WEBVTT physics/ap-physics-1-2/fullerton
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Hi everyone. I am Dan Fullerton and I am thrilled to welcome you back to Educator.com.
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Today's topic is simple harmonic motion.
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Our objectives are going to be to sketch and analyze a graph of displacement as a function of time for an object undergoing simple harmonic motion, to write down an appropriate expression for displacement of the form A-cos(ωt) or A-sin(ωt), where ω is going to be that angular frequency.
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We will state the relations between displacement, velocity, and acceleration and determine the points in the motion where these quantities are minimum, 0, and maximum for an object undergoing simple harmonic motion, determine the total energy of an object in simple harmonic motion and sketch graphs of kinetic and potential energies as functions of time or displacement, and determine the period of oscillation for an ideal pendulum as well as a mass on a spring with a horizontal mass and a vertical mass.
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That is what we are going to try and accomplish here.
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All right. What is simple harmonic motion?
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Simple harmonic motion is nature's typical reaction to a disturbance.
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It is all over in this world in many, many places.
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When you disturb something it typically responds with simple harmonic motion.
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That could be something as simple as walking past a tree -- if you brush a branch, the branch gets a force done and starts oscillating back and forth.
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Simple harmonic motion -- it is all over .
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A displacement which results in a linear restoring force results in simple harmonic motion.
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And the simple items we are going to focus on for today are going to be an ideal pendulum, a pendulum where the string has no mass and a mass on a spring, going back and forth due to that spring.
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Let us start with a review of springs.
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When a force is applied to a spring, the spring applies a restoring force and a spring can be compressed or it can be stretched.
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When the spring is in equilibrium, it is unstrained; it is in its happy position; it is just thrilled to be there.
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The factors affecting the force of a spring -- well the spring constant (k) is how tough it is to compress or stretch the spring.
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The bigger the spring constant, the tougher it is to compress or stretch and that is measured in Newton's per meter (N/m) and the displacement is always measured from equilibrium.
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We could make a graph of the force of a spring vs. the displacement and when we do that, for an object obeying Hooke's Law, we should get a straight line where the slope of that line which is rise over run, gives us our spring constant.
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Or we could make the same type of graph -- force of a spring versus displacement get the same basic shape and if we want to get the work done in compressing and stretching that spring, all we have to do is come back and take the area under that curve and the area will be the work.
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Now as we analyze that work, the work that we do on it must be the potential energy stored in the spring, so that is 1/2 base times height or 1/2 our base is going to be (x) and our height is going to be the force.
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But if (F) is (kx), then that is going to be 1/2 x × our force, because it is a spring, (kx), or you could say that the potential energy stored in the spring is 1/2 kx².
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That is where that formula comes from -- one way we can derive the energy stored in a compressed or stretched spring.
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All right. Let us talk about oscillations.
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Repeated motions back and forth are called oscillations -- something going back and forth, back and forth is oscillating.
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Now one revolution or one round trip, one complete cycle all mean the same thing and the period of the oscillation is the time it takes for one complete cycle or one complete revolution.
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That is going to be an important vocabulary word, period.
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Frequency, on the other hand, (F), is the number of cycles per second and it is measured in 1/seconds or a unit known as Hertz (Hz).
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Period (T) is the number of seconds divided by the number of cycles to give you the time for each cycle.
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Frequency is the number of cycles divided by number of seconds to give you the number of cycles per second and they are very closely related because Period is 1/frequency and therefore frequency is 1/Period.
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Let us take a look at the spring block oscillator.
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Imagine that we have a mass and we are going to connect it by a spring to some immovable object like a wall.
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If we pull it one way and let it go, it is going to go back and forth and back and forth and oscillate.
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That is a spring-block oscillator.
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Factors affecting the period of its oscillation are the mass of its block (m) and the spring constant (k).
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As we analyze this in a little bit more detail, we will call this the x = 0 position, the happy position, the equilibrium position -- maximum displacement of A or -A.
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Now the period of a spring-block oscillator, period of a spring STs is going to be 2π times the square root of the mass divided by the spring constant.
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Now the frequency of that spring-block oscillator which is always 1/period is just going to be 1/2π square root then of k/m.
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Now we could also rearrange this a little bit to say that 2π times the frequency equals the square root of k/m.
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Where there is 2π times the frequency is often times called the angular frequency (ω).
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Angular frequency = 2πF, therefore we could write the angular frequency (ω) is the square root of k/m.
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A couple of definitions to go along with our spring-block oscillator system.
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All right. Let us take a look at an example with this system.
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We have a block of mass 5 kg and it is attached to a spring, whose constant is 2,000 N/m.
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Find the period of oscillation, the frequency, and the angular frequency.
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Well, let us start with the period.
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Period for a spring-block oscillator -- we know is 2π square root m/k or 2π square root -- mass is 5 kg, (k) is 2,000 N/m or when I plug that into my calculator, I get about 0.314 s for my period.
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Let us find the frequency.
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Frequency is 1/Period, so that is going to be 1/0.314 = 3.18, 1/seconds or 3.18 Hz.
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And the angular frequency (ω) is 2π times the frequency, 2πF or 2π × 3.18 Hz, or 20 and the units are rad/s, although radians are not an official unit -- 20 rad/s will be our angular frequency.
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Let us take a look at how we might analyze this in even more detail.
