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Lecture Comments (37)

1 answer

Last reply by: Professor Dan Fullerton
Thu Jun 30, 2016 9:22 AM

Post by Peter Ke on June 29 at 06:07:53 PM

For example 6, when you say equilibrium position do you mean by when the pendulum faces directly down. If so doesn't it mean that the KE = maximum while PE = 0, so v=sqrt(2gh) does not apply? I think I am confuse please explain.

Thank you!

3 answers

Last reply by: Professor Dan Fullerton
Sat Apr 30, 2016 12:49 PM

Post by Shikha Bansal on April 29 at 06:40:51 PM

For example 2b)the total energy was equal to the spring potential energy only, but for 2d) it was equal to the spring potential and the kinetic energy. Is it because .5 meters is the max it can go (since velocity is zero there)? If so, how did you know ,5 m is the maximum displacement?
Also, for 2c) wasn't the 5 Joules of total energy for the situation where the displacement was .5 meters? How come you used it when the displacement was 0 meters (aka equilibrium position)?
Furthermore, for 2c) I get that the kinetic energy is max at the equilibrium position, but shouldn't the spring potential energy equal 0? So why did u write that kinetic energy was equal to the spring potential energy?
Thanks for all the help!

1 answer

Last reply by: Professor Dan Fullerton
Thu Mar 31, 2016 3:38 PM

Post by Zhe Tian on March 28 at 03:55:59 PM

Just a technical question but are angular velocity and angular frequency the same?

4 answers

Last reply by: Professor Dan Fullerton
Tue Aug 11, 2015 8:54 PM

Post by Anh Dang on August 10, 2015

In Circular Motion vs SHM, I don't get why you use sin for the equilibrium  position and cosine for the max displacement.  Also, why do you use the y coordinates for sin and not x?

2 answers

Last reply by: Jason Wilson
Tue Dec 9, 2014 5:32 PM

Post by Jason Wilson on December 6, 2014

Example 3:  I may be simply typing incorrectly on the calculator or possibly missing an entire step.  The inverse cos -1(.1/.2) all over 2pi: is coming back on my calc as 60/6.28 = 9.55 what am I missing...I am tired I am cramming right now. :)

3 answers

Last reply by: Professor Dan Fullerton
Sat Aug 30, 2014 8:39 PM

Post by Jungle Jones on August 30, 2014

In example 2, Hooke's Law says that F=-kx, but you just used F=kx to find the spring constant. I understood that the negative sign symbolizes that it is a restoring force, but can you explain why you didn't use the negative sign here?
And do never use the negative sign when doing calculations involving the law?

1 answer

Last reply by: Professor Dan Fullerton
Sat Aug 30, 2014 8:38 PM

Post by Jungle Jones on August 30, 2014

In example 1, I did omega = sqrt(k/m) and got 20, but I got the units of 1/(s^2).
k = 2000 N/m
m = 5 kg
N = kg m/(s^2)
so N/m -> kg/(s^2)
and that divided by 5 kg becomes 1/(s^2)
But it should be just 1/s, so what did I do wrong?

4 answers

Last reply by: Professor Dan Fullerton
Tue Jul 15, 2014 2:40 PM

Post by Jamal Tischler on July 15, 2014

I tried to calculate the spring oscilation period. I used the conservation of energy and kinematics equations for the average acceleration. 1/2kA^2=1/2mv^2 =>v^2=kA^2/m.
v^2=2aA+v0^2=2aA =>kA^2/m=2aA => a= kA/2m.
v=a*t. sqrt(kA^2/m)=kA/2m*t => t= 2 sqrt(m/k). This only the motion from the amplitude to the x=0 position. The farward 3 parts are the same because the kinetic energy is constantly transformed into elastic potential.
T=4*t=8 sqrt(m/k)
The formula said T=2*pi sqrt(m/k). pi doesnt equal 4.
Can you please explain this better ?

1 answer

Last reply by: Professor Dan Fullerton
Thu Mar 13, 2014 9:59 AM

Post by Lin Jiang on March 13, 2014

For example 5, why we needn't to find the equilibrium position. 1/2KX^2, I think x should be how far it pull from equilibrium.

0 answers

Post by Professor Dan Fullerton on February 17, 2014

I use an energy analysis, realizing that at the extremes, the kinetic energy is zero (for an instant the block is still), and all its energy is potential.  So the potential energy in a spring is 0.5kx^2.  When it's at equilibrium, however, there is no stretch in the spring, so PEs=0, and all the energy must be kinetic due to conservation of energy.  So 0.5kx^2=0.5mv^2.  Solve for the velocity!

0 answers

Post by sadia mussa on February 17, 2014

how do you get the formula for speed at equilibrium position as well as finding the last question for speed in the example for harmonic oscillator analysis because I don't seem to quite understand it, thanks

1 answer

Last reply by: Professor Dan Fullerton
Mon Dec 9, 2013 9:12 AM

Post by Constance Kang on December 9, 2013

in example 1,  since w= rot(m/k) then  if i do rot (5/2000) instead of 2pi f, how come i got a different answer ?

3 answers

Last reply by: Professor Dan Fullerton
Tue Apr 30, 2013 7:24 AM

Post by Saki Amagai on April 29, 2013

For example 6, it says "find the tension in the string at the equilibrium position". I think that the net force must equal zero at the equilibrium position; since in the previous slides, it mentioned that force and acceleration all equal zero at the point... Shouldn't "T = mg" and not "T - mg = mv^2/r"?

