M Grell, PHY221: Topics in classical physics. Contents

Transcription

1 Intro: What is classical physics? 1. Dimensional analysis Units vs. dimensions Dimensional analysis Dimensionless groups 2. The harmonic oscillator Simple and damped harmonic oscillator Driven oscillator Coupled oscillators 3. Waves General description, wave equation Waves on strings, in fluids, and in solids Intensity of waves Transmission and reflection of waves Dispersion of waves Lightwaves in matter M Grell, PHY221: Topics in classical physics Contents 4. Fictitious forces Coordinate systems and frames of reference Outer product, right hand rules, pseudovectors Inertial and rotating frames of reference Fictitious forces 5. Mechanics à la Lagrange Lagrange in a nutshell The Lagrangian formalism Cyclic coordinates Using the Lagrangian: Simple examples Using the Lagrangian: Advanced examples The 2- body problem Recommended reading: Fowles Analytical Mechanics, or the newer edition, Fowles and Cassiday, Analytical Mechanics Goldstein, Classical Mechanics For dimensional analysis, e.g. J F Douglas, An introduction to dimensional analysis for engineers, Pitman 1969

2 What is Classical Physics? There is (at least) three ways of defining classical physics: Via the areas of physics it does or doesn t apply too like engineering mechanics, electrical engineering, geometric and wave optics, thermodynamics, and relativity ; but NOT quantum mechanics, quantum optics, solid state physics,.. Or, by philosophical debate on underlying concepts and assumptions on the nature of reality which we will avoid here. Or, we can just state a number of assumptions on which this course is based: - 3-dimensional space with no curvature. - Time, t, ticks away evenly in the same way for all ( absolute time ). - Space and time are separate phenomena that provide an arena for physical events, but are not themselves involved. - There is no limit to the velocity a body can have, and mass is independent of a particle s velocity. - All bodies at the same time have precisely defined location and momentum. - Bodies interact through forces only. You will have noticed that we exclude not only quantum phenomena, but also relativity. Relativity IS part of classical physics, but just like optics or thermodynamics, it gets a course of its own. The weirdest of the assumptions is the final one. Exercise: What quantum mechanical interactions between particles are NOT forces? Practically, these are justified assumptions for bodies that are much bigger and heavier than elementary particles, and move much slower than the speed of light. This still covers a lot of real world, e.g. all mechanical engineering.

3 1 Dimensional analysis Units vs dimensions A key difference between equations in physics, and equations in mathematics, is that physical equations usually relate quantities that have units as well as numbers. Mathematically, = 5, end of story. However, if you have 3 apples and 2 pears, you neither have 5 apples, nor 5 pears. You cannot sensibly add or equate quantities that have different units (strictly: different dimensions- see later). This gives us a consistency check on all physical equations we may have derived, or recalled from memory: If the equation adds, subtracts, or equates, quantities of different units we must have gone wrong. Exercise: Can we multiply or divide quantities of different units? Exercise: You remember the equation for the centrifugal acceleration is EITHER a cf =ωr 2, OR a cf = ω 2 r. ω is the angular velocity, r the distance from the axis of rotation. Which equation is correct? This consistency check is probably the single most powerful tool you have at your disposal when sitting exams. Always carry units as well as numbers through your calculations, and check if units come out right. If not, there s a mistake somewhere, and often you may get a clue, as well- if you try to calculate a length, but units work out as m 2, or m -1, you probably forgot a root or an inversion somewhere. A generalisation of the concept of units is that of dimension. Note dimension can have different meanings even within physics- the one you encounter here may be different from what you are familiar with. Dimension in the sense used here does NOT mean one of the 3 directions of classical space. Instead, best look at an example: Exercise: Spot the odd- one- out in the following list: meters, cm, nanometers, kilometres, feet, seconds, inches, miles. Apart from the second, all are units of length. We therefore assign the quality or dimension length [L] to any quantity that is measured in a unit of length, whichever unit that may be. You may prefer imperial or metric, micrometers or miles, all of these still have something in common, which the second has not. That something is the dimension L. Similarly, we introduce the dimension time [T] to anything measured in seconds, hours, days, years, or any other unit of time, and the dimension mass [M] to anything measured in kg, micrograms, etc. (careful with imperial units here, they don t clearly distinguish between mass and weight. A good reason to avoid them altogether). From these three dimensions, we can make up the dimensions of less basic quantities such as velocity, or force. Velocity is distance/time, so we assign to it dimension L/T or LT -1. Again, we may prefer km/h, or m/s, as units of velocity, but dimension will still be L/T. Note that differentials make no difference here: If we define velocity as v = Δx/Δt, or v = dx/dt, it still has units m/s (or km/h), and dimensions L/T. Acceleration has dimensions LT -2, Force is given by F = ma, so multiply dimensions of mass and acceleration to get dimensions of force, work (energy) is W = Fs (force x distance)- when you know either a defining equation, or a quantities units, you can work out the dimensions. Exercise: What are the dimensions of acceleration, force, energy, density? More dimensions are required when e.g. electrical charges are involved, but we will limit ourselves to non- electrical phenomena here. With the concept of dimension, an equation that equates 2 physical quantities with different units can still be correct, as long as it equates quantities with the same dimensions: 1 m/s = 3.6 km/h is a

