WEBVTT physics/ap-physics-c-mechanics/fullerton
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Hello, everyone, and welcome back to www.educator.com.
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I'm Dan Fullerton and in this lesson we are going to talk about oscillations and simple harmonic motion.
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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.
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Recognizing simple harmonic motion when expressed in differential equations form.
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Calculating the kinetic and potential energies of an oscillating system.
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Analyzing problems involving horizontal and vertical masses attached to springs.
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Finding the period of oscillations for systems involving combinations of springs and deriving the expression for the period of a pendulum,
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both an ideal pendulum and now for the first time, a physical or real pendulum.
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Simple harmonic motion is motion which the restoring force is directly proportional to the displacement of the object.
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The more you displace it, the more restoring force there is trying to bring it back to its initial position.
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The reason this is so important is, in general, nature's response to a disturbance is some sort of simple harmonic motion.
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You can see it all over the place, the blade of grass watch it come back up, simple harmonic motion.
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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.
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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.
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We can even see it in the atoms of an object.
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When they are compressed or stretched, it will vibrate back in simple harmonic motion.
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Let us take a look at simple harmonic motion and start off with an analogy to circular motion.
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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,
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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.
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That is our x coordinate and there is our y coordinate.
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We are going to compare this to a system of the spring attached to a wall with the mass on the end.
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We are going to pull the mass of displacement a from its equilibrium, our happy position, and let it go back and forth.
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What is really amazing about this analogy is, if you we are to pull this to a and let it go,
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you can compare its x position to what you would get is this object goes around the circle its x position.
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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.
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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.
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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.
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But if I rewrite that a little bit, if I separate our variables, we can write that ω dt = d θ.
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If I integrate both sides from some T = 0 to some final time T, integral from θ = 0 to some final θ,
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that implies what angular velocity is constant in uniform circular motion so the left hand side becomes ω T.
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The right hand side just becomes θ, θ is given by ω T, that will come useful later.
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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,
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from the perspective of Newton's law F = ma and that force is restoring force – kx.
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We also know that acceleration is the second derivative of x with respect to T².
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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.
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We have a second order differential equation or the second derivative of a function + a constant × that function gives you 0.
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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.
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The general form of our solution, x is a function of time is given by a, our amplitude cos ω T.
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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.
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Let us take a bit further, as we look at position velocity and acceleration.
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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.
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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.
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It is going to be equal to the derivative of cos is the opposite of the sin.
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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.
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Our maximum velocity is going to be ωa.
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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,
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which is going to be the derivative with respect to T of – ω A sin ω T.
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The derivative of the sin is the cos function.
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We are going to get –ω² A cos ω T.
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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.
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We can keep going with these to find position velocity acceleration.
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As we are doing this, oftentimes we are talking about frequency in period.
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Probably, it is worth bringing this up again.
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Frequency is the number of cycles or revolutions per second.
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Its units are 1/s or hertz.
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Period T is the time for 1 cycle or 1 complete revolution and the units are seconds.
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We can find period when we know frequency.
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Period is 1/ frequency or frequency is 1/period.
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If we talk about angular frequency, angular frequency is the number of radians per second.
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It corresponds to the angular velocity for an object traveling in uniform circular motion.
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Note that angular velocity and angular frequency is not the same thing but they do have a strong correspondents.
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Angular frequency ω is 2π × frequency in Hz or 2 π ÷ period.
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You can also write that period is 2 π ÷ angular frequency.
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Let us do an example, an oscillating system is created by releasing an object from a maximum displacement of 0.2 m.
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The object makes 60 complete oscillations in 1 min, determine the objects angular frequency.
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Angular frequency is 2 π × the frequency in Hz which will be 2 π × 60 Hz.
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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.
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What is the objects position at time T = 10 s?
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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.
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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.
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At what time is the object at x = 0.1 m?
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We have to realize it is going to be oscillating back and forth.
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There could be more than one answer here but let us solve for one answer.
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That x = a cos 2π T is our function which implies then that the cos of 2 π T = x / a
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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.
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We substitute in our values that will be the inverse cos of 0.1/0.2 ÷ 2 π or I get 0.167 s.
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Alright, let us take a look at a little bit more detail of a mass on a spring.
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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
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and released allowing it to oscillate back and forth between - a and a.
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You are going to assume no thing or no loss of energy to friction.
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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.
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Or the frequency of our spring is 1 /period and which is going to be ½ π √K/M.
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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.
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Therefore, ω your angular frequency is the √k/m.
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Let us analyze a spring block system.
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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.
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Determine the period of oscillation, the frequency, and the angular frequency for the block.
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Let us start, it asks for period first so let us start there.
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A period of our spring is 2π √M/K, which is going to be 2π √5kg /2000 N/m or about 0.314 s.
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Next, it wants the frequency.
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Frequency is 1 /period which will be 1/0.314 s or about 3.18 Hz.
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Finally, the angular frequency for the block, ω is 2π × frequency which is 2π × 3.18 Hz or 20 radians/ s.