Simple Harmonic Motion

  • Simple harmonic motion is nature's typical response to a disturbance.
  • Simple harmonic motion is motion in which a restoring force is directly proportional to the displacement of the object.
  • Linear restoring forces exerted on objects displaced from equilibrium result in simple harmonic motion.
  • Restoring forces can result in oscillatory motion.
  • Many systems in simple harmonic motion exhibit an ongoing transformation of kinetic and potential energy.

Simple Harmonic Motion

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
  • Objectives 0:08
  • What Is Simple Harmonic Motion? 0:57
    • Nature's Typical Reaction to a Disturbance
    • A Displacement Which Results in a Linear Restoring Force Results in SHM
  • Review of Springs 1:43
    • When a Force is Applied to a Spring, the Spring Applies a Restoring Force
    • When the Spring is in Equilibrium, It Is 'Unstrained'
    • Factors Affecting the Force of A Spring
  • Oscillations 3:42
    • Repeated Motions
    • Cycle 1
    • Period
    • Frequency
  • Spring-Block Oscillator 4:47
    • Mass of the Block
    • Spring Constant
  • Example 1: Spring-Block Oscillator 6:30
  • Diagrams 8:07
    • Displacement
    • Velocity
    • Force
    • Acceleration
    • U
    • K
  • Example 2: Harmonic Oscillator Analysis 16:22
  • Circular Motion vs. SHM 23:26
  • Graphing SHM 25:52
  • Example 3: Position of an Oscillator 28:31
  • Vertical Spring-Block Oscillator 31:13
  • Example 4: Vertical Spring-Block Oscillator 34:26
  • Example 5: Bungee 36:39
  • The Pendulum 43:55
    • Mass Is Attached to a Light String That Swings Without Friction About the Vertical Equilibrium
  • Energy and the Simple Pendulum 44:58
  • Frequency and Period of a Pendulum 48:25
    • Period of an Ideal Pendulum
    • Assume Theta is Small
  • Example 6: The Pendulum 50:15
  • Example 7: Pendulum Clock 53:38
  • Example 8: Pendulum on the Moon 55:14
  • Example 9: Mass on a Spring 56:01

Transcription: Simple Harmonic Motion

Hi everyone. I am Dan Fullerton and I am thrilled to welcome you back to 0000

Today's topic is simple harmonic motion. 0004

Our objectives are going to be to sketch and analyze a graph of displacement as a function of time for an object undergoing simple harmonic motion, to write down an appropriate expression for displacement of the form A-cos(ωt) or A-sin(ωt), where ω is going to be that angular frequency. 0008

We will state the relations between displacement, velocity, and acceleration and determine the points in the motion where these quantities are minimum, 0, and maximum for an object undergoing simple harmonic motion, determine the total energy of an object in simple harmonic motion and sketch graphs of kinetic and potential energies as functions of time or displacement, and determine the period of oscillation for an ideal pendulum as well as a mass on a spring with a horizontal mass and a vertical mass. 0026

That is what we are going to try and accomplish here. 0052

All right. What is simple harmonic motion? 0056

Simple harmonic motion is nature's typical reaction to a disturbance. 0059

It is all over in this world in many, many places. 0064

When you disturb something it typically responds with simple harmonic motion. 0068

That could be something as simple as walking past a tree -- if you brush a branch, the branch gets a force done and starts oscillating back and forth. 0073

Simple harmonic motion -- it is all over .0081

A displacement which results in a linear restoring force results in simple harmonic motion.0084

And the simple items we are going to focus on for today are going to be an ideal pendulum, a pendulum where the string has no mass and a mass on a spring, going back and forth due to that spring. 0088

Let us start with a review of springs. 0101

When a force is applied to a spring, the spring applies a restoring force and a spring can be compressed or it can be stretched. 0106

When the spring is in equilibrium, it is unstrained; it is in its happy position; it is just thrilled to be there. 0113

The factors affecting the force of a spring -- well the spring constant (k) is how tough it is to compress or stretch the spring. 0120

The bigger the spring constant, the tougher it is to compress or stretch and that is measured in Newton's per meter (N/m) and the displacement is always measured from equilibrium. 0126

We could make a graph of the force of a spring vs. the displacement and when we do that, for an object obeying Hooke's Law, we should get a straight line where the slope of that line which is rise over run, gives us our spring constant. 0136

Or we could make the same type of graph -- force of a spring versus displacement get the same basic shape and if we want to get the work done in compressing and stretching that spring, all we have to do is come back and take the area under that curve and the area will be the work. 0155

Now as we analyze that work, the work that we do on it must be the potential energy stored in the spring, so that is 1/2 base times height or 1/2 our base is going to be (x) and our height is going to be the force. 0178

But if (F) is (kx), then that is going to be 1/2 x × our force, because it is a spring, (kx), or you could say that the potential energy stored in the spring is 1/2 kx2. 0195

That is where that formula comes from -- one way we can derive the energy stored in a compressed or stretched spring. 0211

All right. Let us talk about oscillations. 0221

Repeated motions back and forth are called oscillations -- something going back and forth, back and forth is oscillating. 0225

Now one revolution or one round trip, one complete cycle all mean the same thing and the period of the oscillation is the time it takes for one complete cycle or one complete revolution. 0232

That is going to be an important vocabulary word, period. 0243

Frequency, on the other hand, (F), is the number of cycles per second and it is measured in 1/seconds or a unit known as Hertz (Hz). 0247