4 sensible equation. Units are different, but dimensions are the same. The equation is dimensionally consistent. Dimensional analysis The technique of dimensional analysis takes the consistency check of equations to the next level. Rather then just checking if an equation can possibly be right, the demand for being consistent with respect to dimension can give us a tool to (sometimes) derive equations from dimensional considerations alone. Let us look at the example for the angular frequency ω of the physical pendulum. From proper reasoning (setting up an equation of motion, solving it under the assumption of small amplitude), we know ω = (g/l) 1/2. Instead, let s try the following: We simply list the quantities we think may affect ω: length of the pendulum (l), mass of the pendulum bob (m), acceleration due to gravity (g). Now, we assume we can make up the correct equation for ω simply by taking the relevant quantities l,m,g to (unknown) powers, and multiply these powers: (eq. 1.1) ω = l a m b g c Now, we determine the unknown powers a,b,c by the demand that the equation has to be dimensionally consistent. We simply replace the quantities ω,l,m,g by their dimensions, and write eq. 1.1 as a dimensional equation: (eq. 1.2) T 1 = L a M b L T 2 c = L (a+c) M b T 2c wherein we have used the dimension of acceleration, L/T 2. Now, from comparing left- and right side of the eq. 1.2, we derive a set of equations for a,b,c. No factor L does appear on the left side, which means it has power zero it then must have power zero on the right side, as well : a + c =0. Same for M, so b = 0. T has power -1 on the left side, so it must have -1 on the right side, too: -2c = -1. We have the following set of equations for a,b,c, which is easy to solve: (eq. 1.3) a + c = 0 b = 0 2c = 1 b = 0;c = 1 2 ;a = 1 2 Substituting a,b,c back into eq. 1.1, we get: (eq. 1.4) ω = l 1/2 g 1/2 = g l Which we know to be the correct equation, at least in the limit of small amplitudes, from the proper derivation. Like pulling a rabbit out of a hat. In particular, we have shown that the pendulum period does NOT depend on the mass of the bob. Galileo. One fly in the ointment, with dimensional analysis we can never derive any factors in equations that have no dimension (or, unit). So, there could be a factor 2, or π, or 5 in front of (g/l) 1/2, which we can t find by dimensional reasoning. The unknown numerical factor happens to be one in the present example, but we won t be that lucky every time.

5 In fact, the problem is a bit more serious than that. There is one more variable, the amplitude of the pendulum, which may well affect ω. However, we have not even mentioned it for the purpose of dimensional analysis. Exercise: Why not? Pendulum amplitude, φ max, is an angle. The units of an angle are radian (rad), but if you recap the definition of rad, you ll find it is a length divided by another length- hence it has no units or dimensions at all. Quantities that have no units are called dimensionless. That does not mean that φ max cannot affect ω, but it means that dimensional analysis cannot tell us if or how. So, we should re- cast eq (1.4): (eq. 1.5) ω = f (φ max ) g l Wherein f(φ max ) is an unknown function, which we cannot determine from dimensional reasoning. All we know is that, f(φ max ) has to be dimensionless, just like its argument, φ max. As it happens here, for small amplitude, f(φ max ) 1, f(φ max ) thus becomes invisible for small amplitudes. We have seen in a nutshell what dimensional analysis does: It splits a problem into a dimensional, and a non- dimensional part. It then solves (or, as we will see, sometimes only partly solves) the dimensional problem. Dimensional analysis is therefore also sometimes called nondimensionalisation. Dimensionless groups Dimensional analysis is particularly useful in situations where a full theory is not yet established. (take note- that is almost a euphemism for research ). A practically very important application of dimensional analysis is in fluid dynamics, which is e.g. concerned with the drag force, F, experienced in the movement of objects in water or air, or the flow of fluids through pipes. Predicting the experienced drag force on an object of known size and shape is exceedingly difficult many say, harder than quantum mechanics. While a general solution remains elusive, dimensional analysis can significantly simplify the amount of experimental work required to study drag. Other than in the example for the period of the pendulum, we will not be able to solve even the dimensional part of the hydrodynamic drag problem completely by dimensional analysis, but we will be able to reduce the complexity of the problem by combining some of the relevant quantities into so- called dimensionless groups. The punchline is that the number of dimensionless groups will be smaller than the number of relevant quantities. Hydrodynamics assumes a totally submerged body (or, completely filled pipe) in an incompressible fluid, which is approximately true for liquids. The relevant quantities that determine drag force are the density of the fluid, ρ, the velocity between fluid and craft (or pipe), v, the viscosity of the fluid, η, and the size of the object, l. For compressible fluids (e.g., air: aerodynamics ), the compressibility κ also comes into it, but for simplicity, we will work the example of the incompressible fluid here. Exercise: List the dimensions of force, density, velocity, viscosity, and size. We assume the drag force to be a function of all the relevant quantities in the form: (eq.1.6) F = Al a ρ b η c v d

6 Wherein A is a dimensionless factor that will depend on the (dimensionless!) shape of the craft or object. Eq. 1.6 translates into the dimensional equation (eq. 1.7) MLT 2 = L a [ML 3 ] b [ML 1 T 1 ] c [LT 1 ] d 1 = b+ c 1 = a 3b c+ d 2 = c d These are 3 equations for 4 unknowns, so there cannot be a complete solution. However, it is possible to express 3 of them in terms of a single, final unknown. Of course, it is somewhat arbitrary which power you leave as the remaining unknown, and different choices lead to different dimensionless groups. Go with intuition: Here, c offers itself. Exercise: Why pick c as the final unknown, not a,b, or d? It is then easy to express a, b, and d all in terms of c: (eq 1.8) b =1 c a = 2 c d = 2 c Now, substitute a,b, and d by their respective expression in terms of c in eq. 1.6: (eq.1.9) F = Al 2 c ρ 1 c η c v 2 c F ρv 2 l 2 = A ρvl c η Where we have moved all known powers of ρ,v,l to the F side of the equation and kept the unknown powers on the other side. If all is well with our technique, (ρvl/η) should be dimensionless, otherwise there would be trouble raising it to unknown power c- we might end up with bizarre dimensions, what if c = 17? If (ρvl/η) is dimensionless, then the left- hand side of eq. 1.9 must also be dimensionless. Exercise: Check directly that both ρvl/η, and the left- hand side of eq. 1.9, are both dimensionless. We have re- written eq. 1.6 in different, simpler terms, with variables lumped into so- called dimensionless groups, sometimes just called numbers: The Newton number Ne = F/ρv 2 l 2, and the Reynolds number Re = ρvl/η. Note we started with force a function of four unknown powers, but finished with number (Ne) as unknown power of only one variable, Re : (eq. 1.9) Ne = A Re -c.

