- [Instructor] Let's say you've got a mass
connected to a spring and the mass is sitting
on a friction, the surface.
If the mass is sitting at a point
where the spring is just at the spring's natural length,
the mass isn't going to go anywhere
because when the spring is at its natural length,
it is content with its place in the universe.
It neither pushes nor pulls.
It has no spring energy.
This is like me most days in the summer.
So we call this point where the spring neither pushes
nor pulls, the spring's natural length.
For a mass on a horizontal spring,
this is gonna be the equilibrium position.
What we mean by equilibrium position is the point
where the net force on the mass is zero.
So for a mass on a horizontal spring,
the equilibrium position is at the point
where the spring is at its natural length
because the spring wouldn't be pushing
to the right or the left.
If you just put the mass there at that point,
it would just stay there forever at rest.
That would be boring.
So let's say we pull the mass to the right
at distance d.
If we do this, we give this spring spring potential energy.
If we release the mass from rest while the spring has
spring potential energy, the spring's gonna pull the mass
back to the left and that mass is gonna move
through the equilibrium position with the sum speed.
We can figure out what that speed is
just by using conservation of energy and it's not that hard.
The potential energy, the spring with start with
would be 1/2 k, the spring constant, times d,
the amount of spring has been stretched, squared.
There would be no kinetic energy to start
because we release the mass from rest.
As this mass flies to the left,
it would start gaining kinetic energy, the spring energy
would start turning into kinetic energy.
When the mass gets to the equilibrium position,
the d would be zero.
So at that point there would be no spring energy
and all of the spring energy would have turned
into kinetic energy and you get this simple relationship
that says all the spring energy equals
all the kinetic energy at the equilibrium position.
So if we solve for v, we can get that the v of this mass
at the equilibrium position would be the square root
of k over m times d squared.
You could call this d out
because d squared and square root.
but this is the idea.
This is the speed you would get of the mass
passing through the equilibrium position.
Let me ask you this.
What if it was a vertical spring?
And this mass is sitting here.
We come find a vertical spring with a mass hanging on it.
We're like hey, I wanna pull this down,
the distance d.
If I pull it down to distance d,
when this mass reaches the equilibrium position again,
will it also be going root k over m times d squared?
Or, would it be going at some different speed
because now it's hanging vertically?
Well it turns the spring constant's the same
and you pull it down from the point
where the mass is hanging.
This exact same procedure is gonna hold over here
and you can find the speed in the exact same way
and that should be surprising.
That should not be obvious.
Because when the mass is hanging over here,
you don't just have a spring force.
You got a gravitational force.
You don't just have a spring energy and kinetic energy.
Think about it.
This mass is moving up and down.
You got changes and gravitational potential energy.
So why don't we have to taken into account
the gravitational potential energy
when we're doing conservation of energy in this equation?
Well that's what I wanna prove to you
in the rest of this video.
If all you wanted was the result, if all you,
if you're good, if you're like,
"Man, all right, I can do the same thing above cases,
"don't even tell me anything else."
You're good.
But I suggest you watch the rest of it
because knowing why you can ignore the m g h
in this calculation gives you better insight
into what we really mean by this d and this h and this v
as well as what we really mean by the equilibrium position
and that will conceptually aid you
if you get a problem that's more challenging.
So let's prove this and figure out why we can get away
with ignoring this m g h right here.
So let me get rid of all these.
Let's start fresh.
Let's just say we have a spring hanging from the ceiling
of spring constant k.
Let's say this spring is light.
If it was heavy, it might pull itself down by its own weight
so I'm gonna assume this is very light spring
and it's hanging right here.
There's no mass connected to it initially
so it's just hanging at its natural length.
It's neither pulling up nor pushing down
as you see it right here
because it's at its natural length
but we connect the mass m to it.
When we do that, we lower the mass with our hand.
We don't just let it fall and start oscillating.
We first lower the mass.
We connect it and lower it.
We find the point where the mass is just gonna stay at rest.
That would be the new equilibrium position.
So this right here is essentially
our new equilibrium position.
In other words, that's the point where the net force
on the mass would be zero.
But this time that's not at the spring's natural length.
The way it was on the horizontal spring.