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Another example, rank the following horizontal spring resting on frictionless surfaces in terms of their period from longest to shortest.
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When we look at typically of different masses and spring constant, the way I start this is I would probably look
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at the relationship and recognize first before looking for period.
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The period is 2π √M/K so we could then make a table of information.
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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.
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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.
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Their spring constants, A is 500 N /m, B is 50, C is 2000, and D is 1000.
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The ratio M/K it is pretty easy to calculate.
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It is going to be 0.02, 0.14, 0.001, and 0.005.
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We are looking for the periods from longest to shortest, since period is proportion to √M/K,
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we are going to get the longest period where we have the greatest M /K.
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I would rank these for period, our biggest M/KB, then A, D, C will have the shortest period.
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As we talk about simple harmonic motion, the general form of simple harmonic motion is something where
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we had the second derivative of function with respect to time + the constant ω² × that original function equal to 0.
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When we do this, we find that our general solution is a cos ω T + some phase angle, some phase shift having to do
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with where you are starting your cos function or sin function.
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We are not going to worry a whole lot about that here in this course and note that ω here is your ω there.
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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.
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You oscillate back and forth between maximum and minimum displacement.
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For using the sin function where you got a phase shift of 90°, you still have the same basic shape.
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You are still oscillating for maximum A to – A.
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Let us take a look at energy of simple harmonic motion.
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When an object undergoes simple harmonic motion, kinetic and potential energy are both peering with time.
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Although, the total mechanical energy, kinetic + potential remains constant.
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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
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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,
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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.
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You have done that work on it that must be the potential energy in the spring.
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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
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is going to be ½ K × x, a cos ω T², which will be ½ K A² × √cos ω T.
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Let us go take a look now at velocity.
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Velocity is a function of time, is the derivative of x with respect to T, which is –ω A sin ω T.
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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.
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Or as we multiply through there, kinetic energy will be ½ M ω² a² sin ω T, square the sin ω T.
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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.
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Let us put this all together.
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We got our kinetic energy, we have got our spring potential energy, let us put that together to find your total energy.
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E is kinetic + potential which is going to be our kinetic is ½ ka² sin² ω T + our potential ½ ka² cos² ω T.
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Remember, our trigonometry sin² θ + cos² θ is equal to 1.
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We can then factor that out to say that our total energy then ½ ka² × (sin² + cos² )is just going to be ½ KA².
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No dependence on time because we have a constant total mechanical energy.
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Alright, let us go back to our horizontal spring oscillator again.
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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.
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We have md² x/dt² = - kx or d² x/ dt² = k/ mx² or + k /mx² equal 0.
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We have mentioned before that was ω².
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Our general solution for the horizontal spring oscillator system now looks like x(t) = A cos ω T + some phase angle.
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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,
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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.
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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.
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What about a vertical spring block oscillator?
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We have a mass, we are going to hang from the spring and we are going to let it settle to an equilibrium position
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and then we are going to pull it down some amount A and displace it.
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How do we deal with that where we also have this included effective gravity?
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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.
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We are calling down the positive y direction so when I write my Newton’s second law equation.
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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.
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If you are letting this hang down to the equilibrium position here, that means at this point MG - ky equilibrium must equal 0
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because at that point there is no net force and no acceleration.
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Or we could then write that the equilibrium point on the y axis is just MG/k.
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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.
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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.
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If we multiply through here, that means MG - ky equivalent –ka = our net force.
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Let me just fill them in here.
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Take a look, MG - ky at the equilibrium point is 0, we have the same thing right there net 0.
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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.
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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.
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It just got so much more simple.
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We will do that as sampled problem here shortly.
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We can also have a couple of springs where they are lined up in different ways, combinations of springs.
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If they are in series, such as this, we have k1 attached to k2, attached to a mass, we can analyze that and
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it is nice to figure out how we could treat this as if it was just one equivalent spring.
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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
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which implies then the x 1 is going to be equal to k2 /k1 x2.
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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.
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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.
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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,
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that we could write that - k2 x2 = - k equivalent × x2.
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We will factor that out and I'm left with k2 /k1 + 1, which implies then that k2 = equivalent k × k2 /k1 + 1
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which implies then we could write this as 1/k equivalent = 1/k1 + 1/k2.
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You can follow that same form for any number of springs.
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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.
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Kind of like capacitors in series or resistors in parallel if you have done ENM.
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You can maybe guess already what springs in parallel are going to look like.
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They are going to look like resistors in series or capacitors in parallel.
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But let us take a look and show that derivation, for springs in parallel now we have k1 and k2,
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both attached to our mass M which is going to be displaced some amount x.
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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.
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Pretty easy to see, k equivalent is k1 + k2.
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We just add up our individual spring constants to get our equivalent spring constant when they are in parallel.
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We have spent some time on springs and we will come back to those 2.
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We also see simple harmonic motion and pendulums.
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Let us review what we know about the pendulum already.
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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.
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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.
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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.
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We are down at the lowest points, when it is down there, it is kinetic energy.
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We have talked about that energy transformation.
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We can look at that in a little more detail though.
