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The Harmonic Oscillator I

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • The Harmonic Oscillator 0:10
    • Harmonic Motion
    • Classical Harmonic Oscillator
    • Hooke's Law
    • Classical Harmonic Oscillator, cont.
    • General Solution for the Differential Equation
    • Initial Position & Velocity
    • Period & Amplitude
    • Potential Energy of the Harmonic Oscillator
    • Kinetic Energy of the Harmonic Oscillator
    • Total Energy of the Harmonic Oscillator
    • Conservative System

Transcription: The Harmonic Oscillator I

Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.0000

Today, we are going to start our discussion of the harmonic oscillator.0004

We are going to spent several lessons on it so let us go ahead and get started.0007

The harmonic oscillators are pretty much exactly what you remember from classical physics.0012

Basically, it is some system that is moving back and forth.0016

It is oscillating just like this, back and forth, back and forth.0019

We wanted to discuss the quantum mechanical equivalent of that.0024

Let us start with, when radiation interacts with matter,0029

it causes changes in the energy levels of the particles making up that matter.0047

The whole field spectroscopy is a study of that, the interaction of electromagnetic radiation with matter.0074

At different frequencies we get different responses from the molecules or from the atoms,0082

Whatever it is that we happen to be irradiating.0086

Let us take a molecule like, take a diatomic molecules like hydrogen fluoride.0091

We have that, the bond of this.0108

We have the H + over here, we have the F - over here, and this molecule vibrates back and forth.0113

This goes this way, and this way, this way and this way.0122

This idea of a harmonic oscillator is a really nice model for the vibration of a diatomic molecule and the vibration in a bond in general.0126

L so imagine a spring with two masses going like this was pretty much the same thing.0136

We can study this harmonic oscillator, extract some information from it,0141

and come up with a quantum mechanical version of it and it is going to give us a lot of information.0145

Irradiation of this molecule by infrared radiation causes the molecule to vibrate at different frequencies.0154

Vibrate and oscillate is going to be synonymous.0185

In other words, it is going to change the energy level of the molecule.0196

This oscillation back and forth is harmonic motion.0204

Harmonic motion just refers basically to sin and cos.0221

Sin and cos repeat themselves, they are harmonic, periodic, repeating just back and forth.0228

That is what harmonic means.0235

We want to examine this harmonic behavior and we want to examine the quantum mechanical harmonic oscillation.0239

We are going to begin with a classical harmonic oscillator and then move on to the quantum mechanical harmonic oscillation.0248

We will get a lot of information by actually studying the classical harmonic oscillation.0254

We wish to examine the Q in harmonic oscillation and we begin with the classical harmonic oscillator.0262

Let us go ahead and start on a new page.0301

Consider the following system.0305

We have basically a setup like this and this is a wall.0307

We are going to fix one hand and we are going to have a spring with a mass on it and it is going to be a just a spring in its normal unstretched,0313

uncompressed string, what we called the equilibrium length of the spring.0327

We are going to call that S of 0.0331

There is a mass, it has mass M.0334

S of 0 is the distance of the undisturbed spring.0339

Again, we are making it simple for ourselves at first by holding one end fixed and just having one mass oscillating back and forth.0358

It is going to oscillate around some equilibrium position.0366

That is how it for all science, we start with the simplest case and0368

then we move up in degrees of complication until we arrive at something which matches what we want, our real world.0372

It is always like that, start with the simplest case and move forward from there.0379

We can stretch the spring or compress the spring at distance S.0387

We can stretch or compress the spring at distance S.0394

Let us go ahead and take a look at some of those.0412

If we stretch the spring, here we stretched it and now this distance is S.0416

Or we can go ahead and compress it, push the spring in.0427

We have taken it and we have actually pushed it in.0432

This is the distance S.0440

No gravitation is acting on it, there is no gravitational force.0445

The only force that is acting on this mass is the spring.0452

I will pull the spring this way it will want pull it back that way.0458

I push the spring in, this spring is the one that is going to push back the other direction.0460

There is no gravitational force acting on the mass.0465

So far so good.0475

This spring force is the only force acting on the mass.0481

Hooke’s law, you remember hopefully.0499

Remember having heard from your physics course.0503

The force acting on a mass attached to a restoring medium, in this particular case the restoring medium is the spring.0509

