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

You can even go and plot these if you wanted.1839

Period vs. Length of your pendulum and we are probably going to get some kind of roughly this shape, where T is proportional to √L.1841

If you want to linearize your graph, T and √L, to get something that is fairly linear.1850

If we did that, what happened to our slope?1862

Our slope is just going to be 2 π ÷ √G.1867

Ever want to know the acceleration due to gravity on the Moon?1874

Here is a great way to find that out.1877

Go up to the moon, take a bunch of pendulums, different lengths, measure their periods compared to their length, plot them and take the slope.1879

Slope = 2π /√G, you could then go calculate G, the acceleration due to gravity on the Moon.1887

How do we get this 2π √L /G?1897

Let us see if we can derive that.1900

Here, we have a simple pendulum, an ideal pendulum.1904

We have a mass M on the string about some point P, we have a very light string, and all of the masses here at M.1907

There is no friction just the ideal case.1914

If I look at the forces here, I have the force of gravity MG down on my mass and I am also going to define distance1918

from our point there q to our mass as the position function from point B.1927

Their length is L.1935

If we look at this from the torque perspective, torque about point P is our P crossed with F, which is going to be our P crossed with MG our force.1938

Which implies then that the magnitude of our torque about point B is going to be MGL sin θ,1955

which implies then that - MGL sin θ = our moment of inertia about point P × angular acceleration.1968

Why that negative? If you take a look as θ is getting smaller, we are going to larger values of angular acceleration.1978

So we have to take that into account there.1985

We are going to use what is known as the small angle approximation.1988

For small angles of θ, they are under about 15°.1992

The sin θ is a very close the θ itself.1995

This will allow us to simplify our analysis and write this as - MGL θ = the moment of inertia about point B × α.1999

We also know that α is the second derivative of θ.2011

We can write this as - MGL θ is equal to our moment of inertia P × the second derivative of θ with respect to T.2018

We have a second order differential equation.2030

Let us put into that standard form so we can see a little bit more clearly, d² θ / dt² + we will have MGL / the moment of inertia about point P × θ = 0.2032

By the way, if you remember our form dist is what we called ω².2052

The solution to our problem, our general solution is θ is going to be equal to a cos ω T.2059

The moment of inertia of a single mass at the end of the string, our moment of inertia about point P is going to be ML².2067

That means that ω² will be MGL /ML² IP which is going to be G /L which implies then that ω = √G /L.2076

If we want that period, period is 2π / ω which is going to be 2π ÷ √G/L or 2π √L /G.2093

We derived the formula for the period of a simple pendulum.2110

That is easy, what about real pendulums?2116

Alright, let us try that.2120

Here, we have a non ideal pendulum, a pendulum now it is made out of a rod where it has a center of mass in the middle.2122

We are going to have a pivot point P there and the distance between our pivot point and our center of mass we are going to call D.2129

We can use the same basic analysis in order to find its period.2136

Net torque = R cross F, we know θ is going to be equal to a cos ω T.2140

We have already done that analysis or ω² if we go through all of that again is MGD /our moment inertia about point P,2150

which implies then that ω is √MGD /our moment of inertia about point P.2161

The difference here is we have a different moment of inertia.2169

To find our moment of inertia about point P, all that is going to be the moment of inertia about our center of mass.2175

We can use the parallel axis theorem to shift our pivot point some distance and D away from the center of mass, that will be + the mass of our pendulum bar × D².2181

The mass of our rod rotated about its center is ML² /12 + MD².2196

If period is 2π / ω that is going to be 2π / ω.2208

Here, our √MGD/√IP, this is going to be √IP is ML² /12 + MD² /MGD,2214

which implies then that our period is going to be equal to 2π.2231

We can factor an MN out of all of that and I would have L² /12 + D² /GD or2235

if I wanted to make that a little bit prettier, we can write that as 2π × √L² + 12 D² ÷ 12 GD.2247

If you want to test this out, I lot of my students quite regularly is to take a meter stick.2264

Find its center of mass, drill a hole in it, near one of the end some distance from that center of mass measure that2270

and then put it on a hook and swing it back and forth.2278

Let it go at about 10, 20 times, measure the entire time for 20 revolutions, 20 periods, and divide by 20,2281

and see how that does compared to when you plug it into the equation and you will find it is very accurate.2287

Alright, let us go back to our spring block system for a summary here.2295

If we want to take a look at our spring block system with displacement velocity, force acceleration, potential energy in the spring2299

and kinetic energy and a bunch of different positions as we go from A-B to A-C and back again is we are oscillating some displacement x and –x.2307

Sometimes it is useful to graph all of these just because you see them come up so often.2316