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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,
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which is going to be MGL × (1 - cos θ).
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Here all its energy is kinetic energy, back up here, we have all potential energy.
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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.
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½ MV² must equal MGL × (1 - cos θ).
00:28:34.500 --> 00:28:52.700
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.
00:28:52.700 --> 00:29:03.600
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.
00:29:03.600 --> 00:29:14.100
We have our maximum acceleration, we have our maximum gravitational potential energy that are kinetic energy 0 and our speed is 0.
00:29:14.100 --> 00:29:26.100
Here, where the highest point of kinetic energy, the force on our mass is 0, acceleration is 0, gravitational potential energy 0,
00:29:26.100 --> 00:29:31.800
but we have our maximum kinetic energy and our maximum velocity.
00:29:31.800 --> 00:29:41.000
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.
00:29:41.000 --> 00:29:50.200
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.
00:29:50.200 --> 00:29:53.300
That total always = e total K + U.
00:29:53.300 --> 00:29:57.600
But you have that shift across the different types of energy.
00:29:57.600 --> 00:30:07.400
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,
00:30:07.400 --> 00:30:16.200
where once again we have this as a restoring force leading this into a discussion of simple harmonic motion.
00:30:16.200 --> 00:30:19.900
Let us look at the period and the frequency of a pendulum.
00:30:19.900 --> 00:30:22.600
We have our pendulum like we have talked about previously.
00:30:22.600 --> 00:30:29.100
The period of the pendulum is 2π √its length ÷ G assuming it is an ideal pendulum.
00:30:29.100 --> 00:30:39.100
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.
00:30:39.100 --> 00:30:41.600
You can even go and plot these if you wanted.
00:30:41.600 --> 00:30:50.600
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.
00:30:50.600 --> 00:31:02.300
If you want to linearize your graph, T and √L, to get something that is fairly linear.
00:31:02.300 --> 00:31:07.400
If we did that, what happened to our slope?
00:31:07.400 --> 00:31:13.900
Our slope is just going to be 2 π ÷ √G.
00:31:13.900 --> 00:31:16.800
Ever want to know the acceleration due to gravity on the Moon?
00:31:16.800 --> 00:31:18.700
Here is a great way to find that out.
00:31:18.700 --> 00:31:27.300
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.
00:31:27.300 --> 00:31:36.700
Slope = 2π /√G, you could then go calculate G, the acceleration due to gravity on the Moon.
00:31:36.700 --> 00:31:40.400
How do we get this 2π √L /G?
00:31:40.400 --> 00:31:44.300
Let us see if we can derive that.
00:31:44.300 --> 00:31:47.200
Here, we have a simple pendulum, an ideal pendulum.
00:31:47.200 --> 00:31:54.100
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.
00:31:54.100 --> 00:31:57.800
There is no friction just the ideal case.
00:31:57.800 --> 00:32:07.000
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
00:32:07.000 --> 00:32:15.000
from our point there q to our mass as the position function from point B.
00:32:15.000 --> 00:32:17.900
Their length is L.
00:32:17.900 --> 00:32:34.900
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.
00:32:34.900 --> 00:32:48.000
Which implies then that the magnitude of our torque about point B is going to be MGL sin θ,
00:32:48.000 --> 00:32:58.100
which implies then that - MGL sin θ = our moment of inertia about point P × angular acceleration.
00:32:58.100 --> 00:33:05.000
Why that negative? If you take a look as θ is getting smaller, we are going to larger values of angular acceleration.
00:33:05.000 --> 00:33:08.100
So we have to take that into account there.
00:33:08.100 --> 00:33:11.700
We are going to use what is known as the small angle approximation.
00:33:11.700 --> 00:33:15.600
For small angles of θ, they are under about 15°.
00:33:15.600 --> 00:33:19.200
The sin θ is a very close the θ itself.
00:33:19.200 --> 00:33:31.300
This will allow us to simplify our analysis and write this as - MGL θ = the moment of inertia about point B × α.
00:33:31.300 --> 00:33:38.300
We also know that α is the second derivative of θ.
00:33:38.300 --> 00:33:50.000
We can write this as - MGL θ is equal to our moment of inertia P × the second derivative of θ with respect to T.
00:33:50.000 --> 00:33:52.600
We have a second order differential equation.
00:33:52.600 --> 00:34:11.900
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.
00:34:11.900 --> 00:34:19.200
By the way, if you remember our form dist is what we called ω².
00:34:19.200 --> 00:34:27.200
The solution to our problem, our general solution is θ is going to be equal to a cos ω T.
00:34:27.200 --> 00:34:36.000
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.
00:34:36.000 --> 00:34:53.600
That means that ω² will be MGL /ML² IP which is going to be G /L which implies then that ω = √G /L.
00:34:53.600 --> 00:35:10.400
If we want that period, period is 2π / ω which is going to be 2π ÷ √G/L or 2π √L /G.
00:35:10.400 --> 00:35:16.600
We derived the formula for the period of a simple pendulum.
00:35:16.600 --> 00:35:20.100
That is easy, what about real pendulums?
