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Lecture Comments (25)

1 answer

Last reply by: Professor Dan Fullerton
Thu Sep 1, 2016 5:41 AM

Post by El Einstein on September 1 at 01:35:53 AM

Hello Professor Fullerton, I am confused on example VII (minute 45). Why did you set U=K? Isn't U=0 when at the equilibrium. And why did you set it equal to 5J? Can you explain this part in steps for me please.

1 answer

Last reply by: Professor Dan Fullerton
Sat Jan 9, 2016 12:44 PM

Post by Shehryar Khursheed on January 9 at 11:52:48 AM

On example IV when youre deriving the SHM of a pendulum, can you please explain why there is a negative on the net torque: -mglsin(theta)? You said that as the angle gets smaller, the angular acceleration gets larger. That seems counter-intuitive to me. Thanks!

1 answer

Last reply by: Professor Dan Fullerton
Mon May 11, 2015 9:14 AM

Post by Joshua Bowen on May 10, 2015

Im trying to understand more about wave motion when it comes the the Sinusoidal Wave function, what is the difference between these functions:

x = Acos(wt) (and the derivatives of the function which equal the velocity, and acceleration)

and

y = Asin(kx-wt)

1 answer

Last reply by: Micheal Bingham
Fri Apr 24, 2015 10:20 AM

Post by Micheal Bingham on April 24, 2015

@ About 31:29, why is the slope of the T - sqr(l) Graph, 2pi/sgr(g) ?

2 answers

Last reply by: Professor Dan Fullerton
Sun Feb 15, 2015 10:30 AM

Post by Thadeus McNamara on February 14, 2015

@63:47, does W ALWAYS equal change in KE?

3 answers

Last reply by: Thadeus McNamara
Wed Feb 18, 2015 1:03 PM

Post by Thadeus McNamara on February 14, 2015

33:05 i still am not sure why mglsintheta is negative

1 answer

Last reply by: Professor Dan Fullerton
Sun Feb 15, 2015 10:27 AM

Post by Thadeus McNamara on February 14, 2015

@29:16, why is F at a maximum at the top, and why is F at a minimum at the bottom?

1 answer

Last reply by: Professor Dan Fullerton
Sun Feb 15, 2015 10:25 AM

Post by Thadeus McNamara on February 14, 2015

thanks for the previous answers. i have another question. @24:56, why did u write f=-k2x2 over the arrow? how is Feq equivalent to F2? You said it was because of Newton's third law but I dont know what you mean by that.

1 answer

Last reply by: Professor Dan Fullerton
Sat Feb 14, 2015 5:50 PM

Post by Thadeus McNamara on February 14, 2015

also during that same slide, why does w represent both angular velocity and angular frequency. what is angular frequency?

1 answer

Last reply by: Professor Dan Fullerton
Sat Feb 14, 2015 5:47 PM

Post by Thadeus McNamara on February 14, 2015

@4:56 is that the position, velocity, and acceleration of SHM problems or for circular problems?

1 answer

Last reply by: CHARINA TECSON
Wed Jan 21, 2015 10:55 PM

Post by CHARINA TECSON on January 21, 2015

Din example 1, did you ever convert minutes to seconds?

Oscillations

  • Simple Harmonic Motion (SHM) is motion in which a restoring force is directly proportional to the displacement of an object.
  • Nature’s response to a disturbance is often SHM.
  • Angular frequency (ω) is the number of radians per second, corresponding to an angular velocity for an object traveling in uniform circular motion.
  • When an object undergoes simple harmonic motion, kinetic and potential energy both vary with time, although total energy (E=K+U) remains constant.
  • The period of a spring-block oscillator is proportional to the square root of mass divided by the spring constant.
  • In analyzing a spring block system, if you find a new equilibrium position when the block is hanging on the spring, taking into account the effect of gravity, you can then treat the system with only the spring force to deal with, oscillating around the new equilibrium point.
  • The reciprocal of the equivalent spring constant for springs in series is equal to the sum of the reciprocals of the individual spring constants.
  • The equivalent spring constant for springs in parallel is equal to the sum of the individual spring constants.
  • The period of a pendulum is proportional to the square root of the length of the pendulum divided by the acceleration due to gravity.

Oscillations

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.

