Physics 3B Final
Chapter 15: Oscillations The Simple Pendulum
A simple pendulum consists of an object of mass m suspended from a light string of length L attached to a frictionless pivot. The string makes an angle θ with the vertical. The maximum angular displacement of the simple pendulum system from the vertical is called the angular amplitude θₘₐₓ... (i) For small angle oscillations, θ ≤ 15°, one may use the approximation sin(θ) ≈ θ, and also (ii) From geometry, x = lθ → ... hence the equation of motion becomes ... (small angles θ) compare with the general equation ... → ... so the motion is simple harmonic. Thus, from the equation ... → ... (period for small angle oscillations). (iii) For large angle oscillations, the period is given by a more complicated expression, namely ...
Chapter 15: Oscillations Simple Harmonic Motion: Spring-Mass System Know Hooke's law, and the required conditions for applying related equations. * Know the equation for calculating the restoring force, Fₛ. - Know relevant variables - what they stand for, and their units. What is the equilibrium position? Know how to calculate acceleration for a spring-mass system. * When is the acceleration of the object zero? When is it max? Define: * Amplitude * Period * Frequency * Angular frequency Know relevant variables and units for all. Know the equation relating frequency and period. Know the equation relating frequency to angular frequency. Know the equation relating period and angular frequency. Understand spring pictures.
Consider an object of mass m attached to a spring oscillating on a frictionless horizontal surface. Hooke's law says that the restoring force Fₛ exerted by an ideal spring is: Fₛ = -kx where.. k = spring constant x = displacement from the equilibrium position The minus sign in Hooke's law indicates that the restoring force is directed toward the equilibrium position. Note that the "restoring" force is linear in the displacement x from the equilibrium (unstretched) position. When an object is under the effect of a linear restoring force, the motion the object follows is described as "simple harmonic motion" or SHM. The spring-mass system itself is called a "simple harmonic oscillator". What is the equation that describes the motion of a simple harmonic oscillator? Applying Newton's second law of motion along the x-axis yields... Note carefully that the acceleration of the object is linear in x, that is, it is NOT constant. You must NOT use those 3 kinematic equations you learned in Physics 3A. The acceleration is zero when the object is at the equilibrium position (x=0), while the acceleration is a maximum when the object is farthest away from the equilibrium position, that is, at the amplitude (x = ± A), precisely where the object stops! A = amplitude = the magnitude of the maximum displacement from equilibrium. T = period = time required to complete one oscillation. f = frequency = the number of oscillations per second (in Hertz) ω = angular frequency (in radians per second) From the definitions of frequency and period, one can see that f = 1/T while the frequency f is related to the angular frequency ω by ...
Chapter 15: Oscillations Simple Harmonic Motion: Spring-Mass System Know the expression for energy stored in a spring. * Know what the variables of the expression tell you. Describe how a spring responds to you stretching it: what is the force exerted by the spring always attempting to do? * Relate this to potential energy and x. Describe the direction of Fₛ. What is this force always attempting to do? * When is x = 0? (+)? (-)? * What is the (-) sign for Fₛ showing? * If you stretch the spring, what does it do? * If you compress the spring, what does it do?
Expression for energy stored in a spring: Fₛ = -kx. * (-) sign tells you the force is to bring the spring back to its natural length. If you attach a block to a spring, and there's no friction on the surface between the block and the surface, the spring has its natural length and the block just sits. If you grab the block and pull it to the right, the spring will argue and pull back. The force exerted by the spring is always attempting to bring the spring back to where it was when it was happy, back when no potential energy was stored in it, when x = 0. If you push the spring to the left, compressing it, Fₛ is always towards x = 0, attempting to bring the spring back to its initial length. x = 0 when the spring has its natural length, x = (+) when the spring gets longer (stretched), x = (-) when the spring gets shorter (compressed). The (-) sign for Fₛ is just to show you the direction of the force exerted by the spring - if you stretch it, it pulls back. If you compress it, it pushes forward.
