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I have a pendulum.
A pendulum, basically it is made of a very light string attached to
an object at the end. This end is fixed to the ceiling, that end is attached to a very
small object of mass m. The length of the pendulum is l. Now, you pull the pendulum
to one side, let go of it, and it starts to swing. It goes back down and goes back up,
goes back down. The time it takes for one complete back and forth swing is called the
period, p. Ok it is measured in seconds, of course. Forward, backward, the time is p.
Forward backward, another time, p. Ok, now, we want to know how p is determined by those
other values such as m and l. So we want to determine the formula for the period, p, of
the pendulum.
This calculation can be done in a much later chapter: chapter 15, when
we study physics 1B, one semester after this when we study what's called simple harmonic
oscillation, we can actually calculate this problem. The beauty of our analysis is that
we don't know anything, pretty much. We're only in chapter 1, yet we still want to know
the formula for p. If we have a way of doing that, it would be wonderful because that means
we have a way of bypassing all the detailed physics, so we can get to the core of the
problem. How does one do that? Well, dimensional analysis! That's the magic bullet. Ok, so
let's look at the steps of dimensional analysis.
The first step, look at that. Identify which
quantities affect your result. Our result is p of course, that's what we're looking
for. We have the first known, it's a function of what quantities? You say l, the length
of the string certainly matters. and what about the mass of the bob, ok, let's put it
in here. What else? Well you know, the reason why you pull the pendulum to one side; it will swing because gravity pulls it down. Without gravity,
it's not going to be able to go down,
it will just stay there. Gravity pulls it down, so that's why it swings downward
Ok, so what determines the strength of gravity? Well, you will learn in the next chapter that's called the acceleration of gravity, g.
So these are the three quantities that determine the period p. Well to be precise, there is
another quantity which also comes into play, and that is this angle theta. You pull it
to one side, now do you pull it by ten degrees, forty degrees, fifty degrees, whatever, in
principle that makes a difference as well. But the problem is, if you think that p depends
on theta as well, then that is off, we cannot use dimensional analysis anymore. What's the
problem? The problem is what is the dimension of theta? Theta is an angle. What is the definition
of an angle? You cut a circle, right, theta is defined as the arc length of the radius,
that's what theta is. By definition, theta has no dimension, alright, therefore it is
just a number, invisible in dimensional analysis. I cannot tag it with a dimension, therefore
if this thing depends on theta we cannot do it anymore. Well fortunately, it turns out
from an experiment that as long as theta is small, let's say five degrees or less, then
the period p is not very sensitive about theta. In other words, you can change theta from
one degree to two degree to three degrees, and p is essentially the same. So, as a good
approximation, say theta does not matter as long as the angle is not too large. So, our
first step: These are the three relevant quantities. What was our second step? Check to see if
these quantities can be distinguished with their dimensions. In other words, they must
have different dimensions. Ok let's see now, what's the dimension of p? That's t right.
Dimension of l, length of course. Dimension of m, that's capital M. Dimension of g? g
is the acceleration, it is measured in the SI unit meters per second squared as we will
learn in the next chapter, so it's length over time squared. In other words, it's L
to the first power times T to the power of negative two. We'll learn that in the next
chapter. All we need to know now is that is the dimension. We don't have to know details
about g. Ok, now, look at what we have now. Indeed, they're totally different. Therefore,
that satisfies second condition. I can use their dimensions to tell them apart. They
are indeed distinguishable by dimensions alone. Alright, we're now ready to go to the third
step. You want to right down A as a general function of x, y, and z like this, so let's
do that. So we say, our function in question is p. I'm going to say p is a function of
l, m, and g. There is a constant C (I have no idea what that constant is, but I'm not
going to worry about it), l to a certain power, m to a certain power, and g to a certain power.
Ok, I'm going to raise l to the power alpha, I'm going to raise m to the power beta, and
g raised to the power gamma. This is the most general way of combining l, m, and g together
to get p. Alright, what do I do next? Let me equate the dimensions of both sides.
When finding the dimension of p, we can forget about C (it drops out because there is no
dimension), so it is l to the power alpha, m to the power of beta, and g to the power
of gamma. Now, let me write the dimensions of p, l, m, and g in terms of the basic dimensions
T, L, and M. These are the only three independent dimensions. Ok so for the dimension [p] is
of course that of time, T.
What is that is equal to? The dimension of [l] is just capital
L raised to the power alpha. Next, dimension of the mass m; that's just capital M raised
to the power of beta. Next, dimension of g; now that is L, T to the power negative two,
but don't forget you raised the whole thing to the power of gamma. So this is L to the
power of gamma and T to the power of negative two gamma. So this tells us T to the first
power must be equal to L to the power alpha times M to the power beta times L to the power
gamma times T to the power negative two gamma. Let me do some algebra on the right side.
What I'm going to do is combine all the powers of L together, see there are two powers of
L, alpha and gamma. See if I did so I would be getting this: T to the first power equals
L to the power alpha plus beta. I'm sorry, alpha plus gamma...let me rewrite it
T to the first power equals L to the power alpha plus gamma and then M to the power beta and then T to the power negative
two gamma. Now remember, L, M, and T are totally independent. So to make this equality valid,
both sides should have the same power for L, M, and T.
Now look at L first. On the right
hand side I have alpha plus gamma. What about the left hand side? Well there is no L, which
means L to the power of zero. So zero must equal to alpha plus gamma, that's for L. The
powers of L must be equal on both sides, because that's how we equate the dimensions for L.
What about M? Well again, the left side has no M in it so it's zero. The right side has
a power beta, and that is for M. And what about T?