It could be something else, it could be a rubber band.0533

Anything that will, as you pull it, tries to pull it back.0534

As you push it, it tries to push it back, restores it back to its equilibrium position.0538

The force acting on a mass attached to a restoring medium is directly proportional to the displacement.0542

The displacement is the extent to which it actually pulled away from its equilibrium position.0559

From here to here, that is the displacement right there.0564

That is the displacement from its equilibrium position.0570

It is the S – S0.0573

It is directly proportional to the displacement.0576

That is Hooke’s law and mathematical statement is the force is actually equal to - K × S - S0.0580

If I stretched it, S is bigger than S0.0594

The stretched is bigger than the equilibrium position so S – S0 is positive.0598

The force we put a negative sin in front of it because now the force is actually pulling the mass back that way.0603

If I push it in, our S is smaller than S0.0608

This ends up being negative and this negative sin in front actually makes the force go in the positive direction.0613

It is opposite.0622

I will pull this way, the force wants to pull it back that way.0623

I push this way, the force wants to push it back that way.0626

That is why the negative sin is there.0629

This is Hooke’s law.0631

In the case that the force is constant that is called the force constant or the spring constant.0633

A large K means you have a very stiff spring and a small k means you have a very loose spring, very easy to push and pull.0653

We have F is equal to - K × S - S0.0676

Let us go ahead and draw this out.0684

This is the equilibrium position of the spring.0687

If I had it stretched out, this distance would be my displacement.0691

This is my S - S0.0702

The force, if I stretch it, the force is going to want to pull it back that way.0704

Of course, the compressed, I have pushed it in from its equilibrium position.0708

This is the equilibrium position.0715

We have right there, that is the S, that is the S0, S - S0, that is this displacement.0718

The force is going to be pushing that way.0730

Trying to push it back to its equilibrium, the restoring force, the restoring medium.0733

We have this, we also know from Newton’s second law that force = the mass × the acceleration.0738

It is equal the mass × the acceleration is the second derivative of the displacement.0744

The displacement is S.0752

The first derivative of that is going to be the velocity.0755

The second derivative is going to be the acceleration.0757

We can write D² S DT².0760

Let us let X be the net displacement.0766

Net displacement means this distance, the actual distance I have pulled or push away from the equilibrium position.0777

This is X and this is X, net displacement.0784

In other words, it is S - S0.0788

Mathematically, we are going to let X equal S – S of 0 because we are ultimately concerned with.0791

S0 is the initial, S is the final, the difference between them is the displacement.0798

S = S - S0 means that S = X + S0.0804

Now F and F, I'm going to set them equal to each other.0812

I have MD² S DT² = -K × S - S0.0817

We said the X is equal to S - S0.0829

I’m going to put X in for here and the D² of S DT², I’m going to take the derivative of this.0832

DS DT = DX DT, this is a constant.0842

It is just the equilibrium position so derivative is just 0.0848

Of course, if I differentiate twice, it is just going to be D² S/ DT² = D² X/ DT².0852

That is it, nothing strange going on here.0860

When I put this in for this, I get M D² X DT² = -KX.0863

Or more appropriately M D² X DT² + KX is equal to 0.0876

This is the differential equation for a harmonic oscillator.0889

This is the equation that we solved in order to find the function X of T.0892

When we find X of T, what that tells us that any time T is going to tell us exactly how far this mass is from the equilibrium position, where is it.0900

That is what we are doing.0913

This is the differential equation we solved.0914

The general solution of this equation, I’m not going to go through the actual solution of it.0917

I will just go ahead and give it to you here.0921

The general solution of this differential equation.0925

I need a little more room here.0937

X of T = C1 cos of ω T + C2 × the sin of ω T.0942

Where ω is actually equal to the force constant ÷ the square root mass or to the ½ power.0954

Instead of putting this, we just use the ω.0963

Now X of T, as we just said, X of T tells me how far the mass is from the equilibrium position at any time T.0967

You have this mass equilibrium position, if I pull it and let it go, it is going to just bounce back and forth, oscillate.1002

This equation tells me where, how far from the equilibrium position it is, X at time T.1013

That is all I have done.1020

Let us go ahead and take this mass and that is the equilibrium position.1025

What I have done is I have actually hold it, I have stretched it out, and I’m going to let it go.1038

When I let it go, it is going to oscillate back and forth, back and forth, back and forth.1045