7 Finally, we slightly bend the rules and assume that Ne may be a general (and potentially, complicated) function of Re, not just a power (a point that even the textbooks on dimensional analysis gloss over, so we won t dwell on it). We arrive at 1.10: (eq.1.10) Ne = AΦ(Re) Dimensional analysis cannot tell us anything about A, which is down to the shape of the craft, hence has no dimensions. But A will depend on shape only, not on size, velocity, or else. Also, dimensional analysis has not, and cannot, give us the unknown function, Φ. But it does significantly reduce the effort required if we want to measure Φ with systematic experiments, because we have reduced the number of independent variables. If we have measured the hydrodynamic behaviour of, say, a water pipeline, we then don t have to repeat it for an oil pipeline of the same shape (it would be a cylinder usually). The relationship between Ne and Re will be the same, only you have to calculate Ne and Re with different density, viscosity, velocity, size. A most useful aspect of dimensional analysis is that it allows experimentation with down- scaled models, rather than full- size objects. As long as the shape of the object is kept the same, dimensionless equations such as eq remain valid; just use different length l to calculate Ne and Re. The potential savings in effort and cost are obvious, think e.g. of ship or aircraft design. Dimensional analysis tells you how to translate experimental conditions, and results, between model and full size. Of course, the inverse, square, root, or other power of a dimensionless quantity will again be dimensionless. Therefore, there is some ambiguity in the definition of a dimensionless group like the Reynolds number. Re here happens to emerge in the form Re = (ρvl/η) in our analysis, and we took it as it was - but one could also settle for 1/Re, or Re 2, or, say 2πRe, which would also be dimensionless, and an unknown power of Re is also an unknown power of Re 2, or Re -1 (just, a different one, but still unknown). There is room for manoeuvre here. Some dimensionless groups are convention for historic reasons, but often a particular form of dimensionless group is chosen because it has a clear physical interpretation. An example is the Mach number in aerodynamics, which can be interpreted as the ratio of the speed of an object to the speed of sound. However, Mach number does not immediately emerge in that form from dimensional analysis. Instead, a different dimensionless group emerges, Mach number is a power of that group. Fr and Oh Wikipedia lists a table with about 100 entries under dimensionless quantity. Although here introduced as classical physics, any scientific or engineering equation must be dimensionally consistent, and hence, there will be scope to apply dimensional analysis e.g. in quantum mechanics. Exercise: Show that the Schrödinger equation is dimensionally consistent. 2 important examples: The drag experienced by ships is different again from both hydrodynamics and aerodynamics, as it is largely caused by the surface waves a ship generates as it moves. William Froudé pioneered the testing of ship models in the 19 th century. Because he did not know dimensional analysys yet, Froudé experimented with models of same shape but different size and found a law of comparison (scaling). Later, his law was supported by dimensional analysis, which introduced a dimensionless number that essentially compares the vessel s speed to the speed of waves it generates. This number is now called Froudé number (Fr), although Froudé did not strictly formulate his law of comparison in dimensionless terms. In 1936, W Ohnesorge wrote on the "Formation of drops by nozzles and the breakup of liquid jets", wherein he combined a liquid s viscosity, density, droplet size, and surface tension into a dimensionless group now known as Ohnesorge number, Oh. Oh is highly relevant e.g. for the design of inkjet printers, and Diesel injection nozzles.

8 2 The harmonic oscillator Linear harmonic oscillator Oscillations are common in both classical and quantum mechanical systems. We will here discuss mechanical oscillations, but the developed concepts can readily be generalised, e.g. to electrical oscillators. All oscillators contain elements that can store and release energy, e.g. springs (stores potential energy) and masses (stores kinetic energy) or capacitors (store electrical energy) and coils (store magnetic energy). Such oscillators will go on forever, and that is the first case we will discuss. Realistically, however, oscillators will also dissipate energy, e.g. in a mechanical dashpot or in an electric resistor. We will return to that situation later. The simplest oscillator is a body of mass m attached to an ideal spring, with the mass being able to move only in the direction of the spring s long axis, which we choose to call the x- axis note we are at liberty to do that. Such an oscillator is called a linear harmonic oscillator (LHO), sometimes also simple harmonic oscillator. An ideal spring is a spring that has zero mass of its own, and responds to stretching or compression away from its equilibrium length with a restoring Force, F res = -k(x-x 0 ), that points back towards the equilibrium point, x 0 (Hooke s law). k is known as spring constant, and is a characteristic of the spring- it can be very different for different springs. Exercise: What are the dimensions of k? We are at liberty to place the origin of the coordinate system we are using at any point that we find convenient- say, x 0. (Note we do NOT have to choose the point where the spring is anchored as origin!). We can therefore always, without loss of generality, say x 0 = 0, and F res = -kx. Pause here and question how important or general the assumption of a linear force law, F res =-kx, is in the real world, there are few bodies that are literally connected to springs, but there many oscillators, the spring is a model for all sorts of restoring forces. Is it sensible to assume restoring forces are linear- couldn t it be quadratic, root, exponential,. - are we just conveniently picking something that is mathematically easy to handle, at the expense of losing the generality of our approach? Even some springs are deliberately designed to have force laws other than Hooke s law ( progressive springs in vehicle suspension). As a general oscillator, assume a mass that initially rests in a local minimum x 0 of a general potential energy function, V(x). As above, we can place the origin of our coordinate system so that x 0 = 0, and we are at liberty to gauge our potential energy so that V(x 0 =0) = 0 (Note, a force is the negative derivative of a potential energy- hence, adding or subtracting any constant to a potential does not change forces). What restoring force will the mass experience when it is displaced for a small distance from 0? To answer that, we use the first few terms of a Taylor expansion of the potential energy around its local minimum at x 0 = 0 as an approximation: (eq.2.1) V(x) = V(0)+ x dv + 1 dx x=0 2 x2 d 2 V dx x=0 Therein, V(0) is a constant, which we have just gauged to be zero, and dv/dx (x=0) = 0, because we have assumed we have a minimum at x = 0- at a minimum, derivative is 0. The potential energy therefore, in the approximation of small deflection, x, scales quadratically with x. To a first approximation (that is, for small deflection), that is true for every realistic potential energy. A potential energy that scales quadratically with deflection is called harmonic. The negative derivative with respect to x of the potential is the restoring force, hence for the restoring force of the harmonic potential, we find F res = -kx.