Chapter 15: Oscillations Simple Harmonic Motion Mathematical Representation of SHM: Know how to find x(t) for the given situations, and know the related graphs: 1. Suppose that at t=0, the object starts from rest at the amplitude x=A 2. Suppose that a t=0, the object starts from rest at x=-A 3. Suppose that at t=0, the object starts with vₘₐₓ at the equilibrium position x=0 toward positive values of x 4. Suppose that at t=0, the object starts with vₘₐₓ at the equilibrium position x=0 toward negative values of x
Given that the acceleration equals ... we can then rewrite the equation of motion as ... In mathematics, one learns that an equation of the form ... has a general solution x(t) given by ... where A and ⦽ are constants of the motion determined from the "initial conditions". The quantity ⦽ is called a phase angle and its value depends on where and in which direction the object is moving at t=0. (a) Suppose that at t=0, the object starts from rest at the amplitude x=A. Then... (b) Suppose that a t=0, the object starts from rest at x=-A. Then... (c) Suppose that at t=0, the object starts with vₘₐₓ at the equilibrium position x=0 toward positive values of x, that is, stretching the spring. Then... (d) Suppose that at t=0, the object starts with vₘₐₓ at the equilibrium position x=0 toward negative values of x, that is, compressing the spring. Then...
Chapter 15: Oscillations Energy in Simple Harmonic Motion Know the four graphs we can use to determine the position of the spring, x, as a function of time, t: x(t).
How do we get x as a function of time? What is the equation going to be? What you write down is going to depend on where the block is at t = 0, or when you begin keeping track of time. Depending on where it is, it's a different expression that you write down. Remember four graphs - for final, there's going to be one of those four graphs and you'll need to figure out and then get going!!! Case #1 - we're interested in x, what is x? How far is the block away from its position of happiness? From the equilibrium position? We call this x, and we want to know x as a function of time. When we started rotational motion, theta, for something with constant and with zero angular acceleration, this is omega t. Case #2 - omega t instead of theta, and x instead of y. +A instead of +1, and -A instead of -1. What is x as a function of time here? x = -Asin(ωt) Case #3 - Notice A is the maximum value of x. It is the farthest away the block is going to be from the equilibrium position, from x = 0. * x = Acos(ωt). Case #4 - x = -Acos(ωt) * A represents the maximum value of x, its amplitude, it's the farthest away the spring will be from the end that it wants to be at, x = 0, the farthest it will be from the equilibrium position (with the block attached to it).
Chapter 15: Oscillations Energy in Simple Harmonic Motion If "happy", describe the motion of the spring. Describe the relationship between motion and potential energy - if there's motion, what does this say about potential energy? If there's no motion, what does this say about potential energy? When the spring has its happy length, its natural length, describe potential energy of the spring. If we stretch/compress the spring, describe what happens. Describe this in terms of energy. Know the formula for calculating elastic potential energy of the spring. Describe the energy stored in the spring when the spring has its natural length. In this state, what is x? (x = ___). What is the force exerted by the spring called? Describe the amount of energy stored in the spring, the more you stretch the spring. Why is this? What is the spring constant (k) a measure of? * The ___ the spring, the higher the k. * Units?
If happy, the spring isn't pulling or pushing (not fighting with you). You don't see any motion when the spring is happy. If there is motion, there must have been energy stored somewhere that's going to be converted into motion. Because there's no motion when the spring is happy, there's no potential energy. When the spring has a happy length, it's natural length, there's no potential energy stored in the spring. If we fix one end and stretch the spring, and then let go, it's going to go "UGH" and yank itself back to its initial natural length. As it goes back, there's motion, there's kinetic energy that must've come from somewhere. So when the spring is stretched, there's energy stored in the spring → potential energy. If we compress the spring and let go, the spring pushes back → motion → kinetic energy. The kinetic energy is from energy stored in the spring, from when the spring was compressed (potential energy). If the spring is stretched from its natural length, it has potential energy. If the spring is compressed from its natural length, it has term-14energy stored in it, has potential energy. When the spring gets deformed, energy is stored in the system → Uₛ = (1/2)kx² No energy is stored in the spring when the spring has its natural length, when x = 0, so Uₛ = 0. Stretched/compressed = energy is stored in the spring (elastic potential energy). Force exerted by the spring = elastic force. * Elastic potential energy is always positive, because x gets squared. The more you pull the spring, the more energy is stored in the spring, because the x is bigger. Measure of stiffness of the spring = k. * The stiffer the spring is, the higher the k value, spring constant. - Units: N/m - k is always a (+) value.