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Here we are going to show our spring-block oscillator, mass (m), and we are going to look at it at three different positions -- at its equilibrium, position (A), at maximum displacement to the right (B), back to (A) to its minimum displacement or maximum displacement on the left (C), and back to (A).
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So it is going to go back and forth, (A) to (B) to (A) to (C) to (A) and back and forth and it is going to displace a distance (x) to the right or (-x) to the left.
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Well, if we want to look and see if we have different displacements, at point (A), we have 0 displacement, that is its equilibrium.
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When it is at (B), its displacement is (x) and when it gets all the way to (C), its displacement is (-x).
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Let us take a look at velocity now .
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When it is at (A), it is going to have maximum velocity because all of its energy is going to be kinetic, so this will be maximum velocity here at (A).
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Over here at (B), it is not going to have any velocity.
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That is where all of its kinetic energy is converted into spring potential energy, so that will be 0 and (C) is the same way just on the other side.
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There is no velocity at the end points.
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It goes back and forth and for a split second it stops, turns around, speeds up, slows down, slows down, slows down, stops, speeds up, speeds up, speeds up, speeds up, slows down, stops, and back and forth as it goes on its oscillating path.
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Let us take a look then at the force.
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Well, while it is at point (A), the force is going to be 0 because there is no force to the spring on it because there is no displacement.
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When it is at (B), it is going to have a maximum force, but it is a restoring force bringing it back the other direction, bringing it back towards its equilibrium position, so we will call that the negative max force.
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And at (C) it is going to have the maximum force back to the right in the positive direction, so that will be maximum.
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Because force = mass × acceleration, or acceleration = force/mass, the acceleration charge should look extremely similar.
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At (A) there is no acceleration, at (B) it has its negative maximum acceleration, and at (C) it has its positive max acceleration.
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Let us look at energy now. Spring potential energy over here is (U).
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At (A), there is no displacement, so the spring potential energy must be 0.
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It will have its maximum spring potential energy here at (B) and (C), so there is a maximum at (B) and (C).
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If we want to look at kinetic energy, well that is going to be basically the inverse, right?
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At (A), it is going to have its maximum velocity so it will have its maximum kinetic energy over there at (A), so maximum kinetic energy at (A), and at (B) and (C), it stopped at those points, it has no velocity for a split second there, so 0 and 0.
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Let us take a look at what would happen if we made a graph of some of these.
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I am going to start by taking a look at the displacement (x) and we are going to look at it as the object goes from (A) to (B) to (A) to (C) to (A) to (B) and back again.
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Let us make this point (A) down all of these graphs, then we will draw another line all the way down here from (A) to (B)...
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...another one here back to (A), and it goes to (C), and back to (A).
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I think you quickly get the idea of what is going on here.
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From (A) of course and back to (B) and so on, on its oscillating journey.
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Well, if we look at displacement -- at (A) its displacement is 0, so anywhere we have (A) we can fill in the dots for 0.
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At (B), its displacement is going to be positive (x), so it must come up like that and come back to (A).
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At (C), its displacement must be negative (x) -- back to (A) and so on.
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A graph of the displacement versus time.
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If we want to do that for velocity, however, let us take a look and see what we know here.
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For velocity, we know =at (B) and (C) we have 0, so anywhere we have (B) or (C) we must have 0 velocity.
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At (A) we have maximum velocity and initially it is going to the right, so we will make that a positive velocity as it comes through.
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When it comes back to (A), it has velocity going in the opposite direction at the same magnitude and back up and you can quickly see the pattern here for velocity of our spring-block oscillator.
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Let us go to force and acceleration, let us put those over here -- a nice purple color, maybe for our force.
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If we want to look at force, we know anywhere we have an (A), the force is 0.
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So 0 here, 0 here, 0 here, and we have a maximum force when we are at (C), so let us fill that in -- maximum force over here at (C) and the negative maximum force when we are at (B).
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So at (B) over here, we have negative maximum force, negative maximum force, and we can quickly plot our force versus time graph looking something like that.
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And acceleration is just going to follow that -- force = mass × acceleration, so almost a mirror image of the graph, just different values but same shape.
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Now, let us take a look and finish off our graph by looking at potential energy and kinetic energy versus time.
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Let us look at potential energy in the spring first over here...
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...there is (T), we have (A) here, goes to (B), back to (A), to (C)...
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...from (C) we go back to (A), and from (A) we go back to (B).
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All right. Potential energy in the spring is 0 at any of the (A)'s.
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So there you go, a 0, 0 , 0; when it is at (B) it is at a maximum, so we will fill in our points there, and when it is at (C), it is also at a maximum.
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Potential energy is scalar; it does not have a direction, so our graphs can look kind of like this.
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When we look at kinetic energy, let us go there right underneath it.
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Kinetic energy is going to have values of 0 at (B) and (C) when it is not moving.
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So for (B), (C) -- (B) is at 0 and it is going to have maximum value when it is at (A), so it is going to look almost like the inverse of the potential energy graph.
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Something kind of like that, and if we were to add the kinetic and potentials, they would add to a constant value because we are neglecting friction in this problem, using all conservative forces so we have conservation of mechanical energy.
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Let us take a look at solving some problems with this and we will start off with a detailed harmonic oscillator analysis.
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A 2 kg block is attached to a spring. A force of 20 N stretches the spring to a displacement of half a meter (0.5 m). 0996 Find the spring constant.