9 Exercise: Relate k in F res = -kx to the potential energy. (Hint: Look at the Taylor expansion of the potential). Why is k always positive? This explains the importance of the so- called harmonic oscillator. For small amplitudes, every oscillator is harmonic, that is why F res = -kx is much more than just a convenient assumption. Now, we apply the Newtonian equation of motion, F = ma, to our model spring, with m the mass of the body and a, the acceleration, equal to d 2 x/dt 2. This leads to the following differential equation: (eq.2.2) m d2 x(t) dt 2 = kx(t) d 2 x(t) dt 2 + k m x(t) = 0 Exercise: What are the dimensions of k/m? Differential equations are notorious throughout physics, and often very hard to solve (no worries, not in this case). A differential equation describes a function in this case, x(t). The set of functions that answer the description are called the solutions of the differential equation. While it is often a hard, and mathematical rather than physical, task to find all solutions of a differential equation, it is easy to test if a candidate is or isn t a solution: Enter into the equation, see if the = sign holds true. That means you have found a particular solution to the diff. eqn. There often are several different (i.e., linearly independent) particular solutions, e.g. eqn. 2.2 has two. The general solution of a differential equation is a function containing several parameters, that encompasses all particular solutions as special cases. In a specific situation, the parameters of the general solution have to be chosen to find a particular solution that is consistent with the initial and/or boundary conditions of a system, e.g. its position and velocity at t = 0. Eqn. 2.2 is an example of a linear and homogeneous differential equation. Linear means that the function x and all its derivatives enter the equation linearly, not with a power or root or log or else (Note, the square in d 2 x/dt 2 stands for second derivative, not first derivative squared!). Homogeneous means the right- hand side is zero, rather than a function of the variable, t. Eq. 2.2 is known as second order differential eqn, because the highest derivative of the function is the second derivative. A diff. eqn. has as many linearly independent particular solutions as its order is, i.e. eqn. 2.2 has two. Linear homogeneous diff. eqn.s are among the most benign. It is a property of linear differential equations that the linear combination of solutions again is a solution. If a set of n linearly independent particular solutions to a linear diff. eqn. of order n is known, the general solution of that linear diff eqn. can be constructed by linear combination of all particular solutions. Exercise: From the defining properties of linear diff. eqns, show that linear combinations of solutions are again solutions! You can easily confirm two particular solutions of eqn. 2.2: x(t) = Asin(ω 0 t), and x(t) = Bcos(ω 0 t), wherein ω 0 = (k/m) 1/2 is known as angular frequency of the harmonic oscillator. Exercise: Show that Asin(ω 0 t) and x(t) = Bcos(ω 0 t) are solutions of eqn What are the SI units of A and B? Why is simply sin(ω 0 t) or cos(ω 0 t) a mathematically, but NOT physically, acceptable solution? Exercise: Use dimensional analysis to derive ω 0 = (k/m) 1/2. Initially, consider the possibility that ω 0 may depend on amplitude as well as k, m- dimensional analysis will prove that it does not. As we have confirmed two particular solutions of a 2 nd order linear diff. eqn., we can construct the general solution of the harmonic oscillator:

10 (eq. 2.3) x(t) = A cos(ω 0 t)+ B sin(ω 0 t) This is mathematically equivalent to (eq.2.4) x(t) = X max cos(ω 0 t +ϕ) Exercise: Show that the above eqn.s 2.3 and 2.4 are equivalent, and give the relation between A, B and X max, φ. Note that eq. 2.4 is more convenient, as it allows to directly read the amplitude X max of oscillation. The harmonic oscillator perpetually undergoes periodic (sinusoidal) motion, with an amplitude X max, depending how much energy it had in the beginning, and a phase, φ. The oscillator repeats itself with frequency f = ω 0 /2π, or period T = 1/f. Oscillators are clocks. During oscillation, energy is converted forward and backwards between potential and kinetic energy, with maximum potential energy and zero kinetic energy when x = X max, and maximum kinetic energy and zero potential energy at x = 0. The following relations hold: v max = ω 0 X max (eq. 2.5) a max = ω 0 2 X max W = V max = T max = 1 kx 2 2 max = 1 mv 2 2 max Wherein v max is maximum velocity, a max maximum acceleration, of the oscillator, and W is the oscillator energy, T is kinetic energy, V is the potential energy, (do not confuse V and v here: Pot. Energy vs. velocity). Note that phase angle, φ, is absent from eq Exercise: Derive all of eq. 2.5 from 2.4. Above exercise confirms that eq. 2.4 is the more convenient form of representing the oscillation, because maximum amplitude X max is directly linked to the oscillators energy. Mathematically, that s it- as physicists, we can do better. As much as we can call any minimum of a potential, x 0, the origin of a coordinate system, hence x 0 = 0, we can call any time, t 0, as the beginning of time, hence t = 0. Obviously, not the beginning of all time, but the beginning of our timekeeping of a particular observation. So we can always make it so that our oscillator starts at maximum amplitude, X max (or at zero amplitude, but non- zero velocity but I prefer the previous convention: Pull your body away from equilibrium to some amplitude, X max, and let go. The moment you let go you call t = 0). By that convention, we always have ϕ = 0. For a single oscillator, ϕ is a rather meaningless concept. It becomes meaningful only when we compare two oscillators, which may or may not be in step with each other. Since it is physically meaningless, ϕ is absent from eq.s 2.5. Damped harmonic oscillator The harmonic oscillator goes on forever, no real oscillator does. Our model misses to take into account damping. Damping is introduced conceptually in the form of a dashpot that displays loss of energy via friction. A real oscillator may not literally contain a dashpot, but e.g. there may be air resistance in mechanical oscillators, or ohmic electrical resistance in an electrical oscillator. The dashpot is assumed to exhibit friction, that is a force, F f, that always points into the opposite direction of the current velocity v = dx/dt - that much is not controversial and in magnitude, is proportional to velocity with a constant c: (eq. 2.6) F f = cv Exercise: What are the dimensions of c?