Chapter 15: Oscillations Energy in Simple Harmonic Motion What forces are acting on the spring? What forces affect the motion of the spring? As the block is getting closer to the equilibrium position, what is happening to the spring? What variables are changing? As the block slides to the left, what is it doing?
If the object is acted on by various forces, this object has various forces acting on it (normal force (N) is being exerted by the horizontal surface on the spring, the force of gravity (mg)). Three forces in total are being exerted on the spring, acting on the object. As the spring moves right and left, the N and mg doesn't affect the motion of the spring. The spring is pointing to the left, and as the block gets closer to the x = 0 position, it's getting smaller and so the force is getting smaller, so the resulting acceleration is changing because the sum of the forces in the x-direction = ma, so -kx = ma, so the object (the block) slides to the left, and as it does, it is accelerating. The acceleration is not constant. The acceleration is = -(k/m)*x - not constant. Therefore, we can't use any 3A kinematic equations.
Chapter 15: Oscillations Simple Harmonic Motion: Spring-Mass System Define "x". Describe how a spring responds as you continue to stretch it. What makes a spring elastic? What does this tell you about the spring? What happens if we pull a spring beyond its elastic limit? What does this tell you about the spring? What are the requirements for a spring to obey Hooke's law?
If we deform the spring (stretch/compress), how far we pull or compress it = x. x has nothing to do with the length of the spring. It has to do with how much you stretch or compressed the spring - the length of which you made the spring unhappy. If we grab the spring at its natural length, and stretch it, the more we pull it, the more unhappy it gets, and the harder it pulls back. If we let go at some point, and the spring regains its natural length, the spring is said to be elastic, and obeys Hooke's law. If the spring obeys Hooke's law, then when that force is exerted on the spring to deform it, when we remove the force, the spring will go back to its natural length. We can pull the spring beyond its elastic limit, such that if we let go, it won't go back to its initial length it once had. Once we pull the spring beyond its elastic limit, as you keep pulling and pulling, that x doesn't obey Hooke's law, so the spring doesn't obey Hooke's law. A spring can only obey Hooke's law if it's elastic. If it is said to be elastic, if you remove the force to deform the spring, it will go back to its original length.
Chapter 15: Oscillations Energy in Simple Harmonic Motion If we stretch the spring 15 cm and hold it, given A, at t = 0, and we let go, how would we solve for the position of the spring at t = 5s? Where is it 5s after we let go of the spring? How fast is it traveling in this instance? In what direction? What's the acceleration? (What is the position (x) as a function of time, velocity as a function of time, and acceleration as a function of time?)
If we grab the block and pull it and hold it, 15 cm, given A, at t = 0, and we let go, and we're interested in the springs position at 5s, because it's oscillating, can we figure out where it is 5s after we let go, and how fast it's traveling? And in what direction? And what the acceleration is? What is the position (x) as a function of time, velocity as a function of time, and acceleration as a function of time? We can set the spring into oscillation in any way we want, and when we say t = 0, at any later time, we should be able to figure out its position, velocity, and acceleration. * Need to obtain x given any time (t). * When asked for velocity, take the derivative of the position with respect to time, or now that we know x, we can plug it in and solve for the speed using the work energy theorem. Now that we know x, because we're told time, so we can solve for x, we can find the speed when it gets to the position (x). If we know x at any instance of time, we can plug it in and find the acceleration. If we forgot about that, once we've found the velocity, by differentiating the position, we can differentiate the velocity with respect to time and figure out the acceleration.
Chapter 15: Oscillations Energy in Simple Harmonic Motion Know how to determine which graph to use to solve for x(t). When is the only time these four cases work or apply?