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Well, we know that F = kx, therefore (k) must equal F/x or 20 N/0.5 m, which is 40 N/m.
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The total energy is going to be the spring potential energy when it is at its maximum displacement or 1/2 kx²...
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...which is 1/2 × 40 N/m (k) and its maximum displacement (0.5m²) for a total energy of 5 J.
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About the speed at the equilibrium position, well at that point we have converted all of that spring potential energy into kinetic.
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So we could start solving this one by saying that the spring potential energy is equal to the kinetic energy at the equilibrium position or 1/2 mv² and that must equal that 5 J.
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Therefore, the velocity = 2 × 5 divided by the mass, or 2 × 5/2 square root of 5 for about 2.24 m/s.
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And how about the speed when it is at (x) = 0.3 m?
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Well to do that, we are going to have to look at an energy analysis again.
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The total energy is the spring potential energy plus the kinetic, therefore the total energy = 1/2 Kx² + 1/2 mv².
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And if I am going to try and find the speed, let us get (V) isolated; I will multiply both sides by 2...
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...2 there and I will subtract the kx², therefore mv² = 2 total energy minus kx².
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Divide by (m) and take the square root to find that the velocity will be 2 × the total energy - kx², all divided by the mass, square root, and finally I can substitute in my values.
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I have 2 × 5 J (total energy) - 40 (spring constant) × 0.3 (x-value), our displacement² divided by 2 (mass) and the square root of that entire thing gives me a velocity, a speed at 0.3 m of about 1.79 m/s.
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And I will do a quick check to see if that makes sense.
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It should be less than the speed at the equilibrium position, the maximum speed, and it is less than 2.24 m/s.
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So that makes sense. Excellent!
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Let us go a little bit further with this one.
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Let us try and find the speed now at x = -0.4 m.
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We can use the same formula we just had, but plug in a different displacement value.
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Velocity is going to be equal to 2 times total energy minus kx² divided by (m), square root...
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...that is going to be 2 × 5 J, (total energy) - 40 (spring constant) × -0.4² (new displacement/2 kg (mass); square root of all of that is about 1.34 m/s.
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As it is a little closer to its full extension, it is slowing down even more -- less than our velocity when we were at 0.3 m. That, too, makes sense.
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Let us find the acceleration at the equilibrium position.
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Well when we were at x = 0, the force must equal 0 by Hooke's Law, and if the force is 0, then Newton's Second Law (F = ma) tells us that the acceleration must also be 0.
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But if we want the acceleration at 0.5 m -- well to do that I am going to use Hooke's law (F = -kx), where (k) again is 40 N/m and a displacement at 0.5 m or -20 N.
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We can apply Newton's Second Law again -- acceleration is force divided by mass, or -20 N/0.2 kg for an acceleration of -10 m/s².
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What does the negative tell you?
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When it is at its furthest positive displacement, the acceleration is back towards its equilibrium position, going in the opposite direction, hence the negative.
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Let us take this one even further. Let us find the net force at the equilibrium position.
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Well at the equilibrium position we already determined the acceleration was 0, so the net force there must be 0 N.
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How about the net force at half a meter?
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Well, we could use Hooke's Law again (F = -kx) or -40 N/m × 0.25 m (displacement) or -10 N.
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Where does kinetic energy equal potential energy?
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Well if our total energy is 5 J, that is going to be the spot where the kinetic energy is 2.5 J and the potential energy is 2.5 J.
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So if that is the case, we could figure that out -- the spring potential energy there is 2.5 J -- that is 1/2 kx², where (x) is what we are solving for.
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Therefore, (x)² = 2 × 2.5/40 (K), and as I solve that then, that is equal to 0.125 and if I take the square root of both sides...
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...(x) then equals the square root of 0.125 or about 0.354 m.
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Pay special attention here -- note at this point where the kinetic energy and potential energy are equal is not midway between the equilibrium and the maximum displacement positions.
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It does not work out that way, you cannot just guess halfway in between the two, you actually have to go through and solve; that is not halfway between the equilibrium and maximum displacement positions.
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So let us take a look at circular motion versus simple harmonic motion and how they are related.
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We have already talked about rotational motion for an object moving in a circle where we have some radius or amplitude of its motion (A) -- the angle θ measured from the horizontal, angular velocity ω and its position vector are, given by its (x) coordinate A-cos(θ), its (y) coordinate A-sinθ.
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Well you could think of that almost as a projection down in one dimension to the spring-block oscillator system.
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As you look at that system, imagine the object moving in a circle.
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If you could shine a light down on it so you were just getting the projection in one dimension, you would see the exact same motion as what you see here in one coordinate.
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Let us take a look at the (x) coordinates here -- (x) = A-cos(θ), but as we learned previously, θ = ω × (T).
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Therefore we could write that as x = A-cos(ωT).
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We also know that ω = 2π/period (T), so we could write that as x = A--cos(2π)/period × time.
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Or going back to this equation, we also know that ω = 2π × the frequency, so we could write that -- replacing ω with 2π × the frequency as x = A-cos(2π), frequency, × time.
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A bunch of different ways of looking at the same simple harmonic motion.
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Instead of having our block start over here at a maximum displacement, what happens if we want to have it start here at x = 0?
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Well we could use the sine function for that.
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If we are going to have the block start at x = 0 at its equilibrium position -- start at x = 0 at time (t) = 0, then we could have the exact same equation -- we are just going to replace cosine with sine.