Period (T) is the number of seconds divided by the number of cycles to give you the time for each cycle. 0261

Frequency is the number of cycles divided by number of seconds to give you the number of cycles per second and they are very closely related because Period is 1/frequency and therefore frequency is 1/Period. 0269

Let us take a look at the spring block oscillator. 0283

Imagine that we have a mass and we are going to connect it by a spring to some immovable object like a wall.0286

If we pull it one way and let it go, it is going to go back and forth and back and forth and oscillate.0295

That is a spring-block oscillator. 0297

Factors affecting the period of its oscillation are the mass of its block (m) and the spring constant (k). 0299

As we analyze this in a little bit more detail, we will call this the x = 0 position, the happy position, the equilibrium position -- maximum displacement of A or -A. 0306

Now the period of a spring-block oscillator, period of a spring STs is going to be 2π times the square root of the mass divided by the spring constant. 0318

Now the frequency of that spring-block oscillator which is always 1/period is just going to be 1/2π square root then of k/m. 0332

Now we could also rearrange this a little bit to say that 2π times the frequency equals the square root of k/m. 0344

Where there is 2π times the frequency is often times called the angular frequency (ω). 0355

Angular frequency = 2πF, therefore we could write the angular frequency (ω) is the square root of k/m. 0366

A couple of definitions to go along with our spring-block oscillator system. 0384

All right. Let us take a look at an example with this system. 0390

We have a block of mass 5 kg and it is attached to a spring, whose constant is 2,000 N/m.0393

Find the period of oscillation, the frequency, and the angular frequency. 0399

Well, let us start with the period. 0406

Period for a spring-block oscillator -- we know is 2π square root m/k or 2π square root -- mass is 5 kg, (k) is 2,000 N/m or when I plug that into my calculator, I get about 0.314 s for my period.0408

Let us find the frequency. 0435

Frequency is 1/Period, so that is going to be 1/0.314 = 3.18, 1/seconds or 3.18 Hz.0436

And the angular frequency (ω) is 2π times the frequency, 2πF or 2π × 3.18 Hz, or 20 and the units are rad/s, although radians are not an official unit -- 20 rad/s will be our angular frequency. 0457

Let us take a look at how we might analyze this in even more detail. 0487

Here we are going to show our spring-block oscillator, mass (m), and we are going to look at it at three different positions -- at its equilibrium, position (A), at maximum displacement to the right (B), back to (A) to its minimum displacement or maximum displacement on the left (C), and back to (A). 0491

So it is going to go back and forth, (A) to (B) to (A) to (C) to (A) and back and forth and it is going to displace a distance (x) to the right or (-x) to the left.0508

Well, if we want to look and see if we have different displacements, at point (A), we have 0 displacement, that is its equilibrium. 0518

When it is at (B), its displacement is (x) and when it gets all the way to (C), its displacement is (-x). 0526

Let us take a look at velocity now .0536

When it is at (A), it is going to have maximum velocity because all of its energy is going to be kinetic, so this will be maximum velocity here at (A). 0538

Over here at (B), it is not going to have any velocity. 0548

That is where all of its kinetic energy is converted into spring potential energy, so that will be 0 and (C) is the same way just on the other side.0551

There is no velocity at the end points.0558

It goes back and forth and for a split second it stops, turns around, speeds up, slows down, slows down, slows down, stops, speeds up, speeds up, speeds up, speeds up, slows down, stops, and back and forth as it goes on its oscillating path.0560

Let us take a look then at the force.0573

Well, while it is at point (A), the force is going to be 0 because there is no force to the spring on it because there is no displacement.0576

When it is at (B), it is going to have a maximum force, but it is a restoring force bringing it back the other direction, bringing it back towards its equilibrium position, so we will call that the negative max force.0585

And at (C) it is going to have the maximum force back to the right in the positive direction, so that will be maximum.0597

Because force = mass × acceleration, or acceleration = force/mass, the acceleration charge should look extremely similar.0604

At (A) there is no acceleration, at (B) it has its negative maximum acceleration, and at (C) it has its positive max acceleration.0612

Let us look at energy now. Spring potential energy over here is (U).0623

At (A), there is no displacement, so the spring potential energy must be 0.0631

It will have its maximum spring potential energy here at (B) and (C), so there is a maximum at (B) and (C).0638

If we want to look at kinetic energy, well that is going to be basically the inverse, right?0647

At (A), it is going to have its maximum velocity so it will have its maximum kinetic energy over there at (A), so maximum kinetic energy at (A), and at (B) and (C), it stopped at those points, it has no velocity for a split second there, so 0 and 0.0652

Let us take a look at what would happen if we made a graph of some of these.0671

I am going to start by taking a look at the displacement (x) and we are going to look at it as the object goes from (A) to (B) to (A) to (C) to (A) to (B) and back again.0675

Let us make this point (A) down all of these graphs, then we will draw another line all the way down here from (A) to (B)...0687

...another one here back to (A), and it goes to (C), and back to (A).0697

I think you quickly get the idea of what is going on here.0714

From (A) of course and back to (B) and so on, on its oscillating journey.0722

Well, if we look at displacement -- at (A) its displacement is 0, so anywhere we have (A) we can fill in the dots for 0.0730

At (B), its displacement is going to be positive (x), so it must come up like that and come back to (A).0738

At (C), its displacement must be negative (x) -- back to (A) and so on.0747

A graph of the displacement versus time.0754

If we want to do that for velocity, however, let us take a look and see what we know here.0759

