Download Mass on a Spring: Oscillations

Mass on a Spring: Oscillations
Abstract
You will place varying masses on a spring and measure the extension of the spring. You will
place varying masses on the spring and measure the period of the oscillations. Finally, you will vary
the amplitude of the oscillations. The data taken will be used to validate a model for the spring
oscillations.
Introduction
Professor William “Wild Bill” Deering once observed that “all the world's a harmonic
oscillator”, meaning that a great many physical processes can be explained in terms of one simple
model: the harmonic oscillator. Over the next two labs, we'll see why that's true. To begin, we'll
consider a linear restoring force. You know what a force is (right?), so we'll discuss the other two
words. NOTE: We've named lots of forces, usually based on the objects interacting. So a “contact
force” is a force that happens when two objects touch. A “frictional force” is the component of a
contact force parallel to the surface. A “gravitational force” is an interaction between any two objects
with gravitational mass. In this case, we're naming a force by the effect it has – to “restore” the
equilibrium.
To understand a restoring force, we must first imagine a point of stable equilibrium. In physics,
equilibrium means “no net force” which also means “zero acceleration” and also “constant velocity”. A
“point of equilibrium” is a place (location in space) at which the force is zero. A book is at equilibrium
pretty much anywhere on a table. If you hang a mass from a string (this is a pendulum) is at
equilibrium when it's hanging straight down. A restoring force is one that pushes the object BACK
toward equilibrium if it's moved. So if you push a book on a table UP, it will be pushed back down by
gravity. But if you push it sideways or DOWN, gravity will NOT push it back to the original point. So
the weight of a book on a table is NOT a restoring force. BUT, if our solar systems moves ABOVE the
plane of the galaxy, the gravitational force of all the other stars will pull it back – in this case, weight
(of the solar system) IS a restoring force. Similarly, if you push a pendulum mass to the side and let go,
the weight will pull it back to equilibrium. Again, in this case, weight is a restoring force. And if you
have a point of equilibrium with a restoring force, this is a point of stable equilibrium, because if you
move the object, it gets pushed back to equilibrium. (Imagine balancing a meter stick on your finger –
if it gets pushed, it will move FARTHER from equilibrium. This is UNSTABLE equilibrium). That a
restoring force leads to oscillations is clear from the following consideration.
Restoring Force = Oscillations Near Equlibrium
Imagine that you pull an object (feeling a restoring force) away from the point of equilibrium
and let go. It is at rest, but feels a force TOWARD the point of equilibrium. The force is IN THE
DIRECTION of the velocity. So the object speeds up TOWARD the equilibrium point. When it arrives
at the equilibrium point, it will have a velocity. Since the force is zero, the velocity will stay constant.
So the object passes the equilibrium point. It then feels a force OPPOSITE the velocity. So it SLOWS
DOWN. Eventually it will stop. But it still feels a force toward the equilibrium point. So now it moves
back toward the equilibrium point, speeding up since the force is now in the direction of the velocity.
And so on. Repeated motion back and forth across the point of equilibrium. Oscillations.
Another point of view is to consider Newton's Second Law of Motion (i.e., the energy
principle), which says that the acceleration is proportional to the force. Suppose that you set up a
coordinate axis (the x-axis) and displace an object away from equilibrium (x=0) in the positive x
direction. Since x is positive and the force is a restoring force, the force will be in the negative x
direction. That is, the acceleration will be negative. In Equation form, this is:
a=
d 2 x F ( x)
(1)
=
2
m
dt
Again, if x is positive, then F(x) will be negative and vice versa. If I now PLOT the position, x, as a
function of time, t, then the acceleration represents the curvature of that graph. So whenever x is
positive, the curvature of the graph x(t) will be negative and whenever x is negative, the curvature will
be positive. Such a graph of x(t) would look something like:
That is, knowing nothing except that the force is a restoring force, we conclude that oscillations MUST
occur. (NOTE: If the force were constant, like for a solar system above the galaxy, what would the
oscillations look like – that is, what curve x(t) has a constant curvature?)
So a pendulum will oscillate (the first decent clock), the solar system will oscillate above and
below the galaxy (mass extinctions every 100 million years), but a book will not oscillate on a table.
The last point is the TIME for one oscillation, called the period of oscillation and symbolized (usually)
as T (yes, this is the same symbol we use for “tension” - sorry 'bout that).
The time of the oscillation is set by how fast an object, released from a point away from the
equilibrium point, can pass through the equilibrium point and get back – that is, it depends upon the
speed. That, in turn, depends upon the acceleration. The acceleration, according to Newton's Second
Law of Motion (AKA, the energy principle) is given by:
a=
⃗
⃗
F
m
(2)
⃗ is the net force (in this case, the restoring force) and m is the inertial mass. A larger
where F
acceleration means a larger velocity magnitude and so a shorter period. That is, if the restoring force is
large, then the acceleration will be large and so the speed will be large and the period will be short.
Large restoring force – short period. On the other hand, if the inertial mass is large, then the
acceleration will be small, the speed will be small and the period will be long. Large mass – long
period. Finally, consider the impact of the distance the system is moved from equilibrium (the
“amplitude” of the oscillations). If you move the object a larger distance, then the time to return will
be longer, IF ALL ELSE IS EQUAL. But often, the restoring force depends upon the distance moved
(for example, the restoring force could INCREASE with distance from equilibrium). In this case, it's
not so clear – the extra distance takes more time, but the bigger force means higher speed means it
takes LESS time. There's one very special case, called a “linear” restoring force.