11 There was a good justification for harmonic forces (F res = -kx). F f = -cv is much less well justified more generally, one should assume F f = -cv n. Different types of friction are known with different powers n. Stokes friction (slow movement in highly viscous medium) indeed shows n = 1, but there are other friction laws, such as Coulomb friction (friction of dry body on dry surface, n = 0), Newton friction (fast movement in low viscosity medium, n = 2), Reynolds friction (between lubricated solid bodies, n = 1/2). For now, we stick with F f = -cv for our further discussion. We have to add the force due to friction into the oscillator s equation of motion, leading to the following extended diff eqn.: (eq. 2.7) m d2 x dx = c 2 dt dt kx d 2 x(t) dt 2 dx(t) 2 + 2γ +ω 0 x(t) = 0 dt Wherein we define the damping factor γ = c/2m, and as before, ω 0 2 =k/m. It looks weird at first to define γ = c/2m, and then have 2γ in the equation, but you ll soon see why. Exercise: Look at eqn why did we insist in the sometimes unphysical assumption n = 1? What type of diff. eqn. is eqn 2.7 as long as n = 1, but not for n 1? The extended differential eqn. is again linear, homogeneous, and 2 nd order, like eq However, the presence of first as well as second derivative in eq. 2.7 means that neither sin nor cos alone can be a solution. Instead of guessing 2 particular solutions, we will employ a standard, systematic approach that is known to solve homogeneous linear diff. eq.ns of all orders. The one- size- fits- all approach to all homogeneous linear diff. eqn.s. is to start with the Ansatz (educated guess) that all particular solutions are exponential, of the form x(t) = Aexp(at), but keeping in mind that there may be several sets (A,a) that solve the eqn- in fact, as many as the order of the diff eqn. is. Note how much easier it is to take the derivative of the exponential rather than of sin/cos: Differentiation means multiply by a. The exponential function itself doesn t change sin/cos do when you take the derivative! Exercise: How can exponentials be related to sin/cos? Exercise: Take 1 st, 2 nd derivative of x(t) = Aexp(at), enter into eqn, 2.7, and cancel what you can, to get an eqn for a. If the guess x(t)=aexp(at) is entered into eqn. 2.7, and you cancel all you can, you find that Aexp(at) is indeed always a solution, as long as a fulfils the following equation: (eq. 2.8) a 2 + 2γa +ω 0 2 = 0 Eq. 2.8 is known as the characteristic equation of the linear diff. eqn Note that the characteristic equation is no longer a differential equation, but a conventional ( algebraic ) equation. The characteristic equation is always of the same order as the diff eqn. was, here 2 nd order = quadratic. Eqn. 2.8 can be solved by the standard method for quadratic equations, yielding two a s:

12 (eq. 2.9) a 1/2 = γ ± γ 2 ω 0 2 Inspection of 2.9 reveals that the a s may well be complex numbers, namely in the case ω 0 > γ which in fact is quite common. In that case, eqn. 2.9 shows that a 1, a 2 will be the conjugate complex of each other. The benefit of having a standardised route to solving all linear homogeneous diff eqn.s far outweighs biting the bullet of complex numbers. Exercise: Brush up on complex numbers, in particular the meaning of conjugated complex, and exp(ix) = cos(x) + isin(x), Solutions of eqn. 2.7 hence may be of different types, depending if the quantity under the root in eqn. 2.9 is larger, equal to, or smaller than zero, that is γ 2 = c 2 /4m 2 (>/=/<) ω 0 2 = k/m. Another way of writing this is c 2 (>/=/<) 4mk. We see, the type of solution depends on the relative magnitudes of squared friction constant, c, that quantifies dissipation of energy, to spring constant k times mass m, that is the two quantities that quantify the amount of energy stored in an oscillator. 1 st case: c 2 /4m 2 = γ 2 > ω 0 2 = k/m (c 2 > 4mk) In this case, both roots a 1/2 of the quadratic eqn. 2.8 are real, and both < 0. This case is known as overdamping. Damping is so strong that the oscillator no longer oscillates, as you will see from entering a 1/2 into the solution Ansatz : (eq. 2.10) x(t) = A 1 exp(a 1 t)+ A 2 exp(a 2 t) With a 1/2 given by 2.9. Since a 1.2 both < 0, the exponentials decay to zero for large times. The only unknowns in the general solution eq are A 1, A 2, which have to be fitted to the system s specific initial conditions. Exercise: Show that the initial conditions x(0), v(0) specify A 1, and A 2 as A 1 =(a 2 x 0 -v 0 )/(a 2 -a 1 ), and A 2 =x 0 -A 1. For v 0 =0, A 1 = a 2 x 0 /(a 2 -a 1 ), A 2 =a 1 x 0 /(a 1 -a 2 ). Note that if v(0)= 0, then x(t) never changes sign: x(0) is the largest deflection in modulus the system will ever have, from then on, it decays- but it never changes sign. (If v(0) 0, x(t) may change sign once, but no more than once!). The overdamped oscillator does not oscillate. The system creeps to zero. This is the case most different from the original, undamped oscillator we had discussed first. Friction, quantified by c, has the upper hand over energy storage, quantified by k and m. 2 nd case: c 2 /4m 2 = γ 2 < ω 0 2 = k/m (c 2 < 4mk) Now, there is a negative number under the root in eq Consequently, a 1 and a 2 are conjugated complex (a 2 = a 1 * ) with Re{a 1 } = Re{a 2 } = -γ. We also introduce the quantity ω d as ω d 2 = ω 0 2 -γ 2, with the index d for damped. This gives the general solution (eq. 2.11) a 1/2 = γ ± iω d and [ ] x(t) = exp( γt) A + exp(iω d t) + A exp( iω d t) This looks confusing: x(t) has to be real- so how do we make sure of that when eq contains imaginary exponents? It is easy to show that eq always returns a real number, as long as A +, A - are both complex numbers themselves, and conjugated to each other: A - =A + *. This also implies that there are only 2 independent parameters in A +, A - (not 4, as it would be in two independent complex numbers). Hence, 2 initial conditions (x(0), v(0)), are again sufficient to determine both A + and A -.