For velocity, we know =at (B) and (C) we have 0, so anywhere we have (B) or (C) we must have 0 velocity.0764

At (A) we have maximum velocity and initially it is going to the right, so we will make that a positive velocity as it comes through.0773

When it comes back to (A), it has velocity going in the opposite direction at the same magnitude and back up and you can quickly see the pattern here for velocity of our spring-block oscillator.0781

Let us go to force and acceleration, let us put those over here -- a nice purple color, maybe for our force.0798

If we want to look at force, we know anywhere we have an (A), the force is 0.0810

So 0 here, 0 here, 0 here, and we have a maximum force when we are at (C), so let us fill that in -- maximum force over here at (C) and the negative maximum force when we are at (B).0815

So at (B) over here, we have negative maximum force, negative maximum force, and we can quickly plot our force versus time graph looking something like that.0832

And acceleration is just going to follow that -- force = mass × acceleration, so almost a mirror image of the graph, just different values but same shape.0848

Now, let us take a look and finish off our graph by looking at potential energy and kinetic energy versus time.0865

Let us look at potential energy in the spring first over here...0877

...there is (T), we have (A) here, goes to (B), back to (A), to (C)...0878

...from (C) we go back to (A), and from (A) we go back to (B).0895

All right. Potential energy in the spring is 0 at any of the (A)'s.0906

So there you go, a 0, 0 , 0; when it is at (B) it is at a maximum, so we will fill in our points there, and when it is at (C), it is also at a maximum.0913

Potential energy is scalar; it does not have a direction, so our graphs can look kind of like this.0925

When we look at kinetic energy, let us go there right underneath it.0935

Kinetic energy is going to have values of 0 at (B) and (C) when it is not moving.0942

So for (B), (C) -- (B) is at 0 and it is going to have maximum value when it is at (A), so it is going to look almost like the inverse of the potential energy graph.0947

Something kind of like that, and if we were to add the kinetic and potentials, they would add to a constant value because we are neglecting friction in this problem, using all conservative forces so we have conservation of mechanical energy.0967

Let us take a look at solving some problems with this and we will start off with a detailed harmonic oscillator analysis.0982

A 2 kg block is attached to a spring. A force of 20 N stretches the spring to a displacement of half a meter (0.5 m). 0996 Find the spring constant.0989

Well, we know that F = kx, therefore (k) must equal F/x or 20 N/0.5 m, which is 40 N/m.0998

The total energy is going to be the spring potential energy when it is at its maximum displacement or 1/2 kx2...1015

...which is 1/2 × 40 N/m (k) and its maximum displacement (0.5m2) for a total energy of 5 J.1026

About the speed at the equilibrium position, well at that point we have converted all of that spring potential energy into kinetic.1041

So we could start solving this one by saying that the spring potential energy is equal to the kinetic energy at the equilibrium position or 1/2 mv2 and that must equal that 5 J.1048

Therefore, the velocity = 2 × 5 divided by the mass, or 2 × 5/2 square root of 5 for about 2.24 m/s.1064

And how about the speed when it is at (x) = 0.3 m? 1083

Well to do that, we are going to have to look at an energy analysis again.1088

The total energy is the spring potential energy plus the kinetic, therefore the total energy = 1/2 Kx2 + 1/2 mv2.1092

And if I am going to try and find the speed, let us get (V) isolated; I will multiply both sides by 2...1107

...2 there and I will subtract the kx2, therefore mv2 = 2 total energy minus kx2.1115

Divide by (m) and take the square root to find that the velocity will be 2 × the total energy - kx2, all divided by the mass, square root, and finally I can substitute in my values. 1128

I have 2 × 5 J (total energy) - 40 (spring constant) × 0.3 (x-value), our displacement2 divided by 2 (mass) and the square root of that entire thing gives me a velocity, a speed at 0.3 m of about 1.79 m/s.1144

And I will do a quick check to see if that makes sense.1165

It should be less than the speed at the equilibrium position, the maximum speed, and it is less than 2.24 m/s.1168

So that makes sense. Excellent!1175

Let us go a little bit further with this one.1177

Let us try and find the speed now at x = -0.4 m.1181

We can use the same formula we just had, but plug in a different displacement value.1185

Velocity is going to be equal to 2 times total energy minus kx2 divided by (m), square root...1191

...that is going to be 2 × 5 J, (total energy) - 40 (spring constant) × -0.42 (new displacement/2 kg (mass); square root of all of that is about 1.34 m/s.1202

As it is a little closer to its full extension, it is slowing down even more -- less than our velocity when we were at 0.3 m. That, too, makes sense.1222

Let us find the acceleration at the equilibrium position.1231

Well when we were at x = 0, the force must equal 0 by Hooke's Law, and if the force is 0, then Newton's Second Law (F = ma) tells us that the acceleration must also be 0.1234

But if we want the acceleration at 0.5 m -- well to do that I am going to use Hooke's law (F = -kx), where (k) again is 40 N/m and a displacement at 0.5 m or -20 N.1249

We can apply Newton's Second Law again -- acceleration is force divided by mass, or -20 N/0.2 kg for an acceleration of -10 m/s2.1267

What does the negative tell you? 1283

When it is at its furthest positive displacement, the acceleration is back towards its equilibrium position, going in the opposite direction, hence the negative.1284

Let us take this one even further. Let us find the net force at the equilibrium position.1298

Well at the equilibrium position we already determined the acceleration was 0, so the net force there must be 0 N.1305