Let us actually work this out.1050

This is the general solution.1051

We want to find some specific cases.1054

The initial position, let me go to blue here.1057

The initial position, in other words X at time 0 is equal to A.1065

A is the distance that I have pulled.1074

I have pulled it to distance A.1077

Let us go ahead and call that A.1079

The initial velocity, velocity is the first derivative.1083

X prime at T = 0.1088

At 0, I have to let it go.1092

When I let it go, that is when it starts.1095

The initial velocity is actually 0.1097

We have C1 cos ω T.1100

We have our equation which is X of T = C1 × cos of ω T + C2 × the sin of ω T.1116

The X of 0 is equal to C1 × the cos of 0, when I put 0 in for T here and here, + C2 × the sin of 0.1128

And we know that X of 0 is equal to A.1143

Sin of 0 is 0 so that term goes to 0.1148

Cos of 0 is 1, therefore, C1 is equal to A.1151

We found the C1, we found that coefficient.1157

Let us take X prime.1164

Let me rewrite the equation up here.1167

X of T = C1 × the cos of ω T + T2 sin ω T.1170

X prime of T that is going to equal -ω C1 × the sin of ω T + ω C2 × the cos of ω T.1179

X prime of 0 = - WC 1 × the sin of 0 + ω C2 × the cos of 0.1194

We said that that actually = 0.1207

This is 0, this becomes C2 W = 0.1209

W is not equal 0 which means that C2 = 0.1219

Now we went ahead and found C2.1223

We have X of T is actually equal to A × the cos of ω T.1226

We found a specific solution under the circumstance of me pulling it and letting it go.1237

This is harmonic motion, you know what a cos function looks like.1248

In this particular case, A is the amplitude and it goes like this, it oscillates back and fourth.1251

This is harmonic motion, it is always going to be in terms of sin and cos.1258

The period of this oscillation is equal to 2 π ÷ ω.1262

The frequency of the oscillation is equal to ω/ 2 π.1273

It is the reciprocal of the period and it is in cycles per second.1283

Cycles per second, otherwise known as a Hertz.1291

We often do not include the cycles, we just put second.1295

It is usually like this, inverse second.1297

This is frequency.1300

Let us take ω / 2 π cycles per second.1309

Let us multiply by how many radians are there in a cycle.1317

If you do one cycle circle it was 2 π radians.1321

It is 2 π radians per cycle.1323

The 2 π and the 2 π cancel and I'm left with radiance per second.1330

Ω, the unit is radians per second, this is the angular velocity.1334

We have our period, our frequency, we have our angular velocity which is ω and A is called the amplitude.1348

A is the amplitude of the oscillation.1356

It is the largest value of the amplitude can actually have.1361

If a stretch something A, when it goes the other direction it can only compress the spring at distance A maximum.1363

It will just go back and forth, AAA –AA – AA, something like that.1370

A is the amplitude of the oscillation.1374

All this information is extractable from this amplitude angular velocity ÷ 2 π gives you the frequency.1380

Take the reciprocal of the frequency will give you the period.1391

Or you can just take 2 π over the angular velocity and that will give you the period.1393

All of this information comes from that.1397

Recall from physics, I will go back to blue here.1402

Recall from physics that a force is equal to - the derivative of the potential energy.1407

DV = - F of X DX.1421

If I want the potential energy it equals- the integral of DF of X DX.1427

Just integrate that equation.1434

Here our F of X is equal to – KX, that is Hooke’s law.1436

The force acting on a mass is directly proportional to the extent to which I actually pull that mass away from its equilibrium position.1446

Our potential energy is equal to - the integral of - KX DX, which means it is equal to ½ KX² + some constant.1456

We can set C equal to 0, it is just a 0. energy.1472

We can go ahead and set it to 0 so what we get is that the potential energy stored1478

When I stretch the spring or compress the spring is equal to ½ KX².1485

This is very important, that is a potential energy of a mass spring system.1491

We have our V of X is equal to ½ KX², that is nice.1504

Our kinetic energy, we know the kinetic energy is equal to ½ the mass × the velocity².1512

Velocity is just a derivative of DX is DX DT.1518

I’m just going to go ahead and call it X prime².1525

X of T, we know that X of T is A × the cos of ω T.1532

When we take the first derivative which is going to be the velocity, it is going to end up being -ω A sin of ω T.1538