If you just know it starts at 0, it's a sine function. * If it goes up → (+). * If it goes down → (-). The only time these four cases work or apply is when t = 0. When we start the timer, the block is in one of the four possible positions. * For example, case #1 - the block is at x = 0 when t = 0. Then x gets more and more (+), so we're stretching the spring. That's if t = 0. * At t = 0, you grab a hammer and hit it, and it flies. What's x as a function of time? When t = 0, we said the block was at x = 0. When we hit it, did it begin to go to x = +A or x = -A? x = +A! Then, we know this is a x = Asin(ωt) and then we can start solving this problem. If at t = 0, we're told we have our hammer and the block is at the equilibrium position, and we hit it, and give it some initial speed that way, what is x as a function of time? Draw the graph. At t = 0, it was at the equilibrium position, and we hit it and it began to go initially towards x = -A or x = +A? x = -A! And then we already know what x is as a function of time! And then we can start solving the problem. What if we're told that at t = 0, we grab the block and say okay, we're going to set it into oscillation by pushing the block some distance A, holding it compressed, and at t = 0, we let go. What's x as a function of time? Well, where did it start at? Where was it at t = 0? At x compressed (x = -A). So, it starts at x = -A → x = -Acos(ωt) and then we can start solving the problem. If instead we grab the block attached to the spring, pull it some distance A, and hold it there, look at our watch and see t = 0, and let go, x as a function of time = ? * It started out stretched, so at x = A → x = Acos(ωt)
Chapter 14: Entropy Reversible and Irreversible Processes A. The Second Law of Thermodynamics
In all of nature there is a one-way drive toward thermodynamic equilibrium. There is no way to undo these one-way drives. (a) Mechanical one-way drive: frictional dissipation of kinetic energy into internal energy as heat. (b) Thermal one-way drive: thermal energy flows spontaneously from hot to cold. (c) Chemical one-way drive: chemicals flow from region of high concentration to regions of low concentration. This is diffusion. This is the general irreversibility statement of the second law of thermodynamics.
Chapter 15: Oscillations Simple Harmonic Motion: Spring-Mass System Understand when a spring is "happy" vs. "unhappy". Know how a spring responds to being stretched or compressed from its initial, natural length.
Let's say we have a spring with a "length". If it has that length, then it's happy. It's happy because it doesn't want to change it's state → in a state of happiness. If we grab the spring, hold one end fixed, and stretch the other end, the spring is unhappy → now has a length it didn't want to have. By Newton's third law, we pulled the spring one way, and the spring wants to pull back. Or, the spring was at the length that it liked - if we pull it to make it longer, the spring doesn't like it - wants to go back → spring is unhappy, it wants to go back to its happy length → pulls back. If we compress the spring, the spring now has a different length → unhappy. Spring will push back, trying to achieve its happy length.
Chapter 15: Oscillations Simple Harmonic Motion Spring-mass system: Know the equation for calculating omega (ω) of a spring-mass system. * Know how we can use this in equations solving for x(t). * Know equations relating period to angular frequency, and know how these equations can be altered, using equations for omega (ω). * Know equations relating frequency and period, and how these equations can be altered, using equations for omega (ω). Compare horizontal and vertical springs. * How do the un-stretched position and equilibrium position relate for either springs? * When do equations apply to both springs?
Note from the equations above that for a spring-mass system, ... so that the position x(t) of the object of mass m attached to the spring is given as a function of time as... and hence... Question: What about the case of an object of mass m attached to a vertical spring? It is interesting to note for a horizontal spring, the unstretched position and the equilibrium position coincide. But for a vertical spring, the unstretched position does not coincide with the equilibrium position. One thus needs to be careful when analyzing the motion of a vertical spring-mass system. It turns out that all the equations above apply to both a horizontal and vertical spring if the position x(t) is measured from the equilibrium position.
Chapter 15: Oscillations Energy in Simple Harmonic Motion How would we solve a problem to figure out the speed of the spring when given a position (x)? (what would we use to solve the problem?) When a spring is stretched to its max displacement from the equilibrium position, and the block is at rest, what is the energy of the spring? What is the speed? What is x at this position? What will be the energy of the system at the instant the spring passes the equilibrium position? Describe what happens when the spring reaches the equilibrium position. Understand why. Describe the speed at the instant the spring passes the equilibrium position. Explain why.