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So x = A-sin(θ) or A-sin(ωt).
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All it is, is a face shift, a sliding of the graph one way or another, depending on what you are calling your starting point.
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As we look at graphing simple harmonic motion in that system, let us take a look at what happens to our graph of x = A-cos(ωt).
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We will graph the (x) displacement and as we do that, we are going to look at it in terms of radians at 0, at π/2, at π, at 3π/2, 2π and so on.
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I will make those marks on our graph right now -- π/2, π, 3π/2, 2π, and we will copy the same notches down here as well.
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All right, for x = A-cos(ωt), we will do this as a function over here of ωt.
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What are we going to graph?
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Well, when our argument is 0, cosine here at times 0 is going to be 1, so we are going to get (A), so our maximum value here is (A).
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As we get to π/2 right here, well our (x) coordinate is now 0, so we come down here to 0.
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As we get over here to 2 full π, now we are adding negative maximum displacement, -(A).
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Let us mark that on our graph, -(A), and back to 3π/2, 2π again and the cycle repeats.
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We get ourselves -- just like the curve we had expected -- we get our cosine curve.
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Looking at that, if we wanted to use the sine function instead, x = A-sin(ωt), we are going to have the same maximum amplitudes of course -- (A) and -(A).
00:27:50.000 --> 00:28:07.000
But we are going to start -- if we look at the y-coordinate or for the sine function -- we are going to start at 0 for our (y), so 0 at π/2, we are at maximum value here for the (y), so positive (A)...
00:28:07.000 --> 00:28:18.000
...down here to 0 at π, down here to -(A) at 3π/2 and back to 0 at 2π -- so we get our sine curve.
00:28:18.000 --> 00:28:31.000
But as you look at the graphs, notice they are really the same shape; they are just off set by that π/2 amount, so you can use either one depending on which starting point you prefer to work with.
00:28:31.000 --> 00:28:35.000
All right. Let us take a look at an example where we are looking at the position of an oscillator.
00:28:35.000 --> 00:28:43.000
A spring-block oscillator makes 60 complete oscillations in 1 minute or 60 seconds.
00:28:43.000 --> 00:28:47.000
Its maximum displacement is 0.2 m.
00:28:47.000 --> 00:28:55.000
What is its position at time (t = 10 s) and at what time is it at position x = 0.1 m?
00:28:55.000 --> 00:28:58.000
Well let us start up here with question A.
00:28:58.000 --> 00:29:19.000
If x = A-cos(ωt) and we know that ω = 2πF, we could write this as x = A-cos(2πFt).
00:29:19.000 --> 00:29:25.000
If we have 60 cycles in 60 seconds, then we know that our frequency must be 1 Hz.
00:29:25.000 --> 00:29:54.000
So if frequency is 1 Hz and we know that our maximum amplitude (A) is 0.2 m, we can fill in our function a little bit to say that X = 0.2 cos(2π) × 1 (frequency) × 10 s (time) or putting that all together I get about 0.2 m.
00:29:54.000 --> 00:29:57.000
It is at its maximum displacement.
00:29:57.000 --> 00:30:04.000
Moving down here to B, at what time is it at position x = 0.1 m?
00:30:04.000 --> 00:30:27.000
All right, s = A-cos(2πFt), but now we are solving for the time, so this implies then that 2πFt must be equal to the inverse cosine of x/A.
00:30:27.000 --> 00:30:41.000
Or if we want just (t) but itself let us isolate the variable we want to find -- t = the inverse cosine of x/A divided by 2πF.
00:30:41.000 --> 00:31:00.000
Now we can substitute in our variables, so that will be the inverse cosine of 0.1/0.2 or 1/2/2π × 1 Hz (frequency) or about 0.167 s.
00:31:00.000 --> 00:31:12.000
Now it is important to note here for an oscillating system, that is not going to be the only time it is in that position, but it is one answer to that question.
00:31:12.000 --> 00:31:17.000
Let us take a look at a vertical spring-block oscillator system.
00:31:17.000 --> 00:31:33.000
Once the system settles the equilibrium where we are hanging our block from a spring instead of having it on a horizontal surface, we are going to displace the mass by pulling at some amount either +A -- pull it down, let it jump up -- or -A -- lift if up a little bit, drop it and let it oscillate up and down.
00:31:33.000 --> 00:31:38.000
This is a really slick derivation and a neat analysis.
00:31:38.000 --> 00:31:50.000
If we looked at a free body diagram (FBD) when it is at its equilibrium position, we have gravity pulling down and the force of the spring, (ky) pulling it up, where (y) is the equilibrium point.
00:31:50.000 --> 00:32:13.000
So since it is at equilibrium at that point, we could write that the net force in the y-direction is going to be (mg), calling down positive, minus (ky) and since it is at equilibrium, let us say that that is mg - ky (equilibrium point), that must equal 0.
00:32:13.000 --> 00:32:21.000
Therefore we could solve to say that the y-equilibrium point must be mg/k.
00:32:21.000 --> 00:32:44.000
Now when we displace it by some amount (A), the net force in the y-direction, as we pull it down, is going to be mg - K × whatever that (y) would happen to be, which is going to be Y-equilibrium point plus that (A) amount we pulled it down.
00:32:44.000 --> 00:32:56.000
We can distribute that through -- multiply that (k) through -- to find that it is mg - ky-equilibrium position - kA
00:32:56.000 --> 00:33:09.000
Here is the slick part though -- Notice then as we do this that we had up here mg - ky-equilibrium = 0.