13 Exercise: Show that eq will always return a real number as long as A - =A + *. The real nature of eqn 2.11 is clearly visible when it is written in the alternative, mathematically identical form 2.12: (eq.2.12) x(t) = X max (0)exp( γt)cos(ω d t + ϕ) = X max (0)exp( t /τ )cos(ω d t + ϕ) Wherein X max, φ can be related to A +, A -. Exercise: Derive the relation between X max, φ and A +, A -. Eq describes an oscillation, similar to the undamped oscillator (cf. Eqn. 2.4), not creep to zero. The angular frequency of this oscillation is ω d < ω 0, smaller than the angular frequency of the corresponding undamped oscillator. For very weak damping, γ << ω 0, ω d is very close to ω 0. Again, if we assume oscillation starts at maximum amplitude and zero velocity, then ϕ = 0. The main difference between Eq. 2.4 and 2.12 is that in 2.4, the oscillator amplitude always remains at X max, because in the absence of damping, the oscillator doesn t lose energy. The amplitude of the damped oscillator, described by eqn. 2.12, decays over time with a time constant τ = 1/γ, X max (t) = X max (0) exp(-t/τ), because the oscillator loses energy due to damping. We have an oscillation with somewhat lower angular frequency, folded into an exponential decay envelope. This is quite different from the creep observed in the overdamped case. When just referring to a damped oscillator, we usually mean the scenario described by eq Exercise: Show that for weak damping, ω d ω 0 γ 2 /2ω 0 Exercise: Show that the time constant for the loss of the energy stored in the oscillator is half of the time constant for amplitude decay. There is a 3 rd possibility in the characteristic eqn. 2.9, which marks the borderline between the two cases discussed above: 3 rd case: c 2 /4m 2 = γ 2 =ω 0 2 =k/m (c 2 = 4mk) Now, energy storage (4mk) and dissipation (c 2 ) are balanced into a stalemate : The root in eq. 2.9 is zero. This is known as critical damping. Now, eqn. 2.9 has only one solution, which is real, and negative, and a 1 = a 2 = -γ. In quantum mechanics, this is called degeneracy. At first glance that may seem to simplify matters, but it doesn t: The characteristic eqn. provides only one particular solution, but we need two particular solutions to make up the general solution. It takes a bit of mathematical trickery to come up with another, linearly independent particular solution. We won t go there but it can be shown that the general solution in critical damping is given by: (eq.2.13) x(t) = (At + B)exp( γt) With a 1 =a 2 =γ. Exercise: Show that for γ =ω 0, eq.2.13 is a solution for eq.2.7. Again, there are no oscillations, only creep towards zero. The time it takes to approach 0 is as short as possible without oscillation- time constant τ = 1/γ. Practically, critical or near- critical damping often is desirable, as it is the fastest return to equilibrium without oscillations. An important example of near- critical oscillators are vehicle suspension systems. Wheels are linked to springs to soften the blows from potholes etc. However, with springs alone, your car would soon hop along the road like a bouncy ball. Therefore, in parallel to the springs, a car has shock absorbers, that is dashpots with damping, c. Shock absorbers should

14 be strong enough (c large enough) to make your car overcritically damped- but on the other hand, the car should creep back to equilibrium position quickly. Ideally, therefore, you should be precisely at critical damping. Since the mass of the car may change, engineers tend to err on the save side and somewhat overdamp the suspension, but when you overload the car you may cross the critical boundary: Overloaded cars tend to swing a few times after a pothole blow. Don t overdo it. So far the canonical treatment of the damped oscillator, which you find in many textbooks. But note, it all relies in the assumption that damping force is proportional to v. If it is not and there is a number of friction laws with powers n 1 of velocity the resulting diff. eqn is no longer linear, and the standard procedure introduced above does not apply. In the case of weak damping, approximate solutions to the nonlinearly damped oscillator can be found though- no details here. The prediction is that such an oscillator will still undergo decaying harmonic oscillations, but the decay envelope is not exponential. The following table covers a few examples: Table 1 n Shape of envelope 0 Linear 1/2 Parabolic 1 Exponential 2 Hyperbolic Note that the approximate treatment of the weakly linearly damped (n=1) oscillator predicts an exponential shape for the decay envelope- which we know to be correct from the precise treatment. So it seems the method of approximate treatment is reliable. Whatever the decay law, every practical oscillator is somewhat damped, and will not go on forever unless, we drive it. The driven harmonic oscillator In this chapter we introduce the extremely important concept of resonance. This is important well beyond classical physics. Resonance is particular interesting for weakly damped oscillators. We will, in due course, make the assumption of weak damping, γ << ω 0, to keep equations simple. Let s go back to the diff eqn. 2.9 of the free (i.e., not driven) harmonic oscillator. The absence of any external, driving force results in the 0 on the right- hand- side. To describe an oscillator that is driven (or forced ) by a (time- dependent) external force, F ext (t), we introduce F ext (t) on the righthand- side: Eq d 2 x(t) dt 2 dx(t) γ +ω 0 x(t) = dt m F ext (t) (We need to introduce m on the left- hand- side- or, 1/m on the right hand side- to keep the dimensions correct. The left- hand- side is an acceleration, not a force). In mathematical terminology, introducing a function of time on the right- hand side of the differential eqn. means we go from a homogeneous diff eqn to an inhomogeneous diff. eqn. Exercise: Show that for a constant external force, F ext = F 0 F ext (t), the solution of the driven oscillator eqn is exactly the same as that for the damped free oscillator, apart from the fact that the mass no longer oscillates around the origin (the equilibrium of the spring under no force), but around x eq = F 0 /k, the equilibrium of the spring stretched by F 0. The above exercise shall convince you that an external force that is constant is a boring scenario. Also, forces that forever increase will at some point break the spring, while forces that forever decrease will at some point become the same as zero (or constant) force. That s hardly what we