00:35:20.100 --> 00:35:22.100
Alright, let us try that.
00:35:22.100 --> 00:35:29.300
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.
00:35:29.300 --> 00:35:35.800
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.
00:35:35.800 --> 00:35:40.500
We can use the same basic analysis in order to find its period.
00:35:40.500 --> 00:35:50.300
Net torque = R cross F, we know θ is going to be equal to a cos ω T.
00:35:50.300 --> 00:36:01.200
We have already done that analysis or ω² if we go through all of that again is MGD /our moment inertia about point P,
00:36:01.200 --> 00:36:09.600
which implies then that ω is √MGD /our moment of inertia about point P.
00:36:09.600 --> 00:36:14.700
The difference here is we have a different moment of inertia.
00:36:14.700 --> 00:36:21.200
To find our moment of inertia about point P, all that is going to be the moment of inertia about our center of mass.
00:36:21.200 --> 00:36:36.200
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.
00:36:36.200 --> 00:36:47.900
The mass of our rod rotated about its center is ML² /12 + MD².
00:36:47.900 --> 00:36:54.500
If period is 2π / ω that is going to be 2π / ω.
00:36:54.500 --> 00:37:11.300
Here, our √MGD/√IP, this is going to be √IP is ML² /12 + MD² /MGD,
00:37:11.300 --> 00:37:14.900
which implies then that our period is going to be equal to 2π.
00:37:14.900 --> 00:37:27.000
We can factor an MN out of all of that and I would have L² /12 + D² /GD or
00:37:27.000 --> 00:37:44.000
if I wanted to make that a little bit prettier, we can write that as 2π × √L² + 12 D² ÷ 12 GD.
00:37:44.000 --> 00:37:50.500
If you want to test this out, I lot of my students quite regularly is to take a meter stick.
00:37:50.500 --> 00:37:57.700
Find its center of mass, drill a hole in it, near one of the end some distance from that center of mass measure that
00:37:57.700 --> 00:38:00.800
and then put it on a hook and swing it back and forth.
00:38:00.800 --> 00:38:07.100
Let it go at about 10, 20 times, measure the entire time for 20 revolutions, 20 periods, and divide by 20,
00:38:07.100 --> 00:38:15.500
and see how that does compared to when you plug it into the equation and you will find it is very accurate.
00:38:15.500 --> 00:38:19.600
Alright, let us go back to our spring block system for a summary here.
00:38:19.600 --> 00:38:27.000
If we want to take a look at our spring block system with displacement velocity, force acceleration, potential energy in the spring
00:38:27.000 --> 00:38:35.800
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.
00:38:35.800 --> 00:38:41.300
Sometimes it is useful to graph all of these just because you see them come up so often.
00:38:41.300 --> 00:38:49.900
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,
00:38:49.900 --> 00:38:53.000
then we will pull the B to A to C and back again.
00:38:53.000 --> 00:39:19.500
I'm just going to plot some of these lines in here very loosely so that we have a common reference frame.
00:39:19.500 --> 00:39:44.300
I think we will do it there and let us do the same thing here for our couple of energy graphs.
00:39:44.300 --> 00:39:52.900
If this is position A, that will be B, back to A to C to A back to B again.
00:39:52.900 --> 00:40:00.700
The same thing here, A to B back to A, C to A to B again.
00:40:00.700 --> 00:40:03.800
We will have graphs of let us see.
00:40:03.800 --> 00:40:06.200
Let us start with our displacements.
00:40:06.200 --> 00:40:08.900
We will have this as our x.
00:40:08.900 --> 00:40:12.800
At time 0, when its position A our displacement is 0.
00:40:12.800 --> 00:40:20.800
At position B, where there are maximum displacement which we call x or typically capital A.
00:40:20.800 --> 00:40:26.100
Back to A 0 to C to A, back to B.
00:40:26.100 --> 00:40:33.500
Our plot would look something like that.
00:40:33.500 --> 00:40:38.300
In that position A, where our 0 displacement at B our displacement is x.
00:40:38.300 --> 00:40:40.700
At C, it would be –x.
00:40:40.700 --> 00:40:46.900
We can take a look now at velocity which we know is the derivative of x.
00:40:46.900 --> 00:40:54.700
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.
00:40:54.700 --> 00:41:00.200
We can fill those points in and we know that 2 because if we take the derivative of position that will be the slope.
00:41:00.200 --> 00:41:06.800
We have a 0 here, we have a 0 there, we have a 0 slope there, so we have 0 values for velocity there.
00:41:06.800 --> 00:41:17.200
We are going to start at A and the maximum velocity come down to a minimum or maximum speed in the opposite direction
00:41:17.200 --> 00:41:22.400
and end up with a graph that looks kind of like that.
00:41:22.400 --> 00:41:25.800
How about if we take a look at the force?
00:41:25.800 --> 00:41:31.800
If we look at the force, we are going to have, let us fill in our table here toward velocity.
00:41:31.800 --> 00:41:40.200
We have maximum velocity at A, at B we have 0 and at C we have 0.