  1. Intro
    • Objectives
      • Simple Harmonic Motion
      • Circular Motion vs. Simple Harmonic Motion (SHM)
      • Position, Velocity, & Acceleration
      • Frequency and Period
      • Angular Frequency
      • Example I: Oscillating System
      • Mass on a Spring
      • Example II: Analysis of Spring-Block System
        • Example III: Spring-Block ranking
          • General Form of Simple Harmonic Motion
          • Graphing Simple Harmonic Motion (SHM)
          • Energy of Simple Harmonic Motion (SHM)
          • Horizontal Spring Oscillator
          • Vertical Spring Oscillator
          • Springs in Series
          • Springs in Parallel
          • The Pendulum
          • Energy and the Simple Pendulum
          • Frequency and Period of a Pendulum
          • Example IV: Deriving Period of a Simple Pendulum
            • Example V: Deriving Period of a Physical Pendulum
              • Example VI: Summary of Spring-Block System
                • Example VII: Harmonic Oscillator Analysis
                • Example VIII: Ranking Spring Systems
                  • Example IX: Vertical Spring Block Oscillator
                    • Example X: Ranking Period of Pendulum
                      • Example XI: AP-C 2009 FR2
                      • Example XII: AP-C 2010 FR3
                      • Intro 0:00
                      • Objectives 0:08
                      • Simple Harmonic Motion 0:45
                        • Simple Harmonic Motion
                      • Circular Motion vs. Simple Harmonic Motion (SHM) 1:39
                        • Circular Motion vs. Simple Harmonic Motion (SHM)
                      • Position, Velocity, & Acceleration 4:55
                        • Position
                        • Velocity
                        • Acceleration
                      • Frequency and Period 6:37
                        • Frequency
                        • Period
                      • Angular Frequency 7:05
                        • Angular Frequency
                      • Example I: Oscillating System 7:37
                        • Example I: Determine the Object's Angular Frequency
                        • Example I: What is the Object's Position at Time t = 10s?
                        • Example I: At What Time is the Object at x = 0.1m?
                      • Mass on a Spring 10:17
                        • Mass on a Spring
                      • Example II: Analysis of Spring-Block System 11:34
                      • Example III: Spring-Block ranking 12:53
                      • General Form of Simple Harmonic Motion 14:41
                        • General Form of Simple Harmonic Motion
                      • Graphing Simple Harmonic Motion (SHM) 15:22
                        • Graphing Simple Harmonic Motion (SHM)
                      • Energy of Simple Harmonic Motion (SHM) 15:49
                        • Energy of Simple Harmonic Motion (SHM)
                      • Horizontal Spring Oscillator 19:24
                        • Horizontal Spring Oscillator
                      • Vertical Spring Oscillator 20:58
                        • Vertical Spring Oscillator
                      • Springs in Series 23:30
                        • Springs in Series
                      • Springs in Parallel 26:08
                        • Springs in Parallel
                      • The Pendulum 26:59
                        • The Pendulum
                      • Energy and the Simple Pendulum 27:46
                        • Energy and the Simple Pendulum
                      • Frequency and Period of a Pendulum 30:16
                        • Frequency and Period of a Pendulum
                      • Example IV: Deriving Period of a Simple Pendulum 31:42
                      • Example V: Deriving Period of a Physical Pendulum 35:20
                      • Example VI: Summary of Spring-Block System 38:16
                      • Example VII: Harmonic Oscillator Analysis 44:14
                        • Example VII: Spring Constant
                        • Example VII: Total Energy
                        • Example VII: Speed at the Equilibrium Position
                        • Example VII: Speed at x = 0.30 Meters
                        • Example VII: Speed at x = -0.40 Meter
                        • Example VII: Acceleration at the Equilibrium Position
                        • Example VII: Magnitude of Acceleration at x = 0.50 Meters
                        • Example VII: Net Force at the Equilibrium Position
                        • Example VII: Net Force at x = 0.25 Meter
                        • Example VII: Where does Kinetic Energy = Potential Energy?
                      • Example VIII: Ranking Spring Systems 49:35
                      • Example IX: Vertical Spring Block Oscillator 51:45
                      • Example X: Ranking Period of Pendulum 53:50
                      • Example XI: AP-C 2009 FR2 54:50
                        • Example XI: Part A
                        • Example XI: Part B
                        • Example XI: Part C
                      • Example XII: AP-C 2010 FR3 1:00:18
                        • Example XII: Part A
                        • Example XII: Part B
                        • Example XII: Part C
                        • Example XII: Part D
                        • Example XII: Part E

                      Transcription: 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 / a 0561

                      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 position 0619

                      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 look 0783

                      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 where 0880

                      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 do 0895

                      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 integral 0963

                      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 time 1000

                      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 position 1260

                      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 0 1307

                      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 and 1417

                      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 x2 1430

                      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 + 1 1507

                      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 distance 1918

                      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 or 2235

                      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 that 2270

                      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 spring 2299

                      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 direction 2467

                      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 displacement 2683

                      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 /M 2811

                      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 spring 3160

                      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 or 3403

                      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/ IB 3444

                      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 IB 3484

                      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 there 3624

                      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 energy 3782

                      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 velocity² 3828

                      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 parallel 3899

                      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 = 0 3966

                      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 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 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