M Grell, PHY221: ‘Topics in classical physics”

Contents 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 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

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.

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, 3 + 2 = 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 acf=ωr2, OR acf = ω2r. 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 m2, 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

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)

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ω = l ambgc 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)

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T −1 = LaM b LT 2

c= L(a+c)MbT −2c

wherein we have used the dimension of acceleration, L/T2. 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)

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a+ c = 0b = 0−2c = −1

⇒ b = 0;c = 12;a = −12

Substituting a,b,c back into eq. 1.1, we get:

(eq. 1.4)

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ω = l−1/2g1/2 = gl

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.

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, φmax, 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 φmax 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)

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ω = f (φmax )gl

Wherein f(φmax) is an unknown function, which we cannot determine from dimensional reasoning. All we know is that, f(φmax) has to be dimensionless, just like its argument, φmax. As it happens here, for small amplitude, f(φmax) 1, f(φmax) 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 non- dimensionalisation. 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)

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F = Alaρbηcvd

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)

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MLT−2 = La[ML−3]b[ML−1T−1]c[LT−1]d

⇒

1= b+ c1= 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)

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b =1− ca = 2− cd = 2− c

Now, substitute a,b, and d by their respective expression in terms of c in eq. 1.6:

(eq.1.9)

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F = Al2−cρ1−cηcv2−c

⇒

Fρv2l2

= A ρvlη

−c

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/ρv2l2, 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.

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)

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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. 1.10 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 Re2, or, say 2πRe, which would also be dimensionless, and an unknown power of Re is also an unknown power of Re2, 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 19th 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.

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, Fres = -k(x-x0), that points back towards the equilibrium point, x0 (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, x0. (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 x0 = 0, and Fres = -kx.

Pause here and question how important or general the assumption of a linear force law, Fres =-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 x0 of a general potential energy function, V(x). As above, we can place the origin of our coordinate system so that x0 = 0, and we are at liberty to gauge our potential energy so that V(x0=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 x0 = 0 as an approximation:

(eq.2.1)

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V (x) =V (0)+ xdVdx x=0

+12x2 d

2Vdx2 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 Fres = -kx.

Exercise: Relate k in Fres = -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 Fres = -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 d2x/dt2. This leads to the following differential equation:

(eq.2.2)

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m d2x(t)dt2

= −kx(t) → d2x(t)dt2

+kmx(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 d2x/dt2 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(ω0t), and x(t) = Bcos(ω0t), wherein ω0 = (k/m)1/2 is known as angular frequency of the harmonic oscillator.

Exercise: Show that Asin(ω0t) and x(t) = Bcos(ω0t) are solutions of eqn. 2.2. What are the SI units of A and B? Why is simply sin(ω0t) or cos(ω0t) a mathematically, but NOT physically, acceptable solution?

Exercise: Use dimensional analysis to derive ω0 = (k/m)1/2. Initially, consider the possibility that ω0 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 2nd order linear diff. eqn., we can construct the general solution of the harmonic oscillator:

(eq. 2.3)

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x(t) = A cos(ω0t)+B sin(ω0t)

This is mathematically equivalent to (eq.2.4)

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x(t) = Xmax cos(ω0t+ϕ)

Exercise: Show that the above eqn.s 2.3 and 2.4 are equivalent, and give the relation between A, B and Xmax, φ.

Note that eq. 2.4 is more convenient, as it allows to directly read the amplitude Xmax of oscillation. The harmonic oscillator perpetually undergoes periodic (sinusoidal) motion, with an amplitude Xmax, depending how much energy it had in the beginning, and a phase, φ . The oscillator ‘repeats’ itself with frequency f = ω0/2π , 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 = Xmax, and maximum kinetic energy and zero potential energy at x = 0. The following relations hold:

(eq. 2.5)

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vmax =ω0Xmaxamax =ω0

2XmaxW =Vmax = Tmax = 12 kXmax

2 = 12mvmax2

Wherein vmax is maximum velocity, amax 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. 2.5. 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 Xmax 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, x0, the origin of a coordinate system, hence x0 = 0, we can call any time, t0, 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, Xmax (or at zero amplitude, but non- zero velocity – but I prefer the previous convention: Pull your body away from equilibrium to some amplitude, Xmax, 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, Ff, 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)

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Ff = −cv

Exercise: What are the dimensions of c?