00:41:40.200 --> 00:41:47.300
Now looking at force, when we are at A we have 0 force, so anytime we are at A we can fill 0.
00:41:47.300 --> 00:41:53.700
When we are at B, we have maximum force but in the negative direction so we will have a negative there.
00:41:53.700 --> 00:42:00.400
We will have a positive here, when we are at C maximum force back to the right restoring force and so on.
00:42:00.400 --> 00:42:07.400
We can graph our force working something like that.
00:42:07.400 --> 00:42:18.900
And our acceleration of course, should follow the same thing, the same general pattern by Newton’s second law.
00:42:18.900 --> 00:42:20.600
They should tolerate along.
00:42:20.600 --> 00:42:24.500
We will have 0 force and 0 acceleration when we are at point A.
00:42:24.500 --> 00:42:32.200
At point B, we are going to have the negative maximum for force and acceleration going back toward the left.
00:42:32.200 --> 00:42:38.200
At C, we will have our maximum positive force and acceleration.
00:42:38.200 --> 00:42:44.900
And again, if we want to look at our derivative slope relationship, look at the velocity here at A.
00:42:44.900 --> 00:42:52.900
The slope of that is 0 therefore, we have 0 acceleration right there as well and force correlate right along with that.
00:42:52.900 --> 00:42:56.600
How about potential energy and kinetic energy?
00:42:56.600 --> 00:43:04.000
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.
00:43:04.000 --> 00:43:10.300
A for potential energy is 0 there, we will have a 0 at A.
00:43:10.300 --> 00:43:13.900
At B and C, we are going to have our maximum values.
00:43:13.900 --> 00:43:17.300
We do not have a negative because this is a scalar.
00:43:17.300 --> 00:43:27.800
We will have a graph that looks something like this.
00:43:27.800 --> 00:43:39.400
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.
00:43:39.400 --> 00:43:43.100
For kinetic B and C, for that split second, it stopped.
00:43:43.100 --> 00:43:49.500
Kinetic will have 0 here and we will have our maximum when we are at position A.
00:43:49.500 --> 00:43:54.400
We can draw this, we will put our 0 in here to make our drawing a little bit simpler.
00:43:54.400 --> 00:44:04.400
We are just going to come down and up, the complimentary graph to our potential energy function.
00:44:04.400 --> 00:44:13.400
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.
00:44:13.400 --> 00:44:17.700
Let us do an analysis of the simple harmonic later.
00:44:17.700 --> 00:44:27.000
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.
00:44:27.000 --> 00:44:42.800
We can do that using F = kx or k = f /x which is 20 N /0.5 m or 40 N/m.
00:44:42.800 --> 00:44:51.400
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
00:44:51.400 --> 00:45:10.000
or ½ kx² which is ½ × 40 N /m × maximum displacement 0.5 m² or about 5 joules, the speed at the equilibrium position.
00:45:10.000 --> 00:45:16.600
To do that, the potential energy of the spring when it is at its maximum displacement must be equal to the kinetic energy.
00:45:16.600 --> 00:45:31.500
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,
00:45:31.500 --> 00:45:38.700
which is just going to be √5 or about 2.24 m/s.
00:45:38.700 --> 00:45:47.500
If we want the speed at 0.3 m, let us do this from a total energy perspective again.
00:45:47.500 --> 00:45:53.200
That is going to be the spring potential energy + the kinetic energy all has to be that constant 5 joules.
00:45:53.200 --> 00:46:08.600
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,
00:46:08.600 --> 00:46:19.800
which implies that V is going to be equal to the √2 × our total energy - kx² ÷ M,
00:46:19.800 --> 00:46:32.900
which implies that V = 2 × 5 joules - our spring constant 40 in our displacement 0.3 m² ÷ our mass of 2 kg.
00:46:32.900 --> 00:46:45.200
The square root of that whole thing gives me 1.79 m/s.
00:46:45.200 --> 00:46:50.800
Moving on, find the speed when it is at -0.4 m.
00:46:50.800 --> 00:47:02.000
We can use the same formula, the same analysis just with different values, V = √2 × our total energy - kx² /M
00:47:02.000 --> 00:47:15.200
which will be 2 × 5 joules - spring constant 40 N/m × displacement -0.4² ÷ our mass 2.
00:47:15.200 --> 00:47:26.400
Square root of all of that is 1.34 m/s, the acceleration of the equilibrium position.
00:47:26.400 --> 00:47:31.500
Remember, at equilibrium, at x = 0, our force is 0.
00:47:31.500 --> 00:47:36.000
Therefore, our acceleration is 0.
00:47:36.000 --> 00:47:40.800
How about the magnitude of the acceleration at 0.5 m?
00:47:40.800 --> 00:47:50.400
The force is - kx which is going to be -40 N/m × 0.5 m or 20 N - 20 N.
00:47:50.400 --> 00:48:05.500
The acceleration which by Newton’s second law is force/mass is -20 N / 2 kg or -10 m/s².
00:48:05.500 --> 00:48:11.100
Continuing on, find the net force of the equilibrium position.