How about the net force at half a meter?1312

Well, we could use Hooke's Law again (F = -kx) or -40 N/m × 0.25 m (displacement) or -10 N. 1315

Where does kinetic energy equal potential energy?1330

Well if our total energy is 5 J, that is going to be the spot where the kinetic energy is 2.5 J and the potential energy is 2.5 J.1334

So if that is the case, we could figure that out -- the spring potential energy there is 2.5 J -- that is 1/2 kx2, where (x) is what we are solving for.1344

Therefore, (x)2 = 2 × 2.5/40 (K), and as I solve that then, that is equal to 0.125 and if I take the square root of both sides...1356

...(x) then equals the square root of 0.125 or about 0.354 m.1373

Pay special attention here -- note at this point where the kinetic energy and potential energy are equal is not midway between the equilibrium and the maximum displacement positions.1384

It does not work out that way, you cannot just guess halfway in between the two, you actually have to go through and solve; that is not halfway between the equilibrium and maximum displacement positions.1396

So let us take a look at circular motion versus simple harmonic motion and how they are related.1407

We have already talked about rotational motion for an object moving in a circle where we have some radius or amplitude of its motion (A) -- the angle θ measured from the horizontal, angular velocity ω and its position vector are, given by its (x) coordinate A-cos(θ), its (y) coordinate A-sinθ.1411

Well you could think of that almost as a projection down in one dimension to the spring-block oscillator system.1431

As you look at that system, imagine the object moving in a circle.1438

If you could shine a light down on it so you were just getting the projection in one dimension, you would see the exact same motion as what you see here in one coordinate.1443

Let us take a look at the (x) coordinates here -- (x) = A-cos(θ), but as we learned previously, θ = ω × (T).1452

Therefore we could write that as x = A-cos(ωT).1465

We also know that ω = 2π/period (T), so we could write that as x = A--cos(2π)/period × time.1472

Or going back to this equation, we also know that ω = 2π × the frequency, so we could write that -- replacing ω with 2π × the frequency as x = A-cos(2π), frequency, × time.1487

A bunch of different ways of looking at the same simple harmonic motion.1504

Instead of having our block start over here at a maximum displacement, what happens if we want to have it start here at x = 0?1509

Well we could use the sine function for that.1518

If we are going to have the block start at x = 0 at its equilibrium position -- start at x = 0 at time (t) = 0, then we could have the exact same equation -- we are just going to replace cosine with sine.1520

So x = A-sin(θ) or A-sin(ωt).1536

All it is, is a face shift, a sliding of the graph one way or another, depending on what you are calling your starting point.1543

As we look at graphing simple harmonic motion in that system, let us take a look at what happens to our graph of x = A-cos(ωt).1552

We will graph the (x) displacement and as we do that, we are going to look at it in terms of radians at 0, at π/2, at π, at 3π/2, 2π and so on.1567

I will make those marks on our graph right now -- π/2, π, 3π/2, 2π, and we will copy the same notches down here as well.1580

All right, for x = A-cos(ωt), we will do this as a function over here of ωt.1598

What are we going to graph?1604

Well, when our argument is 0, cosine here at times 0 is going to be 1, so we are going to get (A), so our maximum value here is (A).1606

As we get to π/2 right here, well our (x) coordinate is now 0, so we come down here to 0. 1618

As we get over here to 2 full π, now we are adding negative maximum displacement, -(A).1628

Let us mark that on our graph, -(A), and back to 3π/2, 2π again and the cycle repeats.1636

We get ourselves -- just like the curve we had expected -- we get our cosine curve.1645

Looking at that, if we wanted to use the sine function instead, x = A-sin(ωt), we are going to have the same maximum amplitudes of course -- (A) and -(A).1655

But we are going to start -- if we look at the y-coordinate or for the sine function -- we are going to start at 0 for our (y), so 0 at π/2, we are at maximum value here for the (y), so positive (A)...1670

...down here to 0 at π, down here to -(A) at 3π/2 and back to 0 at 2π -- so we get our sine curve.1687

But as you look at the graphs, notice they are really the same shape; they are just off set by that π/2 amount, so you can use either one depending on which starting point you prefer to work with.1698

All right. Let us take a look at an example where we are looking at the position of an oscillator.1711

A spring-block oscillator makes 60 complete oscillations in 1 minute or 60 seconds.1715

Its maximum displacement is 0.2 m. 1723

What is its position at time (t = 10 s) and at what time is it at position x = 0.1 m? 1727

Well let us start up here with question A.1735

If x = A-cos(ωt) and we know that ω = 2πF, we could write this as x = A-cos(2πFt).1738

If we have 60 cycles in 60 seconds, then we know that our frequency must be 1 Hz.1759

So if frequency is 1 Hz and we know that our maximum amplitude (A) is 0.2 m, we can fill in our function a little bit to say that X = 0.2 cos(2π) × 1 (frequency) × 10 s (time) or putting that all together I get about 0.2 m.1765

It is at its maximum displacement.1794

Moving down here to B, at what time is it at position x = 0.1 m?1797

All right, s = A-cos(2πFt), but now we are solving for the time, so this implies then that 2πFt must be equal to the inverse cosine of x/A.1804

Or if we want just (t) but itself let us isolate the variable we want to find -- t = the inverse cosine of x/A divided by 2πF.1827

Now we can substitute in our variables, so that will be the inverse cosine of 0.1/0.2 or 1/2/2π × 1 Hz (frequency) or about 0.167 s.1841