Let us make a graph, we will start here by saying on the 0 point on our time axis, we will call when we have position A,2321

then we will pull the B to A to C and back again.2330

I'm just going to plot some of these lines in here very loosely so that we have a common reference frame.2333

I think we will do it there and let us do the same thing here for our couple of energy graphs.2359

If this is position A, that will be B, back to A to C to A back to B again.2384

The same thing here, A to B back to A, C to A to B again.2393

We will have graphs of let us see.2401

Let us start with our displacements.2404

We will have this as our x.2406

At time 0, when its position A our displacement is 0.2409

At position B, where there are maximum displacement which we call x or typically capital A.2413

Back to A 0 to C to A, back to B.2421

Our plot would look something like that.2426

In that position A, where our 0 displacement at B our displacement is x.2433

At C, it would be –x.2438

We can take a look now at velocity which we know is the derivative of x.2441

As we look at that, we know that at A where maximum velocity but it points B and C we are always going to be at 0.2447

We can fill those points in and we know that 2 because if we take the derivative of position that will be the slope.2455

We have a 0 here, we have a 0 there, we have a 0 slope there, so we have 0 values for velocity there.2460

We are going to start at A and the maximum velocity come down to a minimum or maximum speed in the opposite direction2467

and end up with a graph that looks kind of like that.2477

How about if we take a look at the force?2482

If we look at the force, we are going to have, let us fill in our table here toward velocity.2486

We have maximum velocity at A, at B we have 0 and at C we have 0.2492

Now looking at force, when we are at A we have 0 force, so anytime we are at A we can fill 0.2500

When we are at B, we have maximum force but in the negative direction so we will have a negative there.2507

We will have a positive here, when we are at C maximum force back to the right restoring force and so on.2514

We can graph our force working something like that.2520

And our acceleration of course, should follow the same thing, the same general pattern by Newton’s second law.2527

They should tolerate along.2539

We will have 0 force and 0 acceleration when we are at point A.2540

At point B, we are going to have the negative maximum for force and acceleration going back toward the left.2544

At C, we will have our maximum positive force and acceleration.2552

And again, if we want to look at our derivative slope relationship, look at the velocity here at A.2558

The slope of that is 0 therefore, we have 0 acceleration right there as well and force correlate right along with that.2565

How about potential energy and kinetic energy?2573

As I look at potential energy, we know when we are at A that is going to be 0 so we can fill that in right now.2576

A for potential energy is 0 there, we will have a 0 at A.2584

At B and C, we are going to have our maximum values.2590

We do not have a negative because this is a scalar.2594

We will have a graph that looks something like this.2597

When we look at kinetic energy, that one with our table here, we are going to have maximums at B and C for potential energy.2608

For kinetic B and C, for that split second, it stopped.2619

Kinetic will have 0 here and we will have our maximum when we are at position A.2623

We can draw this, we will put our 0 in here to make our drawing a little bit simpler.2629

We are just going to come down and up, the complimentary graph to our potential energy function.2634

You get a feel for how the graphs of all of these would look in simple harmonic motion and how they all relay to each other.2644

Let us do an analysis of the simple harmonic later.2653

We have a 2 kg block attached to a spring, a force of 20 N stretches a spring to a displacement of ½ m, find the spring constant.2658

We can do that using F = kx or k = f /x which is 20 N /0.5 m or 40 N/m.2667

If we want the total energy, our total energy is going to be the total energy stored in our spring when we are at maximum displacement2683

or ½ kx² which is ½ × 40 N /m × maximum displacement 0.5 m² or about 5 joules, the speed at the equilibrium position.2691

To do that, the potential energy of the spring when it is at its maximum displacement must be equal to the kinetic energy.2710

When it is at 0 displacement, which is ½ MV², all of that must equal 5 joules which implies then that V must be equal to √2 × 5/2,2716

which is just going to be √5 or about 2.24 m/s.2731

If we want the speed at 0.3 m, let us do this from a total energy perspective again.2739

That is going to be the spring potential energy + the kinetic energy all has to be that constant 5 joules.2747

But that is going to be ½ kx² + ½ MV² which implies then that MV² is going to be equal to 2 × our energy total - kx²,2753

which implies that V is going to be equal to the √2 × our total energy - kx² ÷ M,2768

which implies that V = 2 × 5 joules - our spring constant 40 in our displacement 0.3 m² ÷ our mass of 2 kg.2780

The square root of that whole thing gives me 1.79 m/s.2793

Moving on, find the speed when it is at -0.4 m.2805

We can use the same formula, the same analysis just with different values, V = √2 × our total energy - kx² /M2811

which will be 2 × 5 joules - spring constant 40 N/m × displacement -0.4² ÷ our mass 2.2822