So, the only way to solve problems as far as figuring out speed when given a position (x), is to use the work-energy theorem, conservation of energy. When the spring is stretched the most in this case, so the block is at rest, what is the energy? The spring is not moving, because it's at the end point, so speed is 0. Is there potential energy in the system? It's stretched, so it's unhappy. So, Uₛ = (1/2)kx². x = A at that position. What will be the energy of the system at the instant the spring passes the equilibrium position? When it gets there, it wants to stop, but the object is moving. Because of inertia, it will overshoot the equilibrium position, the position of happiness. At that instant, the potential energy of the system = 0, so the kinetic energy has to be its initial kinetic energy at the beginning. All of that potential energy gets converted into kinetic energy, so at the instant the spring passes the equilibrium position, it has its maximum speed, vₘₐₓ. All of it is strictly kinetic energy at this point, because x = 0 there, making Uₛ = 0. So, all the kinetic energy, (all (1/2)mvₘₐₓ²). vₘₐₓ is the maximum speed the block is going to have, and it will happen at the position of happiness of the spring, at x = 0. It will then overshoot and go to A on the other side.
Chapter 14: Entropy Reversible and Irreversible Processes
Thermodynamic processes that occur in nature are all irreversible processes. These are processes that occur spontaneously in one direction but not in the reverse direction. Irreversible processes are all non equilibrium processes, in that the system is not in thermodynamic equilibrium at any point until the end of the process. The second law of thermodynamics determines the preferred direction for such processes. A reversible process is an idealized process that a system is taken through such that the system is always very close to thermodynamic equilibrium within itself and with its surroundings. Reversible processes are equilibrium processes, with the system always in thermodynamic equilibrium. A cyclic process is a sequence of processes that eventually leaves the working substance in the same state in which it started. In an internal-combustion engine, such as that used in an automobile, the working substance is a mixture of air and fuel.
Chapter 15: Oscillations Energy in Simple Harmonic Motion What is the value of x at the equilibrium position? After stretching the spring, what is its speed at the equilibrium position? Describe what will happen to the spring when it's trying to return to its happy length after being stretched. Define amplitude. What is the speed when the spring is at this distance? * If one were to let go of the spring at the amplitude, what would happen? - What is changing as this is happening?
Suppose you grab a block and stretch it to the right (10 cm for example), and you hold it, and let go. The spring will pull back and the block will move to the left and slide on the surface. If there's no friction, once it gets to the equilibrium position, what is the speed? x = 0 at the equilibrium position - what is the speed at x = 0? * Because it's moving due to inertia, it will overshoot. Suppose you have a spring with no friction, and attach it to a mass (m). The mass of the spring is usually ignored, and the spring has a spring constant (k), but its mass is ignored. x = 0, the spring has its natural length, where it's happy. Nothing happens. If you pull the block to the right, stretching the spring, v = 0. This distance is A if v = 0. A stands for amplitude, the maximum displacement from the equilibrium position. When you let go of the spring, the spring will shoot back to the left. BUT, the force exerted by the spring (elastic force), magnitude of the force is kx (the (-) sign is ignored, since we're only talking about magnitude right now). As the block continues to slide to the left, x gets smaller and smaller. So, x is changing. When x is changing, the force the spring exerts is changing. If the force changes, the acceleration is changing, because F = ma.
Chapter 14: Entropy Entropy S Entropy Changes in Irreversible Process
The change in entropy ΔS of a system as the system undergoes a quasistatic process between two equilibrium states is given by ... 1. This is so because the differential form of the entropy function is... and integrating both sides of the equation yields... 2. More generally, one can show that the differential form of the entropy function may also be written as ... (a) For an isochoric (isovolumetric) process of an ideal gas, dQ = ncᵥdT so that... (b) For an isobaric process of an ideal gas, dQ = ncₚdT...
