AP Physics C: Mechanics Oscillations

Hello, everyone, and welcome back to www.educator.com.0000

I'm Dan Fullerton and in this lesson we are going to talk about oscillations and simple harmonic motion.0003

Our objectives include analyzing simple harmonic motion in which the displacement is expressed in the form a sin ω T or a cos ω T.0009

Recognizing simple harmonic motion when expressed in differential equations form.0018

Calculating the kinetic and potential energies of an oscillating system.0023

Analyzing problems involving horizontal and vertical masses attached to springs.0027

Finding the period of oscillations for systems involving combinations of springs and deriving the expression for the period of a pendulum,0032

both an ideal pendulum and now for the first time, a physical or real pendulum.0040

Simple harmonic motion is motion which the restoring force is directly proportional to the displacement of the object.0045

The more you displace it, the more restoring force there is trying to bring it back to its initial position.0052

The reason this is so important is, in general, nature's response to a disturbance is some sort of simple harmonic motion.0057

You can see it all over the place, the blade of grass watch it come back up, simple harmonic motion.0065

Or a snowy tree that swayed down a limb with lots of snow as the snow falls often you see the branch go back and forth.0072

All of these responses have restoring forces proportional to the displacement of the object or at least that is a good starting model in a simple harmonic motion.0079

We can even see it in the atoms of an object.0090

When they are compressed or stretched, it will vibrate back in simple harmonic motion.0092

Let us take a look at simple harmonic motion and start off with an analogy to circular motion.0098

Let us assume that we have some mass moving in a circle of radius a, it is an angular velocity ω, that at a given point in time,0101

its position vector is given by a cos θ a sin θ, where a is the radius or the magnitude, and cos θ that is the angular displacement.0113

That is our x coordinate and there is our y coordinate.0123

We are going to compare this to a system of the spring attached to a wall with the mass on the end.0126

We are going to pull the mass of displacement a from its equilibrium, our happy position, and let it go back and forth.0133

What is really amazing about this analogy is, if you we are to pull this to a and let it go,0141

you can compare its x position to what you would get is this object goes around the circle its x position.0146

As the circle goes around at any given point in time, when it is over here, the mass is going to have that same x component all the way around.0153

You get this nice analogy between what we are already familiar with circular motion and this mass moving on a spring in simple harmonic motion.0160

Let us start out by taking a look at the angular velocity there we know is the time rate of change in the angular displacement.0171

But if I rewrite that a little bit, if I separate our variables, we can write that ω dt = d θ.0180

If I integrate both sides from some T = 0 to some final time T, integral from θ = 0 to some final θ,0187

that implies what angular velocity is constant in uniform circular motion so the left hand side becomes ω T.0198

The right hand side just becomes θ, θ is given by ω T, that will come useful later.0205

If we go to the down, we want to analyze our spring block system, we can look at it from terms of Newton's law,0212

from the perspective of Newton's law F = ma and that force is restoring force – kx.0219

We also know that acceleration is the second derivative of x with respect to T².0225

This means that Md² x dt² = - kx or I could write this in a more common differential equation form d² x / dt² + k /m × x must equal 0.0233

We have a second order differential equation or the second derivative of a function + a constant × that function gives you 0.0252

Only one way I can think of to solve these sorts of things and that is the sin or cos, the only functions where their second derivative added to themselves can give you 0.0260

The general form of our solution, x is a function of time is given by a, our amplitude cos ω T.0270

Where we are going to find ω is √k/m or if we look here in our equation that piece right there, that is ω².0280

Let us take a bit further, as we look at position velocity and acceleration.0292

We started off with θ = ω T and we said that x is a position of time is a cos ω T, where there is our angular frequency ω.0299

Our velocity then is the derivative of x with respect to T which is going to be the derivative with respect to T of a cos ω T.0314

It is going to be equal to the derivative of cos is the opposite of the sin.0328

We are going to get –ω A sin ωt where our maximum velocity is going to occur, while the maximum value of the sin function is 1.0332

Our maximum velocity is going to be ωa.0345

Let us take a look at acceleration which is the derivative of velocity with respect to time or the second derivative of x with respect to time,0350

which is going to be the derivative with respect to T of – ω A sin ω T.0359

The derivative of the sin is the cos function.0371

We are going to get –ω² A cos ω T.0373

We are going to have maximum acceleration when the value of the cos function is 1 so that is going to be ω² A for maximum acceleration value.0381