Square root of all of that is 1.34 m/s, the acceleration of the equilibrium position.2835

Remember, at equilibrium, at x = 0, our force is 0.2846

Therefore, our acceleration is 0.2851

How about the magnitude of the acceleration at 0.5 m?2856

The force is - kx which is going to be -40 N/m × 0.5 m or 20 N - 20 N.2861

The acceleration which by Newton’s second law is force/mass is -20 N / 2 kg or -10 m/s².2870

Continuing on, find the net force of the equilibrium position.2885

We have already said that at equilibrium, the acceleration is 0.2891

Therefore, the force is 0, the net force at 0.25 m though we can do that, F = kx is going to be -40 Nm × 0.25 m or -10 N.2898

And where does the kinetic energy equal the potential energy?2914

The total is 5 that means kinetic has to equal 2.5 joules and potential has to equal 2.5 joules.2918

We can solve that from our spring potential energy equation.2925

That has to equal 2.5 joules which is ½ kx² which implies that x² is going to be 2 × 2.5 joules ÷ our k 40 or 0.125 m².2931

Therefore, x is the square root of that or 0.354 m.2948

It is important to note here that is in not halfway between the equilibrium position and maximum displacement.2957

You do not have equal amounts of kinetic and potential energy when you are at half that displacement.2966

It does not work that way.2971

Another example, let us rank some spring systems here.2975

We have the spring block oscillators with combinations of springs, rank them from highest to lowest in terms of equivalent of spring constant and period of oscillation.2979

It looks like this to would be another great spot for a table.2990

We are going to make a table that has our system A, B, C, and D.2993

We will look at our mass, our equivalent spring constant, and M/k which is proportional to our period is 2π √M /K.2999

It looks like we can fill on our masses first.3023

I have 3, 6, 2, and 4, to find the equivalent spring constant this is going to be 1/20 + 1/5 and reciprocal of that will give us our equivalent or 4.3025

For B, because they are in parallel we just add them, that one was easy 25.3039

For C, we have 2 in series again so that is going to be the equivalent spring constant 1 /k equivalent is 1/15 + 1/15.3044

When we solve that, we will get 7.5 N /m.3054

The last one is in series again, just add them up 20.3058

M/k is 0.75 6/25 about 0.242, 27.5 about 0.27 and 4/20 is 1/5 or 0.2.3063

If we wanted rank these in terms of the equivalent spring constant that is going to be, we will start with B, D, C, and A from highest to lowest.3074

Period is 2π √M/k so we can go in the same order as M/k here which is going to be A, C, B, and final D.3088

Alright, let us do a vertical spring block oscillator system.3104

A 2 kg block attached to an un stretched spring, a spring constant k = 200 N/m as shown is released from rest.3107

Determine the period of the blocks oscillation and the maximum displacement of the block from its equilibrium while undergoing simple harmonic motion.3116

We can start with the period that is going to be 2π √M /k which is 2π × √2kg /200 N/m or 0.63 s.3125

To find the maximum displacement, we will do a little bit more work here.3145

The gravitational potential energy at the block starting point MG Δ y must equal the elastic potential energy stored in the spring and its lowest point.3151

We can solve for Δ y using the gravitational potential energy = stored potential energy in the spring3160

which implies that MG Δ y must be equal to ½ K Δ y².3167

½ kx² which implies if we solve for Δ y that is just going to be and divided Δ y at both sides and get 2 MG /k,3175

that is the entire displacement from the top and the bottom.3188

We want the displacement from its equilibrium position to its maximum displacement so that is going to be half of that.3191

Our amplitude will be Δ y/ 2 which is 2 MG /2k or just MG /K with a mass of 2kg, G of 9.8 m/s² ÷ k 200 N /m.3197

I come up with about 0.1 m for our maximum displacement from its equilibrium position while it is in that simple harmonic motion.3216

Let us rank the periods of some pendulums.3230

We have these 4 pendulums of uniform mass density from highest to lowest frequency.3232

Period is 2π √L /G, so the frequency is 1 /the period that must be 1/2 π × √G /L.3239

If we want the shortest period, if we want the shortest frequency, the smallest frequency, we want the biggest length.3254

If we want the highest frequency, we want the shortest length.3267

The shorter ones are going to go back and forth a lot more quickly than the long ones.3269

This would be D, A, B, C, because frequency is only a function of the length.3274

Alright, let us finish this off by looking at a couple of free response problems from old AP exams.3287

Let us start off with a 2009 exam free response question number 2.3292

As we look at this one, we are given a long thin rectangular bar of mass M, length L.3300