Chapter 14: Entropy Entropy S
The second law of thermodynamics can be stated in terms of the concept of entropy, a quantitative measurement of the degree of randomness (or disorder) of a system. Entropy is used to quantify the dispersal of thermal energy. A system in which thermal energy is concentrated - not spread out - has low entropy. Systems in which thermal energy is more spread out are systems with more entropy. That is, entropy increases as more thermal energy becomes more dispersed. The entropy of a system is maximum when the system is in a state in which thermal energy is maximally dispersed (or shared among all the constituents), and when this state is reached, we say the system is in equilibrium. The one-way irreversibility of macroscopic processes in nature can be described by an inequality law. There is a quantity S called the entropy which has the following properties: (a) Additive: S. = S₁ + S₂ + ... (b) Each Sᵢ is a function of the thermodynamic state of the system. It is a function of temperature and volume. (c) Entropy is not conserved. It can only be created, but never destroyed. Entropy is created whenever a one-way process occurs. Hence, ΔS(ᴛᴏᴛᴀʟ ᴜɴɪᴠᴇʀsᴇ) = 0 (for a reversible process, i.e. not a one-way process) ΔS(ᴛᴏᴛᴀʟ ᴜɴɪᴠᴇʀsᴇ) > 0 (when a one-way process occurs) Here, S(ᴜɴɪᴠᴇʀsᴇ) is the total entropy of the objects involved. Entropy is a state variable (like pressure, volume, and temperature). Thus depends only on the properties of the initial and final equilibrium states of the system. * An isentropic process is one for which ΔS(ᴛᴏᴛᴀʟ ᴜɴɪᴠᴇʀsᴇ) = 0.
Chapter 15: Oscillations Energy in Simple Harmonic Motion What if we wanna know the energy of the spring of the system, somewhere over here (red), when the speed is still moving at speed (v) to the left, and the distance is x? It's not quite at A? * What is the total energy? Explain. * What is the total energy at the equilibrium position? * What is the total energy at a position between the amplitude and the equilibrium position? If given the amplitude, spring constant, and mass of the block, if we were to stretch and let go of the spring, how could we solve for the speed of the spring as it passes the equilibrium position? If given how fast the spring is moving when passing the equilibrium position, how can we figure out the amplitude? If given x, and asked how fast the spring is moving at the instant x = (1/3)A, how would we go about solving this problem?
What if we wanna know the energy of the spring of the system, somewhere over here (red), when the speed is still moving at speed (v) to the left, and the distance is x? It's not quite at A. At this position, it's moving, so it has kinetic energy, and there's also an x, so there's potential energy. The total energy is the sum of those two (kinetic + potential energy). Energy is conserved, because there's no friction. The total energy at the equilibrium position is the total energy at the amplitude. At the position between the amplitude and the equilibrium position, → (1/2)kx² + (1/2)mv². If we know the spring constant, we pull the block so many centimeters from the equilibrium position and hold it, so we know this, and we're given the mass of the block, and we pull the block and hold it, given A, k, mass, and let go of the spring, what's the speed as it passes the equilibrium position? * Set equations equal to each other, and then we can find out how fast the spring is moving at the instant it passes the equilibrium position. OR, given how fast the spring is moving when passing the equilibrium position, we can figure out the amplitude by setting the equations equal to eachothr and solving. * If we take the object and pull it to the right some distance A that's given, and let go and ask when it gets to the other side and x = (1/3)A, we can figure out how fast the spring is moving. Given x and asked how fast it's moving at the instant when x = (1/3)A → set equations equal to each other and solve. There's no reference to time in this equation.
Chapter 15: Oscillations Energy in Simple Harmonic Motion When does a spring have energy stored in it? What is this energy called? Know how to calculate it. * Know relevant variables and what they stand for. Define mechanical energy. * Know the equation we get from conservation of mechanical energy. * How is (x) measured? Describe. Understand the spring pictures.
When a spring is stretched (or compressed), it has energy stored in it. This energy Uₛ is called "elastic potential energy". A spring that has a spring constant k and is stretched (or compressed) a distance x, the elastic potential energy equals: Uₛ = (1/2)kx² In physics 3A, you learned about the work-energy theorem. It says that the total work done by the non-conservative forces acting on a system is equal to the change in the mechanical energy E of the system. The mechanical energy is defined as the sum of kinetic and potential energies (PE = elastic and gravitational). Mathematically, ... → conservation of mechanical energy then yields... In both equations, x is measured from the equilibrium position, which does not always coincide with the unstretched position as mentioned above.
