For more information, please see full course syllabus of Physical Chemistry

For more information, please see full course syllabus of Physical Chemistry

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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
- The Harmonic Oscillator
- 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

- 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

### Physical Chemistry Online Course

### 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 and*0368

*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 stored*1478

*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 and*1774

*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

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