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Okay, now that we have some of the mathematical prerequisites that we need,

we can go on and talk about the simple harmonic oscillator.

Now this a model that's used over and over again for oscillatory systems and

we're going to look at this in the context of just a simple mechanical,

model of the simple harmonic oscillator. Se let's say we have a mass m of certain

amount of inertia connected to a spring with its spring constant k, that is then

attached to a rigid support. Now What I'm going to do is say, the

resting position of that mass. We'll measure it right here on this side

of the block and call that x equals zero. And let's assume that we walk up to this

and give it. A impulsive force at time equals zero.

So you could walk up and give this thing a kick.

Now, the instant you do that, the mass does not move instantaneously, but you do

instantaneously change its velocity by imparting a an impulse to it.

Now what I want to do is then examine the resulting motion of this.

Now, your, your intuition,uh, tells you that the mass is going to go out to some

maximum displacement and then turn around and go back the other way, pass through

the equilibrium position compress the spring so that it's moved as far as it's

going to move to the left, and then repeat that motion, and just oscillate

back and forth. So I want to take a little bit closer

look at that. So, at the initial instant when I first

provide the impulsive force to this mass, I now have, it's still in position x

equals 0 and it has velocity v-max. Now I'm going to make a plot of.

X versus time and the velocity versus time.

And so at this initial time, the position is zero and the velocity is maximum in

what I'm calling the positive direction to the right.

Now, a quarter cycle later. The mass reaches its maximum displacement

from the equilibrium position and the string is stretched as far as it's going

to string stretch. And at that instant, the position is

ex-max, the maximum displacement, and the velocity is zero.

The simple harmonic oscillator or the mass Is just turning around, it reached

its maximum displacement, it turns around then it's going to head back the other

way. And so at that instant the velocity is

zero. So, if I plot that the displacement is

now positive maximum value and the velocity is 0.

Now, the mass is now on its way back through the equilibrium position, and

when it passes the starting point, x equals 0, it has a maximum velocity in

the negative direction. V max if I plot that it's maximum

negative velocity and displacement is 0 then the spring compresses the mass

continues to move to the left until the spring is compressed to the point that

it's able to stop the mass. And at that instant in time I've rearched

the mass has reached a maximum displacement in the negative direction

and the velocity is 0 and so at point D, it's maximum negative displacement

velocity 0. Then the mass turns around and is headed

back in the positive direction. And as it passes the zero point, the

velocity is now back to the maximum velocity.

This is how we started. And so it's, it's back to where it

started, with the same velocity that it started with.

now we've ignored damping in this, and so there hasn't been any loss of energy and

it has the same velocity. At the end of one cycle as it had at the

beginning. Now the, if we connect these points it

turns out that resulting motion of the harmonic oscillator is a sign wave.

And so we just looked a few points in the sign wave.

But if we, when we solved the problem in just a minute, we'll see that the

solution really is a sign wave. And the velocity is a cosine curve, and

so this is what the harmonic oscillator is going to do.

Now, let's go back now and use a little bit of calculus that we know to actually

solve this problem. So now it's, we just need to review a

couple of definitions from mechanics, or hopefully you've had this in high school

physics. If I take the time rate of change of

displacement, x is velocity. So the derivative, with respect to time,

of the position of that harmonic oscillator is going to be its velocity.

And the time rate of change of velocity is the acceleration.

And the, we can combine those and say if I take the time rate of change of the

time rate of change of the displacement. That's what we call the second derivative

of the displacement. And so the second derivative of the

displacement is acceleration. So that's just a little kinematics

review. Now let's go back to the harmonic

oscillator. It's our mass on this spring, and we're

going to set up a coordinate system, and say x is measured from this point on the

left of the, the mass, and increases to the right.

Now, there are two pieces of physics that we need to find the differential equation

for this system. The first one is Newtons second law that

says the force equals the mass times acceleration.

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This second one is Hooke's Law that says, the force of a spring is equal to minus

some constant times the displacement of that spring.

So, if I take and I compress a spring, it pushes back and so the force is always in

the opposite direction of the displacement of the spring.

If I stretch a spring. It tries to pull itself back.

If I compress the spring it tries to push back the opposite direction.

So if I combine Newton's second law, and Hook's law, say that the inertial force,

due to the acceleration of the mass, is always balanced by the spring constant

force. Then I can set MA equal to minus Kx and I

will write the acceleration as the second derivative as the displaced according to

these kinematic expressions over here. So here is the differential equation for

this harmonic oscillator. It says the second derivative with

respect to time of the displacement of this oscillator times some constant, the

mass, is equal to minus some other constant, the spring constant, times the

displacement. Now, what we have to do is find a

solution to this equation. So I'll just, just take and put the mass

on this side, so this is the differential equation that we need to solve.

Now, just to keep the neaten up the writing here, we'll let omega squared

equal K over M. So this is minus omega squared X, so

here's the form of the differential equation that we want to solve.

Now what we have to do is find a solution of function of time that's going to

satisfy this equation. And that means if I whatever that

function is. If I take two derivatives of that

function, with respect to time, then it just has to be minus some constant times

the original function. So if I take a function, take two

derivatives, I have to get back something the same function of time But there may

some constant out front. Now, a one time honored way of solving

differential equations is to take a guess at the solution.

And then if you can then demonstrate that, that Is indeed a solution of this

equation, then no one's going to ask how you got the solution.

They're just glad that you found the solution.

So, so let's take a guess, and let's say that our function of time is just going

to be some constant times sine omega t. Now what I wanted to do then is take two

derivatives of that/ So, if I take the first derivative of this.

The x0 is just a, a constant that is along for the ride.

Now when I take the derivative of sine omega T, I get omega times cosine omega

T. So that's using the chain rule, to, to

get this derivative. Now, I have to take the derivative of

this again, I want the second derivative of X with respect to time.

So the X0 is still out in front. And now when I take the derivative of

cosine omega t, I get minus sine omega t. So this whole expression picks up a minus

sine and I have sine omega t sitting here.

Now, I'm going to take and plug these two expressions Into the equation, so here's

x. I'll put it in there, and here's the

second derivative of x right here. I'll put that in this side.

And let's see if this equation is really true.

And it is, because if I multiply by, this expression by minus omega squared that's

exactly what I have over here. So I've demonstrated that with this

guess, if I take two derivatives with respect to time and then plug it back

into the original equation, indeed, a function of this form does solve that

equation. So we found a solution to the

differential equation. So, the function of time that we're

looking for, for this harmonic oscillator, is x zero sine omega t.

Where omega is the square root of k over m.