00:33:09.000 --> 00:33:25.000
That means this part mg - ky-equilibrium must be equal to 0 and we could rewrite this then as FnetY = -kA.
00:33:25.000 --> 00:33:28.000
There is our big result. What does that mean?
00:33:28.000 --> 00:33:38.000
That is the same analysis you would do for a horizontal spring system with a spring constant (k), displaced horizontally some amount (A) from its equilibrium position.
00:33:38.000 --> 00:33:41.000
We just made this so much simpler.
00:33:41.000 --> 00:33:57.000
In short, to analyze a vertical spring system, all you do is find the new equilibrium position of the system, treat that -- once you have taken into account the effect of gravity -- and treat that as if that is the only force you have to deal with in the system -- just the spring force.
00:33:57.000 --> 00:34:04.000
So once you find its new equilibrium position you could almost pretend that you turned it on your side and it is a horizontal spring oscillator system again.
00:34:04.000 --> 00:34:08.000
You do not have to continue dealing with that force of gravity.
00:34:08.000 --> 00:34:12.000
A really, really slick way to analyze a vertical spring-block oscillator.
00:34:12.000 --> 00:34:26.000
Find that new equilibrium position and then ignore the effects of gravity from there; treat that as your new equilibrium position, just find it using the effect of gravity first.
00:34:26.000 --> 00:34:29.000
Let us take an example to make sure we have this.
00:34:29.000 --> 00:34:35.000
A 5 kg block is attached to a vertical spring, with a spring constant of 500 N/m.
00:34:35.000 --> 00:34:39.000
After the block comes to rest it is pulled down 3 cm and then released.
00:34:39.000 --> 00:34:42.000
What is the period of oscillation?
00:34:42.000 --> 00:35:01.000
Well, period is 2π square root m/k or 2π square root 5/500, or 0.628 s -- that straight forward.
00:35:01.000 --> 00:35:06.000
What is its maximum displacement of the spring from its initial unstrained position?
00:35:06.000 --> 00:35:10.000
Well let us first look at it when it is at rest.
00:35:10.000 --> 00:35:44.000
At that point we have (mg) down for our FBD, and we have (k) times -- let us call that displacement (d) at that point, so that the net force in the y-direction is 0, since it is at equilibrium because it is just sitting there, which implies that KD = mg or D = mg/k, which is 5 kg × 10 m/s²/spring constant (500) or 0.1 m.
00:35:44.000 --> 00:35:48.000
So once you are hanging it there it hangs down 0.1 m.
00:35:48.000 --> 00:35:55.000
Now you are going to go displace it; you are going to pull it down 3 cm from that equilibrium position.
00:35:55.000 --> 00:36:10.000
If you then pull it down 3 cm, your maximum displacement in the y-direction is going to be that 0.1 m that you had from when it came to rest due to gravity...
00:36:10.000 --> 00:36:23.000
...and that extra 3 cm that you added on as you pull it down or 0.13 m or 13 cm.
00:36:23.000 --> 00:36:39.000
A nice straight forward analysis, once you take into account and figure out what its new equilibrium position is with gravity and then ignore the effects of gravity and treat it as a standard horizontal spring-block oscillator system.
00:36:39.000 --> 00:36:43.000
All right. Example 5 -- another fairly involved example.
00:36:43.000 --> 00:36:49.000
We have a 60 kg bungee jumper stepping off a 40 m high platform.
00:36:49.000 --> 00:36:53.000
The bungee cord behaves like a spring, of spring constant 40 N/m.
00:36:53.000 --> 00:36:57.000
Find the speed of the jumper at heights of 15 and 30 m above the ground.
00:36:57.000 --> 00:37:02.000
And as we do this, we are going to have to assume that there is no slack in the system.
00:37:02.000 --> 00:37:07.000
Find the speed of the jumper at heights 15 and 30 m above the ground.
00:37:07.000 --> 00:37:11.000
All right. So first thing, let us draw a diagram here.
00:37:11.000 --> 00:37:23.000
Here is our jumper and the jumper is on a 40 m foot high platform, so down here somewhere is the ground.
00:37:23.000 --> 00:37:35.000
We want the speed of the jumper 15 m above the ground, so we will call that position (A) -- that is 15 m above the ground.
00:37:35.000 --> 00:37:47.000
Position (B) is 30 m above the ground, which means we must have another 10 m here -- 10, 15 m, 15 m.
00:37:47.000 --> 00:37:53.000
All right. The key here is we are going to try and do this through conservation of energy at this point.
00:37:53.000 --> 00:38:16.000
So the gravitational potential energy at the top must be equal to the gravitational potential energy at (A) plus the spring potential energy (A) -- UAG, UAS plus the kinetic energy at (A), 15 m above the ground.
00:38:16.000 --> 00:38:32.000
Or mg × H-initial, the initial gravitational potential energy must equal the gravitational potential energy at (A), mg × 15 + 1/2 k.
00:38:32.000 --> 00:38:45.000
At (A) the displacement is 10 + 5, so the spring is stretched 25 m, 25² (kx²) + 1/2 mv².
00:38:45.000 --> 00:38:47.000
All right. A little bit of math here.
00:38:47.000 --> 00:38:56.000
We could then say that mass (60), G (10), and height (40)...