This time the equilibrium position is this place,
the distance a away from the spring's natural length
because right now it's battling the force of gravity.
In other words the spring force exerted upward k x
minus the gravitational force which is m times g
has got to equal zero in order for this mass
to be an equilibrium.
So we can actually figure out what this distance a
would have to be in terms of given variables.
Since at the equilibrium position, x, the distance
the spring has been stretched has just gonna have to equal,
m g divided by the spring constant k.
This is what a is gonna equal.
So the distance, the mass hangs down
at the equilibrium position from the natural length
of the spring is just gonna be m g over k.
This is a in this diagram.
This is gonna be key.
So we're gonna hold on to this result right here.
Well let's do this.
Let's ask if we take this mass and pull it down
an extra amount b from the new equilibrium position,
well at that point the forces won't be equal.
The spring's gonna be stretched extra.
It's gonna be pulling up with more force
in the force of gravity.
So this mass is gonna accelerate upward.
It's gonna reach this equilibrium position with some speed.
It's gonna shoo pass that.
Now the spring force is less than the force of gravity
and so gravity wins in that case
and it keeps going back and forth.
We're gonna ask ourselves the same question we did before.
If we pull this down, the distance b,
what is the speed of the mass
when it passes through the equilibrium position?
Again we're gonna use conservation of energy to answer this.
So we're gonna say that the initial energy in our system
is gonna equal to final energy in our system.
We're gonna choose two points.
Let's choose initially this point down here.
We release the mass from rest
when it's pulled down at distance b
below the new equilibrium position.
Then our final point is gonna be right here
at the equilibrium position
because that's where we wanna know the speed of the mass.
So let's try to figure out how much energy there is
in the system initially
if I pull this mass down and let it go.
Well initially if I'm just letting this mass go,
the mass is starting from rest.
If the mass starts from rest, it's got no speed
and if it's got no speed, it's got no kinetic energy.
So there's no kinetic energy to start with
if this mass is starting from rest
but there is gonna be spring potential energy
and there's gonna be a lot of spring potential energy.
Because think about it.
Not only is this mass stretching the spring passed
the new equilibrium position by an amount b
but the new equilibrium position itself is stretched
from the spring's natural length a.
This formula, when you have 1/2,
the spring constant times the length
that the spring has been stretched,
that's the total amount the spring has been stretched.
So the total amount the spring has been stretched
from its natural length is gonna be a plus b.
I've got to square that whole term.
This is how much spring potential energy there's gonna be
in the system initially.
So how much gravitational potential energy
we're gonna start with?
Well, that's kind of hard to us
because you can always choose where you want your h
equals zero reference line to be.
In other words, I'm gonna choose this lowest point here
because that's often convenient when choose this to be
the h equals zero reference line.
We'll measure all heights from that point.
This is allowed because it's only really differences
in gravitational potential energy that matter
so you can do this.
You just have to be consistent with your choice.
But with that choice where this is h equals zero,
the height my mass has at the initial position
is gonna be zero.
So that means the gravitational potential energy
which is given by m g h is also gonna be zero
at that initial point.
So in terms of initial energies, this is all I've got.
This is my total initial energy,
just the energy from the spring.
Now we can set that equal to our final energies
so we're gonna have any spring potential energy here
in our final position.
You might think no because the final position
is at the new equilibrium position
but remember this new equilibrium position still displays
from the spring's natural length.
So what I have, I'm gonna have 1/2 k times
the amount the spring has been stretched
from its natural position.
At this new equilibrium position,
the amount that has been stretched is just a.
So I'm gonna write a here
because that's how far the spring has stretched
at this new equilibrium position.
I have to square that
because that's 1/2 k x squared.
We're gonna have kinetic energy.
This mass is gonna gain speed as it flies upward.
It's gonna be moving with some speed
when it gets up to that point.
So the kinetic energy is gonna be 1/2 times the mass,
times the speed the mass has
at the equilibrium position squared.
That's what we want to determine.
But there's also gonna be gravitational potential energy.
We said h is zero down here.
So if the mass is not there,
it's gonna have potential energy due to gravity.
If it's b above this point,
look at it, we pulled it down b
so when it gets back to the equilibrium position,
if this is h equals zero,
that's gonna be h equals b above where it started.