Now it is important to note here for an oscillating system, that is not going to be the only time it is in that position, but it is one answer to that question.1860

Let us take a look at a vertical spring-block oscillator system.1872

Once the system settles the equilibrium where we are hanging our block from a spring instead of having it on a horizontal surface, we are going to displace the mass by pulling at some amount either +A -- pull it down, let it jump up -- or -A -- lift if up a little bit, drop it and let it oscillate up and down.1877

This is a really slick derivation and a neat analysis.1893

If we looked at a free body diagram (FBD) when it is at its equilibrium position, we have gravity pulling down and the force of the spring, (ky) pulling it up, where (y) is the equilibrium point. 1898

So since it is at equilibrium at that point, we could write that the net force in the y-direction is going to be (mg), calling down positive, minus (ky) and since it is at equilibrium, let us say that that is mg - ky (equilibrium point), that must equal 0.1910

Therefore we could solve to say that the y-equilibrium point must be mg/k.1933

Now when we displace it by some amount (A), the net force in the y-direction, as we pull it down, is going to be mg - K × whatever that (y) would happen to be, which is going to be Y-equilibrium point plus that (A) amount we pulled it down.1941

We can distribute that through -- multiply that (k) through -- to find that it is mg - ky-equilibrium position - kA1964

Here is the slick part though -- Notice then as we do this that we had up here mg - ky-equilibrium = 0.1976

That means this part mg - ky-equilibrium must be equal to 0 and we could rewrite this then as FnetY = -kA.1989

There is our big result. What does that mean? 2005

That is the same analysis you would do for a horizontal spring system with a spring constant (k), displaced horizontally some amount (A) from its equilibrium position.2008

We just made this so much simpler.2018

In short, to analyze a vertical spring system, all you do is find the new equilibrium position of the system, treat that -- once you have taken into account the effect of gravity -- and treat that as if that is the only force you have to deal with in the system -- just the spring force.2021

So once you find its new equilibrium position you could almost pretend that you turned it on your side and it is a horizontal spring oscillator system again.2037

You do not have to continue dealing with that force of gravity.2044

A really, really slick way to analyze a vertical spring-block oscillator.2048

Find that new equilibrium position and then ignore the effects of gravity from there; treat that as your new equilibrium position, just find it using the effect of gravity first.2052

Let us take an example to make sure we have this.2066

A 5 kg block is attached to a vertical spring, with a spring constant of 500 N/m. 2069

After the block comes to rest it is pulled down 3 cm and then released.2075

What is the period of oscillation?2079

Well, period is 2π square root m/k or 2π square root 5/500, or 0.628 s -- that straight forward.2082

What is its maximum displacement of the spring from its initial unstrained position?2101

Well let us first look at it when it is at rest.2106

At that point we have (mg) down for our FBD, and we have (k) times -- let us call that displacement (d) at that point, so that the net force in the y-direction is 0, since it is at equilibrium because it is just sitting there, which implies that KD = mg or D = mg/k, which is 5 kg × 10 m/s2/spring constant (500) or 0.1 m.2110

So once you are hanging it there it hangs down 0.1 m.2144

Now you are going to go displace it; you are going to pull it down 3 cm from that equilibrium position.2148

If you then pull it down 3 cm, your maximum displacement in the y-direction is going to be that 0.1 m that you had from when it came to rest due to gravity...2155

...and that extra 3 cm that you added on as you pull it down or 0.13 m or 13 cm. 2170

A nice straight forward analysis, once you take into account and figure out what its new equilibrium position is with gravity and then ignore the effects of gravity and treat it as a standard horizontal spring-block oscillator system.2183

All right. Example 5 -- another fairly involved example.2199

We have a 60 kg bungee jumper stepping off a 40 m high platform.2203

The bungee cord behaves like a spring, of spring constant 40 N/m.2209

Find the speed of the jumper at heights of 15 and 30 m above the ground.2213

And as we do this, we are going to have to assume that there is no slack in the system.2217

Find the speed of the jumper at heights 15 and 30 m above the ground.2222

All right. So first thing, let us draw a diagram here.2227

Here is our jumper and the jumper is on a 40 m foot high platform, so down here somewhere is the ground.2231

We want the speed of the jumper 15 m above the ground, so we will call that position (A) -- that is 15 m above the ground.2243

Position (B) is 30 m above the ground, which means we must have another 10 m here -- 10, 15 m, 15 m.2255

All right. The key here is we are going to try and do this through conservation of energy at this point.2267

So the gravitational potential energy at the top must be equal to the gravitational potential energy at (A) plus the spring potential energy (A) -- UAG, UAS plus the kinetic energy at (A), 15 m above the ground.2273

Or mg × H-initial, the initial gravitational potential energy must equal the gravitational potential energy at (A), mg × 15 + 1/2 k.2296

At (A) the displacement is 10 + 5, so the spring is stretched 25 m, 252 (kx2) + 1/2 mv2.2312

All right. A little bit of math here.2325

We could then say that mass (60), G (10), and height (40)...2327

...must equal 60 × 10, = 600 × 15 + 1/2 × 40 (k) × 252 = 625 + 1/2 × 60 (mass) × v2...2336

or let us see here, that is going to be 24,000 = 9,000 + 12,500 + 30 v2.2355

Or 30 v2 will equal about 2,500, so I get a velocity of about 9.13 m/s at (A), so vA = 9.13 m/s.2373