00:38:56.000 --> 00:39:15.000
...must equal 60 × 10, = 600 × 15 + 1/2 × 40 (k) × 25² = 625 + 1/2 × 60 (mass) × v²...
00:39:15.000 --> 00:39:33.000
or let us see here, that is going to be 24,000 = 9,000 + 12,500 + 30 v².
00:39:33.000 --> 00:39:54.000
Or 30 v² will equal about 2,500, so I get a velocity of about 9.13 m/s at (A), so vA = 9.13 m/s.
00:39:54.000 --> 00:39:59.000
We also need the velocity at 30 m above the ground.
00:39:59.000 --> 00:40:05.000
To do that then, we will follow the same basic idea but we are going to have different values for our height.
00:40:05.000 --> 00:40:27.000
So we will have 24,000, our initial total energy, equal to mg × (height) 30 m + 1/2 k -- at (B) the spring is only stretched 10 m, so 10² + 1/2 mv².
00:40:27.000 --> 00:40:49.000
Therefore 24,000 = 600 (mg) × 30 + 1/2 × 20 (k) × 100 = 2,000 + 1/2 mv².
00:40:49.000 --> 00:41:00.000
Therefore 4,000 = 30v²; and in solving for V, I get about 11.55 m/s.
00:41:00.000 --> 00:41:09.000
Makes sense -- it is going a little bit faster at (B), a little bit slower at (A), hopefully slows down before hitting the ground.
00:41:09.000 --> 00:41:16.000
All right, let us go a little bit further with this problem. Let us see what happens next.
00:41:16.000 --> 00:41:22.000
How close does the jumper get to the ground?
00:41:22.000 --> 00:41:31.000
Well, to do that we are going to have to figure out where the kinetic energy becomes 0.
00:41:31.000 --> 00:41:47.000
So as we do this one, we will say potential energy due to gravity total equals the potential energy due to gravity at some point (C), which is where they stop, plus potential energy due to the spring at point (C).
00:41:47.000 --> 00:41:52.000
No kinetic energy -- it is 0; that is where the person stops.
00:41:52.000 --> 00:42:05.000
Following along with our calculations, 24,000 (total energy) = mgH + 1/2 k, and now how far has that person been displaced?
00:42:05.000 --> 00:42:12.000
That is going to be 40 minus whatever height is left, so 40 - H².
00:42:12.000 --> 00:42:33.000
So that implies then that 24,000 = 600 (mg) × H + 1/2 × 40 (k), so that is going to be 20 × 40 - H².
00:42:33.000 --> 00:42:37.000
That is starting to look like a quadratic equation, so let us get it into that form.
00:42:37.000 --> 00:42:56.000
24,000 = 600H + 20 × -- well, if we square this, we get 1600- 80H + H²...
00:42:56.000 --> 00:43:15.000
...or 24,000 = 600H + 32,000 - 1600H + 20H².
00:43:15.000 --> 00:43:26.000
We rearrange this to fit the quadratic formula, H² - 50H + 400 = 0.
00:43:26.000 --> 00:43:29.000
A couple of ways you can solve that, but the quadratic formula is probably my favorite.
00:43:29.000 --> 00:43:35.000
I come up with a height of 10 m above the ground.
00:43:35.000 --> 00:43:38.000
And let us just test that to make sure we did not make any mistakes here.
00:43:38.000 --> 00:43:47.000
H², 10², 100 - 50 × 10, 100 - 500 = -400 + 400 = 0.
00:43:47.000 --> 00:43:55.000
So how close does the jumper get to the ground? 10 m.
00:43:55.000 --> 00:43:59.000
Let us take a look at the pendulum again.
00:43:59.000 --> 00:44:04.000
Now from the perspective of oscillations and simple harmonic motion.
00:44:04.000 --> 00:44:10.000
Mass (m) is attached to a light string that swings without friction about a vertical equilibrium position.
00:44:10.000 --> 00:44:15.000
For all of these, we are going to assume that this θ is relatively small.
00:44:15.000 --> 00:44:33.000
Well, we have length (L) here again -- as it comes down here where there is all kinetic energy, its height has changed and we have derived a couple of times now that this height (H) is going to be L - L-cos(θ)...
00:44:33.000 --> 00:44:45.000
...or this is L-cos(θ) compared to our entire length of our string (L).
00:44:45.000 --> 00:44:56.000
Here we have all potential energy; here we have all kinetic and this (H) as we just said is going to be L - L-cos(θ).
00:44:56.000 --> 00:45:16.000
If we take a look and start analyzing this with energy involved too though, again assuming a small θ, here we have all potential energy -- U = mgH, which is mgL × 1 - cos(θ).
00:45:16.000 --> 00:45:23.000
Here it is all kinetic energy, and back here again it is all potential.
00:45:23.000 --> 00:45:35.000
And if we wanted to find the velocity of our pendulum when it is at this lowest point, well we could say that the kinetic energy at the bottom must equal the potential energy at its highest point -- the top.
00:45:35.000 --> 00:45:48.000
Or 1/2 mv² = mgH where (H) is L - L-cos(θ), and we can divide now our masses.
00:45:48.000 --> 00:46:05.000
So v² = 2gL × 1 - cos(θ) or just velocity itself -- I take the square root and that will be the square root of 2gL, 1 - cos(θ).
00:46:05.000 --> 00:46:09.000
Going a little further here, let us look at some other quantities that might be of interest.