Linear Restoring Force
In this case, the word linear refers to what happens if you plot the restoring force as a function
of the distance from equilibrium. If the graph is linear, then you know that the restoring force is
proportional to the distance from equilibrium and it's called a “linear restoring force”. In this very
special case, when you move the object FARTHER from equilibrium, the increased speed from the
increased force (which decreases the time) EXACTLY, PRECISELY balances the extra distance (which
increases the time), so that the period is the same, independent of the amplitude. This is one of the
issues we'll look at today – whether the period depends upon the amplitude.
If the force is directly proportional to the distance (from equilibrium), then you can write the
restoring force as:
⃗ = k ⃗x
F
(3)
where k is the proportionality constant, also called the “stiffness”. (You should consider why that's a
good name for it – what does it mean for “k” to be large or small?)
Finally, in this case (the LINEAR restoring force) the period of motion can be written as:
T =2 π
√
m
k
(4)
(I'll show how this can be done at the end, but meanwhile you should make sure the units work). This
shows something we saw before. Namely, that the period of oscillation near a point of equilibrium is a
competition between inertia (m), which makes the period longer, and stiffness (i.e., the strength of the
restoring force), which makes the period shorter. These are the other issues we'll look at today; whether
the restoring force of a spring is really proportional to the distance from equilibrium and whether the
period of oscillation is given by Eq. (4).
The Lab
There are two parts to the lab. First, a static part where you will suspend a spring and hang
known weights from it. By measuring the distance the known weight moves the end of the spring, you
can determine whether the restoring force satisfies Eq. (3). (How do you know the magnitude and
⃗ ∝ ⃗x ? If
direction of the restoring force exerted by the spring?) What is the best way to show that F
Eq. (2) is satisfied, you can find the value of the stiffness, k.
The second part is to set a known mass to oscillate, measure the period and determine whether
the period satisfies Eq. (4). That is, does the period increase with mass as indicated? (What is the best
way to show that?) If it does, you should be able to get another value for k to compare with the
previous one (they should be the same). Also, we've claimed the period should be independent of the
amplitude of the motion (how far it moves as it oscillates). So you should check that – if you set the
mass in motion with a larger displacement from equilibrium, do you get the same period?
Conclusions
⃗ ∝ ⃗x and if it is, what is the proportionality
The goals are all as described previously: 1) Is F
constant (i.e., the stiffness)? And 2) Does the period, T, behave as we expect (Eq. (4)) and if so, what is
the stiffness (and does it have the same value)? And remember that error is an ever present issue and
has to be addressed whenever you decide whether two values are “same” or “different”, or a graph is
linear or not. What happens if you put two springs in series or in parallel? Do they behave as expected?
APPENDIX:
To find the expression for T given in Eq. (3) requires at least two semesters of calculus. But
you can get VERY close with just units. Really. Consider that the period can depend on many things
(phase of the moon, time of year) but it's most likely to depend on only a few things. The most obvious
are the stiffness, k , (which tells how hard the spring is pulling the mass back to equilibrium), the
mass, m , (which tells how much the object resists being pushed around) and the amplitude, A ,
(since a larger distance might make the mass take longer to get back). One conceptual leap is to assume
that the period, T , depends on all these as an exponent (or power) law. (If this is false, we're
doomed from the outset – but you can feel confident that if this wasn't going to work, then I wouldn't
bother you with it here). So in short, the assumption is that the equation for the period looks like this:
T = Aa k b m c
(4)
Where a, b, and c are numbers that we don't know yet. But we can find them because we know that
whatever they are, the units HAVE to work out. And we know the units of everything on the right (e.g.,
A must have units of meter) and we know that at the end, the units on the right have to match the units
on the left (seconds). So if you put units into Eq. (4), we have:
1
a
b
c
[s ] =[m] [ N / m] [kg ]
(5)
Putting in what a Newton is and grouping like terms:
[s ]1=[m]a [ s]
2b
[kg ]b+c
(6)
To stress the unit comparisons, I can write this as:
[m]0 [s ]1 [kg ]0=[m]a [ s]
2b
[kg ]b+c
(7)
So comparing the left and right sides, we get three equations to solve for a, b, and c:
0=a
1= 2b
0=b+c
(8)
which are easy enough to solve, to get:
a=0
b= 1 / 2
c=+1/ 2
(9)
Putting this in Eq. (4), we have:
T = A0 k
1/ 2
m+1 /2 =
√
m
k
(10)
Look familiar? It's ALMOST Eq. (3). It's missing 2 π because 2 π doesn't have any units, and you
can't get unitless things with this method. But 2+ semesters of calculus, just for 2 π , doesn't seem
like a bargain. (To be fair, there's more you can do with calculus than just that).
Finally, note that this DOESN'T prove that Eq. (4) has to be true – just that IF it's true, then it has to
look like Eq. (10). For example, with a pendulum, the period DOES depend upon the amplitude, but
not in the way shown by Eq. (4).