15 mean by driving the oscillator! So we should consider a periodic time- dependent external force which still leaves us a lot of choice, so we need to think what driving forces may be sensible. The choice of external force we will discuss is: Eq F ext (t) = F 0 exp(iωt) That is, a harmonic driving force (written in exponent- i- form for mathematical convenience). Therein, ω is an arbitrary drive frequency, not to be confused with ω 0 or ω d. ω 0, ω d are determined by the properties of the oscillator how big k, m, c are. ω isn t! If you drive your oscillator by an external motor, you may have a dial to choose ω, completely independently of k, m, c. We choose harmonic periodic forces, as they are the most basic periodic function. The free oscillator undergoes harmonic motion, so it appears natural to drive it by one. In fact, the driver may be another oscillator, or a passing wave. More rigorously, one argues on the basis of Fourier s theorem. This basically says that every periodic function can be decomposed into a superposition of harmonic oscillations of different angular frequency, amplitude, and phase. So, if we find a general solution for the driven oscillator eqn. in response to a harmonic external force, we have solved the general problem for an oscillator driven by any periodic external force: We only need to decompose the driving force into its harmonic components à la Fourier, then calculate the response of the oscillator to every harmonic component, and add up all responses. That may be difficult technically, but conceptually, the solution for the harmonic driving force is a complete solution of the problem. Soon, we ll see that due to the nature of resonance we can in fact ignore most of the Fourier components apart from those near what we ll call resonance frequency. So, we describe the driven oscillator by the inhomogeneous, linear differential equation: (eq. 2.16) d 2 x(t) dt 2 dx(t) 2 F + 2γ +ω 0 x(t) = 0 dt m exp(iωt) The following theorem applies to inhomogeneous diff eqn.s: The general solution of an inhomogeneous diff. Eqn. is a particular solution of the full inhomogeneous eqn., plus the general solution of the corresponding homogeneous diff eqn. Exercise: Show that if you have a solution of the full inhg. diff eqn., and add to that a solution of the corresponding homogeneous diff eqn, the resulting function will still be a solution of the full, inhg. diff. Eqn. Physically, that means that whatever particular solution of the inhomogeneous eqn we find, the resulting motion may still be superimposed by the harmonic oscillations of the free (not driven) oscillator, which will be with angular frequency ω d while, as we will see, the particular solution of the driven oscillator will usually not be with ω d. This is an irritation we could do without. And we will: In all that follows, we assume the absence of these free oscillations. We note that free oscillations always decay due to damping, over a timescale τ = 1/γ free oscillations are transient. We simply assume our driven oscillator has been driven for a long time, t >> 1/γ, so that all free oscillations have died away. Keep in mind, however, that shortly after switching the driver on, your oscillator may behave differently- let it swing in first to reach its steady state! Now, all we need is to find a special solution of the above inhg. diff eqn. Easier said then done, you may say so let s try our old trick again, the trial solution or Ansatz : Guess a solution, and show it s correct. Exercise: try to guess a sensible trial solution.

16 (eq 2.17) x(t) = A(ω)exp(i(ωt ϕ(ω)) So, we assume that the response of the harmonic oscillator to harmonic drive is a harmonic oscillation. Also we assume the resulting oscillator amplitude may depend on ω, in a way that hopefully, our equation will allow us to work out, and that there possibly is a phase ϕ between the drive frequency, and the oscillator response. Remember, for the free oscillator, we can dismiss phase as arbitrary- if only we choose the arbitrary point in time we call 0 in a suitable way, we can always make the phase 0. For the driven oscillator, however, phase is meaningful. Exercise: Discuss why phase is a meaningful concept for driven, but not for free oscillators. All these assumptions, I hope, are quite intuitive, but there is one less obvious assumption in the trial solution eqn. 2.16: We prescribe that the oscillator responds with the same frequency ω as the drive frequency not with it s own frequency ω d, or any other. Remember, for the damped oscillator without driver, we did not prescribe a frequency: ω d results from the maths, it s not forcefed into the Ansatz at the beginning! A reasoning why that has to be so is that if the driver is linked to the oscillator by a rigid drive shaft, driver and driven have to oscillate with the same frequency so that the drive shaft doesn t have to stretch. Off course, the driver could be linked with a rubber band, or it could be a wind or a wave or a magnetic field so it may be a bit iffy in those scenarios. The maths works out with this assumption, so go along with it. In the end, those who feel uneasy about prescribing a frequency to the oscillator that it may not like will have the last laugh. So, let s take the first and second derivative of our Ansatz and enter into the inhg. diff eqn. 2.16: x(t) = A(ω)exp(iωt ϕ) dx dt = iωa(ω)exp(iωt ϕ) = iωx(t) d 2 x dt 2 = ω2 x(t) See how helpful it is to write harmonics as exp(iωt)- differentiation to t is multiplying by iω, it doesn t get simpler than that! Now, let s substitute dx/dt, d 2 x/dt 2 into eq. 2.16: (eq. 2.18) [ ] exp(i(ωt ϕ)) = F 0 [ ] A(ω) = F 0 A(ω) ω 2 + 2iωγ +ω 0 2 ω 2 + 2iωγ +ω 0 2 m exp(iϕ) m exp(iωt) The time- dependent term exp(iωt) cancels, and we are left with an algebraic rather than a differential equation- so the Ansatz was a success. Exercise: In eqn, 2.18, there are 2 unknowns- name them! How can one equation be enough to solve for two unknowns? If we apply exp(ix) = cosx + isinx to the right- hand side of eqn. 2.18, it splits into two equations: Both real and imaginary parts have to be equal simultaneously. Equations in complex numbers are two equations formulated in a single line.