We can keep going with these to find position velocity acceleration.0392

As we are doing this, oftentimes we are talking about frequency in period.0397

Probably, it is worth bringing this up again.0400

Frequency is the number of cycles or revolutions per second.0403

Its units are 1/s or hertz.0408

Period T is the time for 1 cycle or 1 complete revolution and the units are seconds.0410

We can find period when we know frequency.0417

Period is 1/ frequency or frequency is 1/period.0419

If we talk about angular frequency, angular frequency is the number of radians per second.0425

It corresponds to the angular velocity for an object traveling in uniform circular motion.0431

Note that angular velocity and angular frequency is not the same thing but they do have a strong correspondents.0435

Angular frequency ω is 2π × frequency in Hz or 2 π ÷ period.0442

You can also write that period is 2 π ÷ angular frequency.0449

Let us do an example, an oscillating system is created by releasing an object from a maximum displacement of 0.2 m.0456

The object makes 60 complete oscillations in 1 min, determine the objects angular frequency.0464

Angular frequency is 2 π × the frequency in Hz which will be 2 π × 60 Hz.0472

Pardon me, 2 π 60 complete oscillations in 1 min, that is 1 oscillation/s × 1 Hz, which is just going to be 2 π radians/s.0482

What is the objects position at time T = 10 s?0497

Position is a cos ω T which implies, since we know that ω is 2 π radians/s and our maximum displacement A is 0.2 m, the amplitude.0501

This implies then that x = 0.2 cos 2π T, which implies then if T = 10 s that x is going to be equal to 0.2 × the cos 2 π × our time of 10 s or about 0.2 m.0521

At what time is the object at x = 0.1 m?0551

We have to realize it is going to be oscillating back and forth.0555

There could be more than one answer here but let us solve for one answer.0558

That x = a cos 2π T is our function which implies then that the cos of 2 π T = x / a0561

which implies then the 2 π T = the inverse cos of x / a or T is going to be equal to the inverse cos of x/ a ÷ 2 π.0576

We substitute in our values that will be the inverse cos of 0.1/0.2 ÷ 2 π or I get 0.167 s.0594

Alright, let us take a look at a little bit more detail of a mass on a spring.0613

Here we have our mass M attached to some spring with spring constant K attach to a wall and its move to a displacement A from its equilibrium position0619

and released allowing it to oscillate back and forth between - a and a.0629

You are going to assume no thing or no loss of energy to friction.0633

If we wanted to know the period of our spring, that is going to be 2 π √ M /K, something we will derive a little bit later.0637

Or the frequency of our spring is 1 /period and which is going to be ½ π √K/M.0648

We can arrange this a little bit further, multiply both sides by 2 π to say that 2 π × the frequency = √K /M is 2 π F is what we call that angular frequency.0659

Therefore, ω your angular frequency is the √k/m.0683

Let us analyze a spring block system.0692

A 5 kg block is attached to a 2000 N/m spring as shown in this place to distance of 8 cm from its equilibrium position before it released.0696

Determine the period of oscillation, the frequency, and the angular frequency for the block.0705

Let us start, it asks for period first so let us start there.0712

A period of our spring is 2π √M/K, which is going to be 2π √5kg /2000 N/m or about 0.314 s.0715

Next, it wants the frequency.0737

Frequency is 1 /period which will be 1/0.314 s or about 3.18 Hz.0739

Finally, the angular frequency for the block, ω is 2π × frequency which is 2π × 3.18 Hz or 20 radians/ s.0752

Another example, rank the following horizontal spring resting on frictionless surfaces in terms of their period from longest to shortest.0773

When we look at typically of different masses and spring constant, the way I start this is I would probably look0783

at the relationship and recognize first before looking for period.0788

The period is 2π √M/K so we could then make a table of information.0793

We will have our 4 systems, we have A, B, C, and D, and we need to know their mass, the spring constant and M/K.0801

As I start to look at these, our masses for A is 10, for B we have 7, C is 2, and D is 5.0825

Their spring constants, A is 500 N /m, B is 50, C is 2000, and D is 1000.0834

The ratio M/K it is pretty easy to calculate.0843

It is going to be 0.02, 0.14, 0.001, and 0.005.0847

We are looking for the periods from longest to shortest, since period is proportion to √M/K,0856

we are going to get the longest period where we have the greatest M /K.0863

I would rank these for period, our biggest M/KB, then A, D, C will have the shortest period.0867

As we talk about simple harmonic motion, the general form of simple harmonic motion is something where0880

we had the second derivative of function with respect to time + the constant ω² × that original function equal to 0.0885