00:46:09.000 --> 00:46:26.000
Here at the highest point, you have the maximum force, you have the maximum acceleration, you have the maximum gravitational potential energy, but you have no kinetic energy and no velocity.
00:46:26.000 --> 00:46:39.000
Here on the other hand, you have no force, (F = 0), your acceleration = 0, your gravitational potential energy = 0 -- assuming that is what we are calling 0 in this problem which should make sense.
00:46:39.000 --> 00:46:45.000
Our maximum kinetic energy occurs there and we have maximum velocity there.
00:46:45.000 --> 00:46:50.000
And what is causing our restoring force to put this in simple harmonic motion?
00:46:50.000 --> 00:46:55.000
Well our restoring force is going to be based on the gravitational force pulling this down.
00:46:55.000 --> 00:47:06.000
If we look right here, we have its weight pulling it down, but that is not what is causing the displacement; it is only a portion of that.
00:47:06.000 --> 00:47:24.000
The portion that is going to be perpendicular to our string, mg-sin(θ) here is what is causing our restoring force.
00:47:24.000 --> 00:47:34.000
So another way that we could graph this in terms of energy, is if we looked at energy on the y-axis versus (x) position on the (x), we know of course the total energy must remain the same.
00:47:34.000 --> 00:47:40.000
We are dealing with conservative forces, conservation of energy, and we are not dealing with friction at this point.
00:47:40.000 --> 00:47:45.000
Here at 0 displacement, everything is kinetic energy.
00:47:45.000 --> 00:47:53.000
So as we draw this U-shape, anything above the U-shape between the E (total line) and our parabola is kinetic energy.
00:47:53.000 --> 00:47:55.000
Anything below it is potential.
00:47:55.000 --> 00:47:58.000
The two always sum up to the total energy.
00:47:58.000 --> 00:48:04.000
So if I wanted to look at another point on the graph -- let us say we wanted to look right here.
00:48:04.000 --> 00:48:15.000
In this case anything up here would be kinetic and down here would be our potential.
00:48:15.000 --> 00:48:19.000
Use the graph that way -- what is above the line is kinetic, what is below is potential.
00:48:19.000 --> 00:48:25.000
Another way of representing the same information; it is that important.
00:48:25.000 --> 00:48:30.000
So what happens when we are talking about this pendulum and we have to start dealing with frequency and period?
00:48:30.000 --> 00:48:35.000
Well, the period of an ideal pendulum is 2π square root of L/g.
00:48:35.000 --> 00:48:38.000
The length of the pendulum is your variable.
00:48:38.000 --> 00:48:42.000
The mass on the end does not matter, the length is what matters.
00:48:42.000 --> 00:48:49.000
In your grandfather clocks at home, the length is all set, that is why those are so big.
00:48:49.000 --> 00:48:54.000
Or the frequency is just going to be 1/period or 1/2π square root of g/L.
00:48:54.000 --> 00:49:01.000
Now for all of these, again, we have to assume that θ is small due to a small angle approximation in the mathematics.
00:49:01.000 --> 00:49:11.000
So as we are looking at this, let us take a look and see what would happen if we tried to graph period versus length.
00:49:11.000 --> 00:49:20.000
If you did that you would probably get something that looks kind of like that because period is proportional to the square root of (L).
00:49:20.000 --> 00:49:31.000
So if you wanted to get a nice linear graph that you could do something with, if you wanted to try and determine something like the acceleration due to gravity, for example.
00:49:31.000 --> 00:49:55.000
You could take a pendulum, graph the period versus the square root of the length and you should get a nice linear graph and if you take the slope of that, the slope is going to be rise/run, which is going to be T/square root of L, which turns out to be 2π/square root of (g).
00:49:55.000 --> 00:50:15.000
So if you want to go to the moon, figure out what the acceleration due to gravity is, make a bunch of different pendulums of different lengths, have them go back and forth, measure their periods, come graph the period versus square root of the length, find the slope and you can calculate the acceleration due to gravity that way.
00:50:15.000 --> 00:50:17.000
Let us take a look at an example.
00:50:17.000 --> 00:50:23.000
We have a 1 kg mass suspended from a 30 cm string that creates a simple pendulum.
00:50:23.000 --> 00:50:29.000
The mass is displaced at an angle of 12 degrees from the vertical equilibrium position.
00:50:29.000 --> 00:50:37.000
First thing we have to do, is if we are going to use any of these formulas, is to make sure that θ is small, and 12 degrees is small enough for our purposes.
00:50:37.000 --> 00:50:40.000
Find the frequency and period of the pendulum.
00:50:40.000 --> 00:50:56.000
Well, the period is 2π square root L/g, or 2π square root (L) 0.3 m/10 m/s² (g)...
00:50:56.000 --> 00:51:10.000
...or about 1.09 s and frequency then is just 1/period or 1/1.09 s, which is going to be about 0.92 Hz.
00:51:10.000 --> 00:51:16.000
All right, find the height of the pendulum above equilibrium when at maximum displacement.
00:51:16.000 --> 00:51:33.000
Well we know the height is L × 1 - cos(θ), so that is going to be 0.3 × 1 - cos(12 degrees) or about 0.0066 m.
00:51:33.000 --> 00:51:37.000
And find the speed of the pendulum at the equilibrium position.
00:51:37.000 --> 00:51:47.000
V = square root of 2GH if we want to use conservation of energy, which is square root of 2gL, 1 - cos(θ)...