We also need the velocity at 30 m above the ground. 2394

To do that then, we will follow the same basic idea but we are going to have different values for our height.2399

So we will have 24,000, our initial total energy, equal to mg × (height) 30 m + 1/2 k -- at (B) the spring is only stretched 10 m, so 102 + 1/2 mv2.2405

Therefore 24,000 = 600 (mg) × 30 + 1/2 × 20 (k) × 100 = 2,000 + 1/2 mv2. 2427

Therefore 4,000 = 30v2; and in solving for V, I get about 11.55 m/s.2449

Makes sense -- it is going a little bit faster at (B), a little bit slower at (A), hopefully slows down before hitting the ground.2460

All right, let us go a little bit further with this problem. Let us see what happens next.2469

How close does the jumper get to the ground?2476

Well, to do that we are going to have to figure out where the kinetic energy becomes 0.2482

So as we do this one, we will say potential energy due to gravity total equals the potential energy due to gravity at some point (C), which is where they stop, plus potential energy due to the spring at point (C).2491

No kinetic energy -- it is 0; that is where the person stops.2507

Following along with our calculations, 24,000 (total energy) = mgH + 1/2 k, and now how far has that person been displaced? 2512

That is going to be 40 minus whatever height is left, so 40 - H2.2525

So that implies then that 24,000 = 600 (mg) × H + 1/2 × 40 (k), so that is going to be 20 × 40 - H2.2532

That is starting to look like a quadratic equation, so let us get it into that form.2553

24,000 = 600H + 20 × -- well, if we square this, we get 1600- 80H + H2...2557

...or 24,000 = 600H + 32,000 - 1600H + 20H2.2576

We rearrange this to fit the quadratic formula, H2 - 50H + 400 = 0.2595

A couple of ways you can solve that, but the quadratic formula is probably my favorite.2606

I come up with a height of 10 m above the ground.2609

And let us just test that to make sure we did not make any mistakes here.2615

H2, 102, 100 - 50 × 10, 100 - 500 = -400 + 400 = 0.2618

So how close does the jumper get to the ground? 10 m.2627

Let us take a look at the pendulum again.2635

Now from the perspective of oscillations and simple harmonic motion.2639

Mass (m) is attached to a light string that swings without friction about a vertical equilibrium position.2644

For all of these, we are going to assume that this θ is relatively small.2650

Well, we have length (L) here again -- as it comes down here where there is all kinetic energy, its height has changed and we have derived a couple of times now that this height (H) is going to be L - L-cos(θ)...2655

...or this is L-cos(θ) compared to our entire length of our string (L).2673

Here we have all potential energy; here we have all kinetic and this (H) as we just said is going to be L - L-cos(θ).2685

If we take a look and start analyzing this with energy involved too though, again assuming a small θ, here we have all potential energy -- U = mgH, which is mgL × 1 - cos(θ).2696

Here it is all kinetic energy, and back here again it is all potential.2716

And if we wanted to find the velocity of our pendulum when it is at this lowest point, well we could say that the kinetic energy at the bottom must equal the potential energy at its highest point -- the top.2723

Or 1/2 mv2 = mgH where (H) is L - L-cos(θ), and we can divide now our masses.2735

So v2 = 2gL × 1 - cos(θ) or just velocity itself -- I take the square root and that will be the square root of 2gL, 1 - cos(θ).2748

Going a little further here, let us look at some other quantities that might be of interest.2765

Here at the highest point, you have the maximum force, you have the maximum acceleration, you have the maximum gravitational potential energy, but you have no kinetic energy and no velocity.2769

Here on the other hand, you have no force, (F = 0), your acceleration = 0, your gravitational potential energy = 0 -- assuming that is what we are calling 0 in this problem which should make sense.2786

Our maximum kinetic energy occurs there and we have maximum velocity there.2799

And what is causing our restoring force to put this in simple harmonic motion?2805

Well our restoring force is going to be based on the gravitational force pulling this down.2810

If we look right here, we have its weight pulling it down, but that is not what is causing the displacement; it is only a portion of that.2815

The portion that is going to be perpendicular to our string, mg-sin(θ) here is what is causing our restoring force.2826

So another way that we could graph this in terms of energy, is if we looked at energy on the y-axis versus (x) position on the (x), we know of course the total energy must remain the same. 2844

We are dealing with conservative forces, conservation of energy, and we are not dealing with friction at this point.2854

Here at 0 displacement, everything is kinetic energy.2860

So as we draw this U-shape, anything above the U-shape between the E (total line) and our parabola is kinetic energy.2865

Anything below it is potential.2873

The two always sum up to the total energy.2875

So if I wanted to look at another point on the graph -- let us say we wanted to look right here.2878

In this case anything up here would be kinetic and down here would be our potential.2884

Use the graph that way -- what is above the line is kinetic, what is below is potential.2895

Another way of representing the same information; it is that important.2899

So what happens when we are talking about this pendulum and we have to start dealing with frequency and period?2905

Well, the period of an ideal pendulum is 2π square root of L/g.2910

The length of the pendulum is your variable. 2915

The mass on the end does not matter, the length is what matters.2918

In your grandfather clocks at home, the length is all set, that is why those are so big.2922

Or the frequency is just going to be 1/period or 1/2π square root of g/L.2929

Now for all of these, again, we have to assume that θ is small due to a small angle approximation in the mathematics.2934

So as we are looking at this, let us take a look and see what would happen if we tried to graph period versus length.2941