Well there is a T to the first power,
so that's 1. The other side is negative two gamma. So, I've arrived at three equations:
one for L, one for M, and one for T, which allows me to solve for these three unknowns:
alpha, beta and gamma! As a matter of fact, this is very easy to solve. Beta is already
equal to zero, that's given. Gamma equals negative one half, but alpha plus gamma equals
to zero which means alpha must be positive one half. Therefore the solution is alpha
equals positive one half, beta equals zero, and gamma equals negative one half. In other
words p, the period of the simple pendulum, the time it takes to swing back and forth,
must equal to a constant C (I'll use a different color to indicate that I can't find it) times
l to the power of positive one half divided by g to the power of one half.
In other words, the period of a simple pendulum is proportional to the square root of its length and inversely
proportional to the square root of gravity. Think about it. What did we really know about
the system? Very, very little. All we knew was that these were the quantities, l, m and
g, that can affect the result, and then we just know the dimensions of each quantity
and match the dimensions of both sides and get this result! We have very little knowledge
of the system, and yet we almost got one hundred percent of the answer. The only thing we don't
know is C. That's like five percent of the answer. But of course in order to get that
five percent, we have to do the other ninety-five percent of the work, and to find that C we
have to finish everything we learn in physics 1A and we study physics 1B chapter 15 and
that is when we find C. It turns out C to be equal to two pi, and that has nothing to
do with dimensional analysis. We cannot just get it out like this, it would be too easy.
This will be done in chapter 15. Ok, but aside from that constant we know everything about
the period of a simple pendulum. For example, we know in order to double the time it takes
to swing back and forth we cannot just double the length of the pendulum; we have to make
it four times as long, because the square root of four equals two. Now let's do a little
discussion about this answer. Once we know we found an answer to a problem, don't just
throw it away and go to the next problem. Rather, I want you to take a look back at
the answer; see if it makes sense to you. Now this particular answer says that the longer
the string, the longer the period p which means the slower it takes for it to go back
and forth. That agrees with common sense, the longer the string the slower the thing
goes back and forth. You can easily do this experiment and use a very short string and
a longer one and see for yourself. Now I don't know if you have been to the rotunda of the
Griffith Park Observatory. You go in the hallway and there is a huge pendulum, like three or
four stories tall. It's called a Foucault pendulum and was designed to verify that Earth
is actually rotating. This pendulum, with such a long arm, takes more than a minute
to go back and forth. So that's the case of an extremely large pendulum with a very large
period. So that makes sense, that p increases with l. Now what about g? g is the acceleration
of gravity; it tells you how strong gravity is. The greater the g value, the faster an
object falls under the influence of gravity. This is telling us that the greater the g
value the shorter the period. Doesn't that make sense? Because with a large, strong acceleration
of gravity the moment the bob goes to the side, gravity being so strong will pull it
down immediately. Doesn't that make the process of swinging faster? Doesn't that shorten the
period p? So you see it makes sense that a greater g results in a smaller period. Alright,
one last thing here: What happened to the mass? What happened to the mass M here. Well
according to our calculations, beta is zero so mass is gone.This side has no mass in it.
Well you cannot quite argue with our algebra because the algebra is correct so beta is
equal to zero. Just because the algebra is correct doesn't mean we necessarily understood
it. So does it make sense that the mass of the bob does not matter? It has nothing to
do with the period. Ok, look at this picture again. This is telling me that I can use a
very heavy one, light one, and yet the period of swing is always the same. Does that make
sense? Even though this is our calculation result, let's see if we can make some sense
out of it. As a matter of fact, this is a special example of something very general.
That is, anything that is powered by gravity, satisfies this special feature that the motion
of that object is independent of the mass. That is why, for example, everything falling
under the influence of gravity has exactly the same acceleration, regardless of the mass.
We'll get to the detail in the next chapter. But, there IS an easy way of comprehending
this. Let's say I have a pendulum (the mass is m) so it swings like this right. Say it
takes two seconds to go back and forth, one second and one second; the period is two seconds.
Now imagine this: I have a second, identical pendulum, that has exactly the same length,
exactly the same mass at the end. These are two exactly the same pendulum. If I pull both
of them to the side by the same angle and let go of it, what's going to happen? Well
they're going to swing, and swing exactly in sync, agree?
Right, they go back and forth
like this exactly in sync. They're movement would be identical. Well if their movement
is identical, let me ask you would it make a difference if I somehow glued the two masses
together like this? No! Because in actuality, if you are moving at 65 miles per hour on
the freeway and in the next lane another car is right next to you and it's doing exactly
the same thing, it's going forward at exactly the same speed 65 miles per hour, then you
can actually stick your arm out and shake hands with the other driver and you're not
going to break your arm right? Because you move in sync right? The same thing happens
with these two bobs, ok they move exactly in sync therefore if I tie them together,
nothing is going to happen. But think about it, if you tie them together isn't it true
that you have just created a pendulum with twice as much mass? See, that thing has twice
as much mass in it, 2m. And yet the motion is the same as that. So what does that tell
you? It tells you that when you double the mass of the pendulum by tying these two things
together you get a pendulum with exactly the same period. Ok, doubling the mass has no
effect on the period of the pendulum whatsoever. Isn't it true then that p is not dependent
on the mass. So it makes sense. Alright, there are many other examples of dimensional analysis;
we are going to use that a lot in the entire study of physics. And this particular problem
is simple enough that without dimensional analysis we can still solve it, except you
have to wait until chapter 15 to do it. And it turns out that in physics research there
are many situations where in the beginning the problem is so complicated that we really
don't have a clue where to start. Often times when that happens, a good way to get things
started is by doing a dimensional analysis so that you know roughly how things depend
on each other. So it is not just a way of solving simple problems, in fact sometimes
it provides an important guideline of solving more complicated problems at least to get
things started. So it is a very important technique which we are going to use throughout
the entire study of physics.