For more information, please see full course syllabus of AP Physics C: Mechanics

For more information, please see full course syllabus of AP Physics C: Mechanics

### 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.

- Intro
- Objectives
- Simple Harmonic Motion
- Circular Motion vs. Simple Harmonic Motion (SHM)
- Position, Velocity, & Acceleration
- Frequency and Period
- Angular Frequency
- Example I: Oscillating System
- 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
- 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 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
- 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

### AP Physics C: Mechanics Online Course

I. Introduction | ||
---|---|---|

What is Physics? | 7:12 | |

Math Review | 1:00:51 | |

II. Kinematics | ||

Describing Motion I | 23:47 | |

Describing Motion II | 36:47 | |

Projectile Motion | 30:34 | |

Circular & Relative Motion | 30:24 | |

III. Dynamics | ||

Newton's First Law & Free Body Diagrams | 23:57 | |

Newton's Second & Third Laws of Motion | 23:57 | |

Friction | 20:41 | |

Retarding & Drag Forces | 32:10 | |

Ramps & Inclines | 20:31 | |

Atwood Machines | 24:58 | |

IV. Work, Energy, & Power | ||

Work | 37:34 | |

Energy & Conservative Forces | 28:04 | |

Conservation of Energy | 54:56 | |

Power | 16:44 | |

V. Momentum | ||

Momentum & Impulse | 13:09 | |

Conservation of Linear Momentum | 46:30 | |

Center of Mass | 28:26 | |

VI. Uniform Circular Motion | ||

Uniform Circular Motion | 21:36 | |

VII. Rotational Motion | ||

Rotational Kinematics | 32:52 | |

Moment of Inertia | 24:00 | |

Torque | 26:09 | |

Rotational Dynamics | 56:58 | |

Angular Momentum | 33:02 | |

VIII. Oscillations | ||

Oscillations | 1:01:12 | |

IX. Gravity & Orbits | ||

Gravity & Orbits | 34:59 | |

X. Sample AP Exam | ||

1998 AP Practice Exam: Multiple Choice | 28:11 | |

1998 AP Practice Exam: Free Response Questions (FRQ) | 28:11 |

### 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 ^{-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

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?