00:48:11.100 --> 00:48:18.200
We have already said that at equilibrium, the acceleration is 0.
00:48:18.200 --> 00:48:34.400
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.
00:48:34.400 --> 00:48:37.900
And where does the kinetic energy equal the potential energy?
00:48:37.900 --> 00:48:44.800
The total is 5 that means kinetic has to equal 2.5 joules and potential has to equal 2.5 joules.
00:48:44.800 --> 00:48:51.000
We can solve that from our spring potential energy equation.
00:48:51.000 --> 00:49:08.600
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.
00:49:08.600 --> 00:49:17.600
Therefore, x is the square root of that or 0.354 m.
00:49:17.600 --> 00:49:25.800
It is important to note here that is in not halfway between the equilibrium position and maximum displacement.
00:49:25.800 --> 00:49:31.200
You do not have equal amounts of kinetic and potential energy when you are at half that displacement.
00:49:31.200 --> 00:49:35.500
It does not work that way.
00:49:35.500 --> 00:49:39.200
Another example, let us rank some spring systems here.
00:49:39.200 --> 00:49:49.800
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.
00:49:49.800 --> 00:49:52.800
It looks like this to would be another great spot for a table.
00:49:52.800 --> 00:49:58.900
We are going to make a table that has our system A, B, C, and D.
00:49:58.900 --> 00:50:22.900
We will look at our mass, our equivalent spring constant, and M/k which is proportional to our period is 2π √M /K.
00:50:22.900 --> 00:50:24.900
It looks like we can fill on our masses first.
00:50:24.900 --> 00:50:39.500
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.
00:50:39.500 --> 00:50:44.400
For B, because they are in parallel we just add them, that one was easy 25.
00:50:44.400 --> 00:50:54.300
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.
00:50:54.300 --> 00:50:58.200
When we solve that, we will get 7.5 N /m.
00:50:58.200 --> 00:51:02.700
The last one is in series again, just add them up 20.
00:51:02.700 --> 00:51:14.300
M/k is 0.75 6/25 about 0.242, 27.5 about 0.27 and 4/20 is 1/5 or 0.2.
00:51:14.300 --> 00:51:28.300
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.
00:51:28.300 --> 00:51:43.800
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.
00:51:43.800 --> 00:51:47.500
Alright, let us do a vertical spring block oscillator system.
00:51:47.500 --> 00:51:56.300
A 2 kg block attached to an un stretched spring, a spring constant k = 200 N/m as shown is released from rest.
00:51:56.300 --> 00:52:05.500
Determine the period of the blocks oscillation and the maximum displacement of the block from its equilibrium while undergoing simple harmonic motion.
00:52:05.500 --> 00:52:25.200
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.
00:52:25.200 --> 00:52:30.900
To find the maximum displacement, we will do a little bit more work here.
00:52:30.900 --> 00:52:40.100
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.
00:52:40.100 --> 00:52:47.400
We can solve for Δ y using the gravitational potential energy = stored potential energy in the spring
00:52:47.400 --> 00:52:55.500
which implies that MG Δ y must be equal to ½ K Δ y².
00:52:55.500 --> 00:53:08.000
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,
00:53:08.000 --> 00:53:11.000
that is the entire displacement from the top and the bottom.
00:53:11.000 --> 00:53:17.100
We want the displacement from its equilibrium position to its maximum displacement so that is going to be half of that.
00:53:17.100 --> 00:53:36.100
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.
00:53:36.100 --> 00:53:49.800
I come up with about 0.1 m for our maximum displacement from its equilibrium position while it is in that simple harmonic motion.
00:53:49.800 --> 00:53:52.300
Let us rank the periods of some pendulums.
00:53:52.300 --> 00:53:58.800
We have these 4 pendulums of uniform mass density from highest to lowest frequency.
00:53:58.800 --> 00:54:13.800
Period is 2π √L /G, so the frequency is 1 /the period that must be 1/2 π × √G /L.
00:54:13.800 --> 00:54:26.900
If we want the shortest period, if we want the shortest frequency, the smallest frequency, we want the biggest length.
00:54:26.900 --> 00:54:29.300
If we want the highest frequency, we want the shortest length.
00:54:29.300 --> 00:54:34.300
The shorter ones are going to go back and forth a lot more quickly than the long ones.
00:54:34.300 --> 00:54:46.800
This would be D, A, B, C, because frequency is only a function of the length.
00:54:46.800 --> 00:54:52.300
Alright, let us finish this off by looking at a couple of free response problems from old AP exams.
00:54:52.300 --> 00:55:00.100
Let us start off with a 2009 exam free response question number 2.
00:55:00.100 --> 00:55:06.000
As we look at this one, we are given a long thin rectangular bar of mass M, length L.
00:55:06.000 --> 00:55:07.900
We are to determine its moments of inertia.
00:55:07.900 --> 00:55:10.800
It has a non uniform mass density.
00:55:10.800 --> 00:55:16.300
First A1, we are going to write the differential equation for the angle θ.