00:51:47.000 --> 00:52:10.000
...we have derived that a couple of times at this point, or the square root of 2 × 10 (g), (l) 0.3 × 1 - cos(12 degrees) or about 0.632 m/s.
00:52:10.000 --> 00:52:16.000
Carrying this one a little further -- Find the restoring force at maximum displacement.
00:52:16.000 --> 00:52:38.000
All right. The force at maximum displacement -- we just said was mg-sin(θ), the component of the weight pulling it back down -- mg-sin(θ), which is going to be its mass of 1 kg × 10 m/s² (g)-sin(12 degrees), or about 2.08 N.
00:52:38.000 --> 00:52:42.000
And how about the tension in the string at the equilibrium position?
00:52:42.000 --> 00:52:46.000
Well at that point we can make our FBD.
00:52:46.000 --> 00:53:01.000
There is our tension; there is our weight -- if it is in equilibrium, those must match -- or net force in the centripetal direction is mAC which implies that (t), which is in the direction toward the center of the circle...
00:53:01.000 --> 00:53:13.000
...tension - mg = mv²/r, which implies then that the tension is mg + mv²/r.
00:53:13.000 --> 00:53:21.000
Therefore, the tension must be (mass) 1 × 10 (g) + 1 (mass).
00:53:21.000 --> 00:53:40.000
Our velocity we found was about 0.362 m/s² over our radius (0.3 m) or about 10.44 N of tension in that string.
00:53:40.000 --> 00:53:47.000
How long should the pendulum be in order to keep perfect time with a period of 1 second?
00:53:47.000 --> 00:53:55.000
Well let us start there and we are going to assume again that it is an ideal pendulum, no friction and everything is perfect, and no mass in the string.
00:53:55.000 --> 00:54:07.000
Period is 2π square root L/g and we want a time, a period of 1 s, so we are solving for (L).
00:54:07.000 --> 00:54:14.000
Let us take t² = 4π² (L/g).
00:54:14.000 --> 00:54:30.000
Therefore L = gt²/4π², which implies then that the length (L) should be G (10 m/s²)...
00:54:30.000 --> 00:54:45.000
...with a period of 1 s²/4π² or about 0.253 m, a quarter of a meter.
00:54:45.000 --> 00:54:49.000
How long should the pendulum be if the period is to be half a second?
00:54:49.000 --> 00:54:53.000
Well, same thing -- let us just plug in a couple of different values here.
00:54:53.000 --> 00:55:08.000
L = gt², so (g) × 0.5²/4π², where g = 10 m/s², which gives us about 0.633 m.
00:55:08.000 --> 00:55:13.000
A lot shorter. A lot shorter -- one-fourth.
00:55:13.000 --> 00:55:16.000
Let us take a look then at a pendulum on the moon.
00:55:16.000 --> 00:55:25.000
How long must a pendulum of period 1 second be on the moon if the acceleration due to gravity on the moon is about 1.6 m/s²?
00:55:25.000 --> 00:55:55.000
Well we can use our same formula, L = gt²/4π², where (g) = 1.6 m/s² on the moon × our period of 1 s²)/4π², which is about 0.405 m.
00:55:55.000 --> 00:56:00.000
All right. Doing great. Hang in there. One last sample problem.
00:56:00.000 --> 00:56:08.000
Mass (m) is placed on a horizontal frictionless surface and attached to a spring with spring constant (k).
00:56:08.000 --> 00:56:13.000
The mass is pulled back a distance (x) and released to oscillate horizontally.
00:56:13.000 --> 00:56:22.000
What is the kinetic energy and potential energy of the mass at a displacement halfway between the equilibrium position and maximum displacement?
00:56:22.000 --> 00:56:26.000
Well, let us draw a picture first of what our situation is going to look like.
00:56:26.000 --> 00:56:31.000
Horizontal spring-block oscillator -- let us color that in there.
00:56:31.000 --> 00:56:45.000
We will put our mass over here (m), some spring with spring constant (k), and we will start this at some displacement 0.
00:56:45.000 --> 00:57:00.000
Here is our (x) and it is somewhere in there we are going to have a point (A), and what we know is at the maximum energy is 1/2 kx².
00:57:00.000 --> 00:57:18.000
The potential energy due to the spring at (A) is going to be 1/2 × (k) -- well (A) is halfway between these two, so that is going to be at x/2².
00:57:18.000 --> 00:57:31.000
That will be 1/2 k × x²/4 or 1/8 kx², which is 1/4 × 1/2 kx².
00:57:31.000 --> 00:57:34.000
Why did I write it that way?
00:57:34.000 --> 00:57:51.000
Well, 1/2 ks² is our maximum spring potential energy, so this then says that UA must equal 1/4 of UMax.
00:57:51.000 --> 00:57:56.000
So at (A) it has 1/4 of its maximum energy, so where is that other 3/4 of the energy?
00:57:56.000 --> 00:58:05.000
That has to be kinetic energy at (A), so that must be 3/4 of the maximum spring potential energy.
00:58:05.000 --> 00:58:15.000
Notice here that halfway between the two, the spring potential energy and the kinetic energy are not the same -- they are not equal.
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You have to go through the steps to go solve for points in between those two.
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Do not take shortcuts.
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Hopefully this is a good start for simple harmonic motion.
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Thank you so much for your time and for watching Educator.com. Make it a great day everyone!