We are to determine its moments of inertia.3306

It has a non uniform mass density.3308

First A1, we are going to write the differential equation for the angle θ.3311

To do this, I'm going to draw it at its extreme just to make it easier to see what is going on.3316

We would not raise it at that height because we would lose our small angle approximation.3321

If we brought it up to the side like this, we will have some MG wherever that center of mass happens to be.3325

We will call x the distance from our pivot to our center of mass and there is a our angle θ.3331

We know net torque = Iα so we can write that our net torque we are going to have - mgx × sin θ must equal our moment of inertia IB × α.3338

Αlpha we know is the second derivative of θ so we can write this now as - MG × x sin θ = IB D² θ / DT².3359

That should cover us for writing the differential equation.3376

Now for part A2, it says apply the small angle approximation to calculate the period of the bars motion.3380

The small angle approximation says for small θ is about under about 15°, sin θ is approximately equal the θ.3387

It would not work as we have not draw on our diagram, we would not actually raise it that high.3398

If we did this with a small angle and let it go, we could then write that - mgx × θ = IB D² θ / DT² or3403

putting this into a more familiar form we can write this as DT D² θ / DT² + it looks like we will have mgx / IB × θ = 0.3417

There is a more familiar form, we have the second derivative of θ + some constant mgx / IB θ where that if you recall is our ω².3432

If we want that period, it looks like we are going to be looking for period is 2π / ω which is going to be 2π ÷ √mgx/ IB3444

which is going to be 2π × √ of the moment of inertia of our bar ÷ mgx.3462

That should cover us for part A2.3474

Let us give ourselves more room here for part B.3478

Describe the experimental procedure you would use to make the additional measurements needed to determine IB3484

and how you would use your measurements to minimize experimental error?3489

When I talked about this and we talked about finding that period of the real pendulum, I displace the bar a little bit of under 15°.3493

It will go back and forth say 10 times and measure the amount of time it takes to go back and forth 10 times.3503

Then divide that total time by 10 to get the period, that would give your T and then you could calculate your moment of inertia.3509

Once you have your period, you know period is 2π √ of the moment of inertia of your bar / mgx.3517

So then T² = 4π² IB / mgx which implies then that IB is just going to be equal to T² mgx / 4π².3527

You could use that in order to find your moment of inertia of the bar.3547

Let us take a look at part C, now suppose you are not given the location of the center of mass of the bar, how could you determine it and what equipment would you need?3553

That should be pretty easy too.3563

What I would do is take something like a razorblade, have the razor blade up vertically and put the bar on top of it and move it until you balance it.3564

Wherever you balance it, make a duct and that is going to be your location of the center of mass of the bar.3576

You could also hang it from different points and look where all the plumb bob lines across.3583

The easiest, I think probably needed to set it on some sort of edge and move it back and forth until you get to balance and that will be your center of mass point.3587

In words, something about balancing it and make sure you list any equipment you would need, pen and some sort of flat edge, pick whatever you want to use for that edge.3598

I think that covers the 2009 question, let us do one more.3609

Let us go to the 2010 exam free response 3.3618

Here we have a skier of mass M pulled up the hill by a rope and the magnitude of the acceleration is modeled by that equation there3624

where you have sin function between 0 and T.3631

After T it is not accelerating anymore, acceleration is 0, greater than T.3635

Alright derive an expression for the velocity of the skiers, a function of time.3642

Assume the skier starts from rest, for part A, if we are given the acceleration we can find the velocity as the integral from some value of T= 0 to some final time T.3647

The acceleration is a function of time which will be the integral from 0 to T of looks like acceleration is a max × sin π t/T.3662

We can pull other constant that is going to be A max × the integral to the sin π t /T.3677

I should not forget my DT here and the integral of the sin is going to be the opposite of the cos.3690

We need to have the argument in here, we are going to have our du which is π / T.3696

We need to have T /π out here to make a ratio of 1.3701

And then we can integrate that to say that V is going to be equal to - a max T/π cos of π t /T.3706

All evaluated from 0 to T which is going to be - a max T / π and we will have that cos π t / T – 1.3724

Or putting that negative through, we can rewrite this as V= A max t / π × (1 - cos π t /T) and that should work for T in the integral from 0 to T.3743

For part B, find the work done by the net force on the skier from rest until it reaches terminal speed there at T.3769

We could use the work energy theorem here.3779

Work is going to be change in kinetic energy, our net torque which is going to be our final kinetic energy - our initial kinetic energy3782

which is ½ MV final² -1/2 MV² initial.3790

V initial is 0 that part it is easy.3796

V final is just going to be V at time T which is going to be, we will have a max T /π × 1 - cos π which is going to be 1 - -1 so that is going to be 2a max capital T /π.3799