So I have to put in m times g times the b value.
This length right here.
Since in moving up this mass gained m g times b
a gravitational potential energy.
So how do we make progress here?
I wanna solve for the speed v
but I've got this mess over here.
Look, I've got a plus b squared
so I better handle that first.
So let's say we do the 1/2 k and then we square this out.
Remember we do FOIL.
So it's First, Outer, Inner, Last.
I'm gonna get a quantity of a squared
and then this cross term,
I'm get plus two times a times b
and then plus b squared.
That's what happens if I square this whole term right here.
These start looking really bad but don't despair.
Something great, something wonderful is about to happen
because I'm gonna set this equal to the right hand side.
So if we multiply out on the left hand side,
we're gonna get a 1/2 times k times a squared
plus a 1/2 times k times this 2 a b term,
plus I'm gonna get another one,
1/2 k times this b squared term.
We can say that that's supposed to equal
everything on the right hand side.
So we can already see something that we can cancel out.
I've got a 1/2 k a squared on each side.
So if we subtract that from both sides,
I can get rid of that.
This 1/2 here is gonna cancel this two
and I'm left with k a b on the left hand side
plus 1/2 k b squared.
But, what is k a?
If you're clever and you look up here,
you're like, wait a minute, I remember what k a was.
K times a, if we just multiply both sides by k here,
it's gotta equal m g because that's what just the statement
of equilibrium that at the equilibrium position
k times a has to equal m times g.
So I can replace k times a over on this left hand side
with m times g.
M g might be like, why would I ever wanna do that?
Why would I wanna replace k times a
and I still multiply by this b?
Why would I replace k a with m g?
Because now that term is gonna cancel with the other m g b
on the other side of this equation.
One half m b squared plus m g b
but I've got m g b on the left
and m g b on the right now.
That's why I replace this k a with m g.
I can subtract that from both sides
and magically I just get the exact same relationship
we had for the horizontal spring
measured from the new equilibrium position.
This is important so let me restate this.
You can either when solving these vertical spring problems,
measure your spring displacement from the natural
unstretched length of the spring like we did right here.
We had to add a plus b
because that was the distance
from the natural unstretched length of the spring
all the way to where the mass was.
You can do that and include gravitational potential energy
and get the right answer.
But what we just saw is that these terms always cancel
so the alternative is that you can just measure
the spring extension, the spring displacement
from the new equilibrium position.
If you do that, you just leave off all mentioned
of gravitational potential energy
and you get the same answer.
You can think of gravity simply as shifting
the equilibrium position down a bit
and then the mass and spring behaving
just as I would on a horizontal surface
as long as you only think about spring displacement
from that new equilibrium position.
So in other words, if you were given a problem,
let me get rid of all these.
If you're just given a problem
and you were told a three kilogram mass
is hanging from a vertical spring
of spring constant 50 newtons per meter
and this line here represents the equilibrium position,
it's just hanging out right there at rest.
You pull this mass down from the point
where it was hanging at rest, 0.3 meter and let go.
You wanna figure out what speed will it be going
when it reaches equilibrium position.
You can just do this.
You can say, all right, down here
it's got spring potential energy,
1/2 k is 50 newtons per meter
times the amount the spring has been stretched
but I'm just gonna consider stretches
from the new equilibrium position.
So I'm just gonna do 0.3.
I'm not gonna worry about the fact
that the spring has actually already been stretched
to get to this equilibrium position.
I'm just gonna put this stretching from equilibrium.
I'm gonna set that equal to the kinetic energy
the mass has at the equilibrium position.
We know that m is three kilograms times v squared.
I'm not gonna include information
about the gravitational potential energy at all
because I only measured the displacements
from the new equilibrium position.
So at this point, I can just solve for my v.
If you solve that algebraically for v,
you cancel out the twos.
You divide both sides by three
and take a square root.
You get to the speed of the mass at equilibrium
is gonna be 1.2 meters per second.
So to recap, even though it seems initially
like vertical springs would be much harder
than horizontal springs because you've got
gravitational forces and gravitational potential energy
to worry about, if you measure the spring displacement
from the new equilibrium position
as opposed to the natural spring length,
you can simply use conservation of energy
without mention of gravitational potential energy at all.