00:55:16.300 --> 00:55:21.100
To do this, I'm going to draw it at its extreme just to make it easier to see what is going on.
00:55:21.100 --> 00:55:25.200
We would not raise it at that height because we would lose our small angle approximation.
00:55:25.200 --> 00:55:31.600
If we brought it up to the side like this, we will have some MG wherever that center of mass happens to be.
00:55:31.600 --> 00:55:37.900
We will call x the distance from our pivot to our center of mass and there is a our angle θ.
00:55:37.900 --> 00:55:59.200
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.
00:55:59.200 --> 00:56:15.800
Αlpha we know is the second derivative of θ so we can write this now as - MG × x sin θ = IB D² θ / DT².
00:56:15.800 --> 00:56:19.700
That should cover us for writing the differential equation.
00:56:19.700 --> 00:56:27.300
Now for part A2, it says apply the small angle approximation to calculate the period of the bars motion.
00:56:27.300 --> 00:56:38.600
The small angle approximation says for small θ is about under about 15°, sin θ is approximately equal the θ.
00:56:38.600 --> 00:56:42.700
It would not work as we have not draw on our diagram, we would not actually raise it that high.
00:56:42.700 --> 00:56:56.700
If we did this with a small angle and let it go, we could then write that - mgx × θ = IB D² θ / DT² or
00:56:56.700 --> 00:57:12.000
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.
00:57:12.000 --> 00:57:24.100
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.
00:57:24.100 --> 00:57:41.900
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
00:57:41.900 --> 00:57:54.000
which is going to be 2π × √ of the moment of inertia of our bar ÷ mgx.
00:57:54.000 --> 00:57:57.700
That should cover us for part A2.
00:57:57.700 --> 00:58:03.800
Let us give ourselves more room here for part B.
00:58:03.800 --> 00:58:09.300
Describe the experimental procedure you would use to make the additional measurements needed to determine IB
00:58:09.300 --> 00:58:13.600
and how you would use your measurements to minimize experimental error?
00:58:13.600 --> 00:58:22.700
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°.
00:58:22.700 --> 00:58:29.400
It will go back and forth say 10 times and measure the amount of time it takes to go back and forth 10 times.
00:58:29.400 --> 00:58:37.400
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.
00:58:37.400 --> 00:58:47.300
Once you have your period, you know period is 2π √ of the moment of inertia of your bar / mgx.
00:58:47.300 --> 00:59:06.700
So then T² = 4π² IB / mgx which implies then that IB is just going to be equal to T² mgx / 4π².
00:59:06.700 --> 00:59:13.300
You could use that in order to find your moment of inertia of the bar.
00:59:13.300 --> 00:59:22.700
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?
00:59:22.700 --> 00:59:24.100
That should be pretty easy too.
00:59:24.100 --> 00:59:35.900
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.
00:59:35.900 --> 00:59:42.900
Wherever you balance it, make a duct and that is going to be your location of the center of mass of the bar.
00:59:42.900 --> 00:59:46.800
You could also hang it from different points and look where all the plumb bob lines across.
00:59:46.800 --> 00:59:57.800
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.
00:59:57.800 --> 01:00:09.400
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.
01:00:09.400 --> 01:00:17.800
I think that covers the 2009 question, let us do one more.
01:00:17.800 --> 01:00:24.100
Let us go to the 2010 exam free response 3.
01:00:24.100 --> 01:00:31.500
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
01:00:31.500 --> 01:00:35.000
where you have sin function between 0 and T.
01:00:35.000 --> 01:00:42.300
After T it is not accelerating anymore, acceleration is 0, greater than T.
01:00:42.300 --> 01:00:47.500
Alright derive an expression for the velocity of the skiers, a function of time.
01:00:47.500 --> 01:01:02.200
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.
01:01:02.200 --> 01:01:17.600
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.
01:01:17.600 --> 01:01:30.500
We can pull other constant that is going to be A max × the integral to the sin π t /T.
01:01:30.500 --> 01:01:35.900
I should not forget my DT here and the integral of the sin is going to be the opposite of the cos.
01:01:35.900 --> 01:01:41.600
We need to have the argument in here, we are going to have our du which is π / T.
01:01:41.600 --> 01:01:46.400
We need to have T /π out here to make a ratio of 1.
01:01:46.400 --> 01:02:04.600
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.
01:02:04.600 --> 01:02:23.100
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.
01:02:23.100 --> 01:02:49.400
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.
01:02:49.400 --> 01:02:58.900
For part B, find the work done by the net force on the skier from rest until it reaches terminal speed there at T.
01:02:58.900 --> 01:03:02.500
We could use the work energy theorem here.
01:03:02.500 --> 01:03:10.300
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
01:03:10.300 --> 01:03:16.100
which is ½ MV final² -1/2 MV² initial.
01:03:16.100 --> 01:03:19.300
V initial is 0 that part it is easy.
01:03:19.300 --> 01:03:48.200
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.
01:03:48.200 --> 01:04:03.400
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
01:04:03.400 --> 01:04:26.500
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.