We can plug that in for our final velocity, this piece we said was 0 which implies then that the work done is going to be equal to ½ M × our final velocity²3828

which is going to be 4 a max² T² /π² or putting all of this together that will be 2 × M × maximum acceleration² × T² /π².3843

And that should cover us for part B.3866

Taking a look at part C, determine the magnitude of the force exerted by the rope on the skier at terminal speed.3872

Here I'm going to look at my free body diagram, we will draw our axis in first.3882

We have the normal force of the skier, we have the force of tension from the rope, and we have the weight of the skier down.3889

If I wanted to draw our pseudo free body diagram, breaking up that MG into components parallel3899

and perpendicular with the axis, I would have still, we have our normal force.3906

We have our force of tension from the rope.3912

We have MG sin θ and of course MG cos θ.3916

I can write my Newton’s second law expression in the x direction.3924

Net force in the x direction which we know is going to be 0 because it is moving at constant speed, at terminal speed is going to be equal to tension force - MG sin θ.3929

Therefore, force of tension in the rope must be MG sin θ.3942

That should cover us for C.3951

Part D, derive an expression for the total impulse imparted to the skier during the acceleration.3955

Impulse is the integral of force with respect to time.3962

Impulse is going to be the integral of FD T which will be the integral our forces mass × acceleration which is going to be the integral from some value T = 03966

to some final value T of M, our acceleration function is given we have a max sin π t /T DT.3979

We will pull other constants again that is going to be equal to, we can pull out M, we can pull out a max integral of sin π t /T DT.3992

We already talk about how we integrate that, we are going to π /T here so we are going to need HT T / π out here in order to integrate that.4011

That is going to be Ma max T / π and then we will have that cos π t /T.4024

Let us see, I will evaluate it from 0 to T.4045

We got to have our negative sign and their 2 the integral of the sin is the opposite of cos.4050

That is going to be equal to - M a max T/ π × cos π, when we substitute T in for t.4054

- the cos 0 which is 1, cos π is -1, -1 is going to be 2.4070

-2 we will have -2 × negative this will give us 2 ma max T /π.4079

One more piece to the puzzle, for part E, suppose the magnitude of the acceleration is instead modeled as the exponential for T greater than 0.4094

On the axis below, sketch the graphs of the force exerted by the rope on the skier for the 2 models from T = 0 to some time T greater than terminal velocity.4107

Label them F1 and F2.4116

First thing I want to do is I'm going to figure out what my functions are.4120

If we look at the original first, F1 is what we will call that our original.4123

We had the FT, tension force in the rope - MG sin θ all had the equal MA which is MA max sin π t/T.4131

Since we are going to plot the force of tension in the rope, FT must equal MG sin θ + MA max sin π t /T.4147

There is our first function, our new function says that FT - MG sin θ is now going to be equal to mass × new acceleration function a max E^-π t / 2 T.4163

Therefore, the function we will be modeling here for the force on the rope is going to be MG sin θ + MA max e^-π t /2 T.4186

Let us try our axis and see if we can graph these.4203

Finish this one up in style.4206

There is our y our force, here is our x our time, we have got some value T and force.4211

Let us label here MG sin θ and now for regional force we know that at time T =0,4225

we are going to have the sin of 0 so we are going to start at just MG sin θ so we can start with our first point there.4237

We are going to have our maximum value halfway to T, when we have sin π /2.4244

That will be at that point we will have MG sin θ + M a max.4252

We could label that point up here as well, that will be important.4258

MG sin θ + MA max right there.4262

If I we are to plot this, we know when we get the T, we are going to have sin π which is going to be back to 0.4269

This will be back to MG sin θ and after that, it is a constant flat line.4278

We have that part and in between we have our sin function which is going to bring us up to a maximum here.4283

It is kind of something that looks like that for initial force.4293

Our new force is MG sin θ + Ma max E to all of this.4300

At time T = 0, e⁰ is 1 so we are going to start at MG sin θ + MA max.4306

This one, we will start up there at that point and it looks like we are going to have the decaying as T gets very big, this is going to become a very large number in the denominator.4313

This piece is going to go to 0 and we are going to decay down the MG sin θ.4324

I would say that this one would look something like that, may be not quite like that.4328

Like that with some sort of exponential decay there so there would be our F2.4334

Hopefully, that gets you a great start on oscillations.4340

Thank you so much for joining us here today at www.educator.com.4343

I look forward to seeing you again soon and make it a great day everyone.4347