When we do this, we find that our general solution is a cos ω T + some phase angle, some phase shift having to do0895

with where you are starting your cos function or sin function.0905

We are not going to worry a whole lot about that here in this course and note that ω here is your ω there.0910

When you go to graph this, if you have a cos function, it is going to look something like this at ω T= 0 × 0, your maximum displacement.0921

You oscillate back and forth between maximum and minimum displacement.0930

For using the sin function where you got a phase shift of 90°, you still have the same basic shape.0934

You are still oscillating for maximum A to – A.0939

Let us take a look at energy of simple harmonic motion.0949

When an object undergoes simple harmonic motion, kinetic and potential energy are both peering with time.0952

Although, the total mechanical energy, kinetic + potential remains constant.0957

If we have something like a mass spring system and we are going to compress it, the work we have to do is going to be the integral0963

from x to some 0 of f(x) dotted with dx, which is going to be the integral from x to 0 of – kx dx,0972

which will be - kx² /2 evaluated from x to 0, which is going to be ½ kx² which is the potential energy stored in the spring.0984

You have done that work on it that must be the potential energy in the spring.0996

If we wanted to look at x as a function of T, we know that a cos ω T which implies then that the potential energy in the spring as a function of time1000

is going to be ½ K × x, a cos ω T², which will be ½ K A² × √cos ω T.1013

Let us go take a look now at velocity.1034

Velocity is a function of time, is the derivative of x with respect to T, which is –ω A sin ω T.1036

Therefore, the kinetic energy of the system is going to be ½ MV² will be ½ × mass × velocity² is going to be - ω A sin ω T².1047

Or as we multiply through there, kinetic energy will be ½ M ω² a² sin ω T, square the sin ω T.1065

We know that ω² is k/m so then k if I replace ω² with K/M, our kinetic energy k is ½ k or spring constant × that amplitude² sin² ω T.1081

Let us put this all together.1100

We got our kinetic energy, we have got our spring potential energy, let us put that together to find your total energy.1101

E is kinetic + potential which is going to be our kinetic is ½ ka² sin² ω T + our potential ½ ka² cos² ω T.1110

Remember, our trigonometry sin² θ + cos² θ is equal to 1.1134

We can then factor that out to say that our total energy then ½ ka² × (sin² + cos² )is just going to be ½ KA².1142

No dependence on time because we have a constant total mechanical energy.1155

Alright, let us go back to our horizontal spring oscillator again.1163

We have mass oscillating a spring, spring constant K, force is mass × acceleration which is – kx, but a = the second derivative of x with respect to T.1168

We have md² x/dt² = - kx or d² x/ dt² = k/ mx² or + k /mx² equal 0.1184

We have mentioned before that was ω².1202

Our general solution for the horizontal spring oscillator system now looks like x(t) = A cos ω T + some phase angle.1205

Again, we are not going to have to deal with that phase angle here much, it is going to be 0 as we set up most of our systems,1217

where ω = √K/M so this works the same solution for a spring block oscillator or the period of the system is going to be 2π /ω which is 2π /√k/m we or 2π √m/k.1223

The same basic solution but now we have used this to derive the period of our spring block oscillator like we said we would a few slides ago.1248

What about a vertical spring block oscillator?1257

We have a mass, we are going to hang from the spring and we are going to let it settle to an equilibrium position1260

and then we are going to pull it down some amount A and displace it.1267

How do we deal with that where we also have this included effective gravity?1270

Let us see, we draw our free body diagram first for a mass, we will have some force of the spring which we will call ky and some force of gravity MG.1276

We are calling down the positive y direction so when I write my Newton’s second law equation.1289

The net force in the y direction that is going to be MG - ky and all of that must equal our mass × acceleration in the l direction.1294

If you are letting this hang down to the equilibrium position here, that means at this point MG - ky equilibrium must equal 01307

because at that point there is no net force and no acceleration.1317

Or we could then write that the equilibrium point on the y axis is just MG/k.1321

We have got that figured out but now we are going to take the block, we are going to pull it down or up and let it go in displacement and see what happens.1332

Our analysis says that the net force in the y direction is going to be MG - k where now our y is y equilibrium + the displacement A.1340

If we multiply through here, that means MG - ky equivalent –ka = our net force.1355

Let me just fill them in here.1367

Take a look, MG - ky at the equilibrium point is 0, we have the same thing right there net 0.1369

The net force in the y direction is just – ka, the exact same analysis we would be doing if this is a horizontal spring block oscillator without the effect of gravity.1379