17 (eq. 2.19) 2 A(ω) ω 0 ω 2 [ ] = F 0 m cosϕ 2ωγA(ω) = F 0 m sinϕ Exercise: From eqns, 2.19, find separate expressions for A(ω) and ϕ. This leads to the following result for A(ω) and ϕ: (eq. 2.20) tanϕ = A(ω) = 2γω ω 0 2 ω 2 F 0 m (ω 0 2 ω 2 ) 2 + 4γ 2 ω 2 Exercise: In the limit ω 0, the harmonic driving force becomes a constant, F 0. Show that above eqn for A(ω) reproduces the previous result for constant force, i.e. A(0)= F 0 /k. Eq is the key equation for the driven oscillator, which we now will discuss. The most important question, of course, is, at what ω is A(ω) as large as possible? Exercise: Find the maximum of A(ω) The standard procedure to find a maximum is to equate da(ω)/dω = 0. You ll appreciate that this can be done, albeit it is rather technical Here, only the result: (eq. 2.21) ω max = ω 0 2 2γ 2 = ωd 2 γ 2 = ωr The property of A(ω) to display a maximum when driven with ω r is called resonance. ω r is called resonance frequency, hence the index, r. For weak damping, it is close, but slightly smaller than, ω d, the frequency of the free oscillator. Resonance is the key phenomenon in the physics of the driven oscillator, sometimes a driven oscillator is even called resonator. Exercise: Recap the meaning of the different ω s: ω 0, ω d, ω r, and ω without index. Do not confuse them! Now we know the location of the resonance peak, at ω = ω r, but how high is the amplitude A max at resonance? To calculate A max = A(ω r ), simply enter ω r into the general equation for A(ω). You will find: F 0 (eq. 2.22) A max = A(ω r ) = m 2γω d (Remember the definition of ω d, the frequency of the freely oscillating damped harmonic oscillator).

18 Eq still contains the somewhat arbitrary F 0. To put A max into perspective, we calculate A(ω r )/A(0), that is the amplitude at resonant drive, divided by amplitude under the effect of the same force, F 0, applied statically. Since A(0)=F 0 /k=f 0 /(mω 0 2 ) (eq. 2.23) A(ω r ) A(0) = ω 0 ω 0 2γω d 2γ = 2 km c = Q Eq is only sensible for sub- critically damped oscillators, and the approximation is valid for weakly damped oscillators, γ << ω 0. As of now, we will usually make the assumption of weak damping γ << ω 0. Eq introduces the quantity Q, which is known as the quality of the oscillator. Eq gives Q as the relative magnitude of the two energy- storing elements, mass (m) and spring (k), to the energy- dissipating element, dashpot (c). However, Q is a key general concept that applies to all oscillators, not just mechanical oscillators. Of course, for an electrical oscillator we can not calculate Q from k,m,c- because it doesn t have a mass or spring or dashpot. Instead, Q can be extracted from measured oscillator data we will see how in due course - without breaking it down into its components and measuring those seperatly. Q is so important precisely because it is not specific to a particular type of oscillator. You cannot directly compare e.g. c in a mechanical oscillator to resistance R in an electric oscillator- to begin with, they have different units. But you can compare Q of mechanical and electrical oscillators. Exercise: Show that Q is dimensionless! At resonance, amplitude is Q times higher than when the same oscillator is subject to the same force applied constantly, rather than with frequency ω r. The energy stored in the oscillator is proportional to amplitude squared, hence the oscillator fully swung in at resonance contains Q 2 times more energy than the same oscillator driven with zero frequency. This gives you a glimpse of the potential destructive power of resonance. It also shows one way of extracting Q from measured data: All you need A(0), and A max, then use Q = A max /A(0). With hindsight, the concept of Quality can be applied to the freely decaying, damped oscillator, as well- according to eq. 2.23, all you need to calculate Q are k, m, c, and all of those are present for the freely decaying oscillator. Q is a property of the oscillator, not the driver. Freely decaying oscillations allow another way of measuring Q very simply, without measurement of k,m,c. Exercise: Show that a freely decaying oscillator undergoes Q/π oscillations until its amplitude has decayed to 1/e of its initial amplitude. Deflect the oscillator to an arbitrary initial value, mark 1/e of this value on a measuring scale, and count the number of oscillations, N, until the oscillator has decayed to 1/e of the original amplitude. Multiply that number by π, and you have measured Q: (eq.2.24) Q = πn The above also tells us about how long we have to wait until a driven oscillator has reached its steady state. It takes N oscillations for transient oscillations to decay, which will take time TQ/π, with T the period of the oscillator. In a similar way, you could look at the decay of an electrical oscillator on an oscilloscope screen, and determine Q simply by counting. The electrical oscillator has no m, k, or c (instead, capacitors, inductances, and resistors). Still, Q is well defined and easily measured, even if you do not exactly know how the oscillator works or what parts it contains. Q of a musical instrument is known as its sustain, e.g. a high sustain guitar will keep ringing long after it has been plucked.