If you did that you would probably get something that looks kind of like that because period is proportional to the square root of (L).2951

So if you wanted to get a nice linear graph that you could do something with, if you wanted to try and determine something like the acceleration due to gravity, for example.2960

You could take a pendulum, graph the period versus the square root of the length and you should get a nice linear graph and if you take the slope of that, the slope is going to be rise/run, which is going to be T/square root of L, which turns out to be 2π/square root of (g).2971

So if you want to go to the moon, figure out what the acceleration due to gravity is, make a bunch of different pendulums of different lengths, have them go back and forth, measure their periods, come graph the period versus square root of the length, find the slope and you can calculate the acceleration due to gravity that way.2995

Let us take a look at an example.3015

We have a 1 kg mass suspended from a 30 cm string that creates a simple pendulum.3017

The mass is displaced at an angle of 12 degrees from the vertical equilibrium position.3023

First thing we have to do, is if we are going to use any of these formulas, is to make sure that θ is small, and 12 degrees is small enough for our purposes.3029

Find the frequency and period of the pendulum.3037

Well, the period is 2π square root L/g, or 2π square root (L) 0.3 m/10 m/s2 (g)...3040

...or about 1.09 s and frequency then is just 1/period or 1/1.09 s, which is going to be about 0.92 Hz.3056

All right, find the height of the pendulum above equilibrium when at maximum displacement.3070

Well we know the height is L × 1 - cos(θ), so that is going to be 0.3 × 1 - cos(12 degrees) or about 0.0066 m.3076

And find the speed of the pendulum at the equilibrium position.3093

V = square root of 2GH if we want to use conservation of energy, which is square root of 2gL, 1 - cos(θ)...3097

...we have derived that a couple of times at this point, or the square root of 2 × 10 (g), (l) 0.3 × 1 - cos(12 degrees) or about 0.632 m/s.3107

Carrying this one a little further -- Find the restoring force at maximum displacement.3130

All right. The force at maximum displacement -- we just said was mg-sin(θ), the component of the weight pulling it back down -- mg-sin(θ), which is going to be its mass of 1 kg × 10 m/s2 (g)-sin(12 degrees), or about 2.08 N.3136

And how about the tension in the string at the equilibrium position?3158

Well at that point we can make our FBD.3162

There is our tension; there is our weight -- if it is in equilibrium, those must match -- or net force in the centripetal direction is mAC which implies that (t), which is in the direction toward the center of the circle... 3166

...tension - mg = mv2/r, which implies then that the tension is mg + mv2/r.3181

Therefore, the tension must be (mass) 1 × 10 (g) + 1 (mass). 3193

Our velocity we found was about 0.362 m/s2 over our radius (0.3 m) or about 10.44 N of tension in that string.3201

How long should the pendulum be in order to keep perfect time with a period of 1 second?3220

Well let us start there and we are going to assume again that it is an ideal pendulum, no friction and everything is perfect, and no mass in the string.3227

Period is 2π square root L/g and we want a time, a period of 1 s, so we are solving for (L).3235

Let us take t2 = 4π2 (L/g).3247

Therefore L = gt2/4π2, which implies then that the length (L) should be G (10 m/s2)...3254

...with a period of 1 s2/4π2 or about 0.253 m, a quarter of a meter.3270

How long should the pendulum be if the period is to be half a second?3285

Well, same thing -- let us just plug in a couple of different values here.3289

L = gt2, so (g) × 0.52/4π2, where g = 10 m/s2, which gives us about 0.633 m.3293

A lot shorter. A lot shorter -- one-fourth.3308

Let us take a look then at a pendulum on the moon.3313

How long must a pendulum of period 1 second be on the moon if the acceleration due to gravity on the moon is about 1.6 m/s2?3316

Well we can use our same formula, L = gt2/4π2, where (g) = 1.6 m/s2 on the moon × our period of 1 s2)/4π2, which is about 0.405 m.3325

All right. Doing great. Hang in there. One last sample problem.3355

Mass (m) is placed on a horizontal frictionless surface and attached to a spring with spring constant (k).3360

The mass is pulled back a distance (x) and released to oscillate horizontally.3368

What is the kinetic energy and potential energy of the mass at a displacement halfway between the equilibrium position and maximum displacement?3373

Well, let us draw a picture first of what our situation is going to look like.3382

Horizontal spring-block oscillator -- let us color that in there.3386

We will put our mass over here (m), some spring with spring constant (k), and we will start this at some displacement 0.3391

Here is our (x) and it is somewhere in there we are going to have a point (A), and what we know is at the maximum energy is 1/2 kx2.3405

The potential energy due to the spring at (A) is going to be 1/2 × (k) -- well (A) is halfway between these two, so that is going to be at x/22.3420

That will be 1/2 k × x2/4 or 1/8 kx2, which is 1/4 × 1/2 kx2.3438

Why did I write it that way?3451

Well, 1/2 ks2 is our maximum spring potential energy, so this then says that UA must equal 1/4 of UMax.3454

So at (A) it has 1/4 of its maximum energy, so where is that other 3/4 of the energy? 3471

That has to be kinetic energy at (A), so that must be 3/4 of the maximum spring potential energy.3476

Notice here that halfway between the two, the spring potential energy and the kinetic energy are not the same -- they are not equal.3485

You have to go through the steps to go solve for points in between those two.3495

Do not take shortcuts.3499

Hopefully this is a good start for simple harmonic motion.3501

Thank you so much for your time and for watching Make it a great day everyone!3504