All we have to do to deal with this vertical problem is find its equilibrium position once its hanging there and then treat it as if it is a horizontal spring block oscillator.1392

It just got so much more simple.1404

We will do that as sampled problem here shortly.1407

We can also have a couple of springs where they are lined up in different ways, combinations of springs.1411

If they are in series, such as this, we have k1 attached to k2, attached to a mass, we can analyze that and1417

it is nice to figure out how we could treat this as if it was just one equivalent spring.1423

Let us take a look here, force is going to be - k1x1 and by Newton’s third law that has to be equal to - k2 × the displacement x21430

which implies then the x 1 is going to be equal to k2 /k1 x2.1443

If we want to treat this as an equivalent spring, F is going to be equal to - some k equivalent × the total displacement of our block x1 + x2.1451

We can take this x1 and put it in here for x1 to find that our force is going to be equal to -k equivalent.1463

Our x1 is k2/k1 x2 + x2 not x² + x2 which implies then, since we also know our force = - k2 x2 Newton’s third law there,1474

that we could write that - k2 x2 = - k equivalent × x2.1497

We will factor that out and I'm left with k2 /k1 + 1, which implies then that k2 = equivalent k × k2 /k1 + 11507

which implies then we could write this as 1/k equivalent = 1/k1 + 1/k2.1527

You can follow that same form for any number of springs.1537

1/the equivalent spring constant is 1/ the first for spring constant + 1/ the second spring constant + 1/the third spring constant, for many springs as you might have.1540

Kind of like capacitors in series or resistors in parallel if you have done ENM.1553

You can maybe guess already what springs in parallel are going to look like.1559

They are going to look like resistors in series or capacitors in parallel.1562

But let us take a look and show that derivation, for springs in parallel now we have k1 and k2,1566

both attached to our mass M which is going to be displaced some amount x.1572

We can analyze that by looking at the force will be k1 x + k2 x, which we can say is k1 + k2 × x or k1 + k2 x is equal to some equivalent k × x.1577

Pretty easy to see, k equivalent is k1 + k2.1601

We just add up our individual spring constants to get our equivalent spring constant when they are in parallel.1604

We have spent some time on springs and we will come back to those 2.1614

We also see simple harmonic motion and pendulums.1617

Let us review what we know about the pendulum already.1620

If we have a pendulum here attached to a light spring, a mass on the end, and we will call this entire distance in the string length L.1623

Our pendulum swings back and forth, it is going to come up some amount H, which we will call L - L cos θ or L1 – cos θ.1632

This again is our length L and we know that when we are in our highest points here, we have our most gravitational potential energy.1648

We are down at the lowest points, when it is down there, it is kinetic energy.1655

We have talked about that energy transformation.1660

We can look at that in a little more detail though.1666

If we take a look at these different points, at their highest point here, our gravitational potential energy is MGH, its height above, its change in y position from here to here,1669

which is going to be MGL × (1 - cos θ).1682

Here all its energy is kinetic energy, back up here, we have all potential energy.1689

When we are done here, if we want to know the speed, the kinetic energy at the bottom has to equal the gravitational potential energy at the top.1697

½ MV² must equal MGL × (1 - cos θ).1705

Or rearranging V² is going to be equal to 2 GL × 1 - cos θ or V will be equal to √2 GL 1 - cos θ, when we are at that point right there.1714

As we also look at this from different perspectives, when we are at our highest point here, we have the maximum force on our mass on the string.1733

We have our maximum acceleration, we have our maximum gravitational potential energy that are kinetic energy 0 and our speed is 0.1743

Here, where the highest point of kinetic energy, the force on our mass is 0, acceleration is 0, gravitational potential energy 0,1754

but we have our maximum kinetic energy and our maximum velocity.1766

A lot of times you will see this depicted in a graph of energy showing a constant total mechanical energy as you go to higher and higher displacements.1772

B distance that is under the graph is our potential energy, the distance from your total energy to the top of a graph is your kinetic energy.1781

That total always = e total K + U.1790

But you have that shift across the different types of energy.1793

Taking a look at over here, we know our force MG is down, the force in the direction of displacement is just going to be MG sin θ,1797

where once again we have this as a restoring force leading this into a discussion of simple harmonic motion.1807

Let us look at the period and the frequency of a pendulum.1816

We have our pendulum like we have talked about previously.1820

The period of the pendulum is 2π √its length ÷ G assuming it is an ideal pendulum.1822

All the mass is at the very end, you have a light string or frequency is 1 /period which is going to be 1/2π √G /L.1829