Let us go ahead and put those back in.1555

This is X of T X prime of T, we are going to put these in to here.1557

Our potential energy is actually equal to ½ K × A cos of ω T² which is going to equal ½ K A² cos² ω T.1564

This is a potential energy of our classical harmonic oscillator.1591

Let us go ahead and calculate the kinetic energy.1594

Kinetic energy is equal to ½ the mass × the derivative².1598

The derivative² was - A ω sin because we take the derivative sin ω T².1605

We end up with, the -² cancels out so we end up with ½ MA² O² sin² ω T.1619

This is our kinetic energy of the harmonic oscillator.1641

The total energy of the system we know is equal to the kinetic energy + the potential energy.1644

Our total energy we just add them up.1651

It is just ½ MA² cos² ω T + ½ M ω².1653

I just switched the ω² and A².1667

Sin² ω T.1671

That is the expression for the total energy of a harmonic oscillator.1673

When I plot these, this is what I get.1676

Equilibrium position that is A, this is –A.1687

A equilibrium –A, it is going to oscillate back and forth like this.1692

Here is what it looks like.1697

We have this up here, let me do this one in red.1705

Let me go ahead and go to black.1722

This right here, this is the potential energy curve.1724

This right here, the red, that is the kinetic energy curve.1730

Notice what is going on.1734

I pull the thing to A, all the energy is potential and there is no velocity.1738

I have not let go, 0 kinetic energy.1745

I release it, all that potential energy starts to go to kinetic.1748

At some point they meet, where the kinetic and the potential are equal.1753

As it passes through the equilibrium position because it is actually at the equilibrium position X is 0.1756

Therefore, the potential energy is 0.1763

However, the kinetic energy is at its maximum.1765

It will start to slow down as it passes through the equilibrium position, it starts to go toward the negative.1768

It will start to slow down so the kinetic energy is going to drop down to 0 and1774

the potential energy is going to rise to its maximum, back and fourth.1778

Notice, the sum of energy is actually constant so the kinetic and potential energy they switch off.1785

The total energy of the system is constant.1794

I will go ahead and do an algebraic so you can see.1796

Let us go ahead and write this out.1800

Let us go ahead and write it in blue.1802

At maximum displacement, the mass has stopped which implies that the kinetic energy is equal to 0.1805

V is at a maximum, here and here.1828

As the mass passes through the equilibrium position, now the potential energy is equal to 0 and K is a max.1840

Here and here the potential is 0, kinetic energy is at a max.1862

At the ends it is the potential energy that is on the max and the kinetic energy is at 0.1868

They switch off.1872

In between, we have some of both.1874

Let us go ahead.1878

We have our energy is equal to ½ M A² cos² ω T.1883

I’m checking my equations here.1904

I’m messing up the sin and cos.1906

I will make sure I have this to maintain the order that I actually used in here.1928

Potential is going to be the K and it is going to be square O².1933

Sorry about.1940

We have got ½ mass A² ω² sin² ω T + ½ K A² cos ω cos² ω T.1942

Kinetic energy + potential energy.1962

Recall that that is equal to K/ M ^½.1965

So ω² is equal to K/ M.1977

I can go ahead and put that into there.1986

Therefore, I have my total energy is equal to ½ M A² K/ M sin² ω T + 1/2 K A² cos² ω T.1993

The M cancels the M, I can go ahead and I combined terms and factor out.2016

I have ½ A² K, ½ A² K.2021

I have 1/2 A² × K × sin² ω T + cos² ω T.2025

You remember from trigonometry, the sin + cos² is equal to 1.2035

Therefore, my total energy is 1/2 A² × K.2038

The total energy of the harmonic oscillator.2045

This is a constant, A is the amplitude as a constant.2048

K is a constant.2051

This is an algebraic representation of the fact that the energy of the harmonic oscillator is constant.2053

It oscillates back and forth between the potential and the kinetic.2062

They tradeoff and one of them is maximized, the other is minimized.2065

The one is minimized, the other is maximized, but it is a constant.2068

Let us see.2075

Now a system where the total energy is conserved is called appropriately a conservative system.2082

The harmonic oscillator that we describe is a conservative system.2108

Energy is a constant, it is transferred between kinetic and potential but it is a constant.2112

It does not go away.2119

Energy is conserved, it is not lost to anything.2122

Thank you so much for joining us here at www.educator.com.2127

We will see you next time for a continuation of the discussion of the harmonic oscillator.2129

Take care, bye.2132