01:04:26.500 --> 01:04:32.000
And that should cover us for part B.
01:04:32.000 --> 01:04:41.900
Taking a look at part C, determine the magnitude of the force exerted by the rope on the skier at terminal speed.
01:04:41.900 --> 01:04:49.200
Here I'm going to look at my free body diagram, we will draw our axis in first.
01:04:49.200 --> 01:04:59.500
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.
01:04:59.500 --> 01:05:06.000
If I wanted to draw our pseudo free body diagram, breaking up that MG into components parallel
01:05:06.000 --> 01:05:12.500
and perpendicular with the axis, I would have still, we have our normal force.
01:05:12.500 --> 01:05:16.000
We have our force of tension from the rope.
01:05:16.000 --> 01:05:24.400
We have MG sin θ and of course MG cos θ.
01:05:24.400 --> 01:05:29.500
I can write my Newton’s second law expression in the x direction.
01:05:29.500 --> 01:05:42.200
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.
01:05:42.200 --> 01:05:51.200
Therefore, force of tension in the rope must be MG sin θ.
01:05:51.200 --> 01:05:55.300
That should cover us for C.
01:05:55.300 --> 01:06:02.400
Part D, derive an expression for the total impulse imparted to the skier during the acceleration.
01:06:02.400 --> 01:06:05.700
Impulse is the integral of force with respect to time.
01:06:05.700 --> 01:06:18.700
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
01:06:18.700 --> 01:06:32.500
to some final value T of M, our acceleration function is given we have a max sin π t /T DT.
01:06:32.500 --> 01:06:51.500
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.
01:06:51.500 --> 01:07:04.200
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.
01:07:04.200 --> 01:07:25.300
That is going to be Ma max T / π and then we will have that cos π t /T.
01:07:25.300 --> 01:07:29.800
Let us see, I will evaluate it from 0 to T.
01:07:29.800 --> 01:07:34.600
We got to have our negative sign and their 2 the integral of the sin is the opposite of cos.
01:07:34.600 --> 01:07:49.700
That is going to be equal to - M a max T/ π × cos π, when we substitute T in for t.
01:07:49.700 --> 01:07:59.000
- the cos 0 which is 1, cos π is -1, -1 is going to be 2.
01:07:59.000 --> 01:08:14.100
-2 we will have -2 × negative this will give us 2 ma max T /π.
01:08:14.100 --> 01:08:27.200
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.
01:08:27.200 --> 01:08:36.600
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.
01:08:36.600 --> 01:08:39.900
Label them F1 and F2.
01:08:39.900 --> 01:08:43.200
First thing I want to do is I'm going to figure out what my functions are.
01:08:43.200 --> 01:08:51.300
If we look at the original first, F1 is what we will call that our original.
01:08:51.300 --> 01:09:07.200
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.
01:09:07.200 --> 01:09:23.500
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.
01:09:23.500 --> 01:09:46.000
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.
01:09:46.000 --> 01:10:02.800
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.
01:10:02.800 --> 01:10:05.700
Let us try our axis and see if we can graph these.
01:10:05.700 --> 01:10:11.200
Finish this one up in style.
01:10:11.200 --> 01:10:25.300
There is our y our force, here is our x our time, we have got some value T and force.
01:10:25.300 --> 01:10:36.800
Let us label here MG sin θ and now for regional force we know that at time T =0,
01:10:36.800 --> 01:10:44.400
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.
01:10:44.400 --> 01:10:52.500
We are going to have our maximum value halfway to T, when we have sin π /2.
01:10:52.500 --> 01:10:58.100
That will be at that point we will have MG sin θ + M a max.
01:10:58.100 --> 01:11:02.400
We could label that point up here as well, that will be important.
01:11:02.400 --> 01:11:09.400
MG sin θ + MA max right there.
01:11:09.400 --> 01:11:18.600
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.
01:11:18.600 --> 01:11:23.600
This will be back to MG sin θ and after that, it is a constant flat line.
01:11:23.600 --> 01:11:32.900
We have that part and in between we have our sin function which is going to bring us up to a maximum here.
01:11:32.900 --> 01:11:40.500
It is kind of something that looks like that for initial force.
01:11:40.500 --> 01:11:46.400
Our new force is MG sin θ + Ma max E to all of this.
01:11:46.400 --> 01:11:52.900
At time T = 0, e⁰ is 1 so we are going to start at MG sin θ + MA max.
01:11:52.900 --> 01:12:03.900
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.
01:12:03.900 --> 01:12:08.600
This piece is going to go to 0 and we are going to decay down the MG sin θ.
01:12:08.600 --> 01:12:14.000
I would say that this one would look something like that, may be not quite like that.
01:12:14.000 --> 01:12:20.600
Like that with some sort of exponential decay there so there would be our F2.
01:12:20.600 --> 01:12:23.400
Hopefully, that gets you a great start on oscillations.
01:12:23.400 --> 01:12:27.000
Thank you so much for joining us here today at www.educator.com.
01:12:27.000 --> 01:01:12.000
I look forward to seeing you again soon and make it a great day everyone.