the value of Xeq in the above equation. Don’t forget
to include in your calculation the mass of the hanger which is 50 grams.
You will need the spring constant later so record it here along with xeq:
k
Xeq
Conservation of Energy
Adopting the coordinate system shown,
c
IL
with x
=
0 where the spring is unstretched, the two forces acting on the mass are
Fgravity
=
mgi
and
Fsprjng
=
—kxi.
According to the work energy theorem, or Conservation of Energy, if the mass is released
from rest at the point where the spring is unstretched, called x = 0, its velocity at any x
will he given by the equation
12
mv (x)
=
mgx
12
—
kx
This follows from doing the integrals of the forces or using the potential energy func
tioris
1
Ugravity = —mgx + C
2
and
12
kx + C
2
USpring
in which case the change in the kinetic energy is set equal to the negative of the change in
the total potential energy. It is difficult to measure the velocity but it is easy to measure the
maximum extension of the spring when the mass is released from rest at x = 0. According
to the above equation
mg
Xmas = 2— = 2
Xeq.
You should derive these equations for yourself and determine experimentally the degree to
which the theoretical and experimental results for the maximum extension agree.
Xrnax
(predicted)
Xmas (measured)
=
Simple Harmonic Motion
When the mass was released from rest it oscillated up and down about the equilibrium
position. The description of this motions follows from Newton’s Law and the fact that the
force of the spring varies linearly with the extension of the spring. This gives the equation
mg
—
kx
=
5
ma
X
2
d
=
This equation will he studied later in the course. At this point you might just verify that
the solution is
X = Xeq
Xeq coswt
—
where
IT
Vm
The period of the oscillation is the time it takes to return to the highest point after being
released from that point. Because cos(6) = 1 when 6 = 2ir this gives
T=.
You should measure this time and again determine how well the theory describes the
experimental results.
T(predicted)
T(measured)
=
=
You should find that the theoretical results and the experimental results agree pretty
well. The fact that the spring is not massless is a fairly small effect and can be ignored.
Also irrelevant is the fact that the spring can not be compressed because the coils are
essentially already in contact without any force being applied to the spring.
3
The Effect of Changing the Mass
You are now to repeat the entire experiment but instead of just putting 150 grams
on the hanger, put an additional 200 grams so that the total mass, including the hanger,
is 400 grams. First repeat the procedure for determining the spring constant by slowly
lowering the masses to find Xeq. Isolating the body and Newton’s Law then gives the the
value of the spring constant. Record the results here:
k
Xeq
Repeat the experiment to determine
Xmas.
It should again be such that
Xmas = 2
Xeq.
Compare the two below:
xrnac(measured)
xmax(theoretical)
=
=
Finally measure the period of the simple harmonic motion and compare the result with
the theoretical one:
T(piedicted)
=
T(measured)
=
=
You should find again that the theoretical results and the experimental results agree pretty
well. There is one real problem. You have measured k twice and probably obtained two
quite different results! But k is supposed to be the spring constant and if it is constant
it can’t have two different values! At the same time the observed oscillatory motion of
the mass is described very well by the equation for simple harmonic motion. This means
that the force exerted by the spring is indeed a linear function of the amount the spring is
stretched. Arty idea for a way out of this apparent paradox?