There was a good justification for harmonic forces (Fres = -kx). Ff = -cv is much less well justified – more generally, one should assume Ff = -cvn. 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 Ff = -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)

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m d2xdt2

= −c dxdt− kx →

d2x(t)dt2

+2γ dx(t)dt

+ω02x(t) = 0

Wherein we define the damping factor γ = c/2m, and as before, ω02=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. 2.7 - 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 2nd order, like eq. 2.2. 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 1st, 2nd 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)

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a2 +2γa+ω02 = 0

Eq. 2.8 is known as the characteristic equation of the linear diff. eqn. 2.7. 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 2nd order = quadratic. Eqn. 2.8 can be solved by the standard method for quadratic equations, yielding two a’s:

(eq. 2.9)

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a1/2 = −γ ± γ2 −ω0

2

Inspection of 2.9 reveals that the a’s may well be complex numbers, namely in the case ω0 > γ – which in fact is quite common. In that case, eqn. 2.9 shows that a1, a2 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 γ2 = c2/4m2 (>/=//=/ ω02 = k/m (c2 > 4mk) In this case, both roots a1/2 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 a1/2 into the solution ‘Ansatz’:

(eq. 2.10)

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x(t) = A1 exp(a1t)+ A2 exp(a2t)

With a1/2 given by 2.9. Since a1.2 both < 0, the exponentials decay to zero for large times. The only unknowns in the general solution eq. 2.10 are A1, A2, which have to be fitted to the system’s specific initial conditions.

Exercise: Show that the initial conditions x(0), v(0) specify A1, and A2 as A1=(a2x0-v0)/(a2-a1), and A2=x0-A1. For v0=0, A1= a2x0/(a2-a1), A2=a1x0/(a1-a2). 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. 2nd case: c2/4m2 = γ2 < ω02 = k/m (c2 < 4mk) Now, there is a negative number under the root in eq. 2.9. Consequently, a1 and a2 are conjugated complex (a2 = a1*) with Re{a1} = Re{a2} = -γ. We also introduce the quantity ωd as ωd2 = ω02-γ2, with the index ‘d’ for ‘damped’. This gives the general solution

(eq. 2.11)

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a1/2 = −γ ± iωdandx(t) = exp(−γt) A+ exp(iωd t) + A− exp(−iωd t)[ ]

This looks confusing: x(t) has to be real- so how do we make sure of that when eq. 2.11 contains imaginary exponents? It is easy to show that eq. 2.11 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-.

Exercise: Show that eq. 2.11 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)

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x(t) = Xmax (0)exp(−γt)cos(ωd t +ϕ) = Xmax (0)exp(−t /τ )cos(ωd t +ϕ) Wherein Xmax, φ can be related to A+, A-. Exercise: Derive the relation between Xmax, φ and A+, A-. Eq. 2.12 describes an oscillation, similar to the undamped oscillator (cf. Eqn. 2.4), not ‘creep’ to zero. The angular frequency of this oscillation is ωd < ω0, smaller than the angular frequency of the corresponding undamped oscillator. For very weak damping, γ

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

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. 2.15

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Fext (t) = F0 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 ω0 or ωd. ω0, ωd 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)

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d2x(t)dt2

+2γ dx(t)dt

+ω02x(t) = F0

mexp(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 ωd – while, as we will see, the particular solution of the driven oscillator will usually not be with ωd. 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.

(eq 2.17)

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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 ωd, or any other. Remember, for the damped oscillator without driver, we did not prescribe a frequency: ωd results from the maths, it’s not force- fed 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:

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x(t) = A(ω)exp(iωt −ϕ)dxdt

= iωA(ω)exp(iωt −ϕ) = iωx(t)

d2xdt2

= −ω2x(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, d2x/dt2 into eq. 2.16:

(eq. 2.18)

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A(ω) −ω2 +2iωγ +ω02[ ]exp(i(ωt −ϕ)) = F0m exp(iωt)

⇒ −ω2 +2iωγ +ω02[ ]A(ω) = F0m exp(iϕ)

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.

(eq. 2.19)

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A(ω) ω02 −ω2[ ] = F0m cosϕ

2ωγA(ω) = F0msinϕ

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)

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tanϕ = 2γωω02 −ω2

A(ω) =F0m

(ω02 −ω2)2 + 4γ2ω2

Exercise: In the limit ω 0, the ‘harmonic’ driving force becomes a constant, F0. Show that above eqn for A(ω) reproduces the previous result for constant force, i.e. A(0)= F0/k. Eq. 2.20 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)

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ωmax = ω02 −2γ2 = ωd

2 −γ2 =ωr The property of A(ω) to display a maximum when driven with ωr is called resonance. ωr is called resonance frequency, hence the index, r. For weak damping, it is close, but slightly smaller than, ωd, 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: ω0, ωd, ωr, and ω without index. Do not confuse them! Now we know the location of the ‘resonance peak’, at ω = ωr, but how high is the amplitude Amax at resonance? To calculate Amax = A(ωr), simply enter ωr into the general equation for A(ω). You will find:

(eq. 2.22)

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Amax = A(ωr) =F0m

2γωd

(Remember the definition of ωd, the frequency of the freely oscillating damped harmonic oscillator).

Eq. 2.22 still contains the somewhat arbitrary F0. To put Amax into perspective, we calculate A(ωr)/A(0), that is the amplitude at resonant drive, divided by amplitude under the effect of the same force, F0, applied statically. Since A(0)=F0/k=F0/(mω02)

(eq. 2.23)

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A(ωr )A(0)

=ω02

2γωd≈ω02γ

=kmc

=Q

Eq. 2.23 is only sensible for sub- critically damped oscillators, and the approximation is valid for weakly damped oscillators, γ

‘Sharpness’ of resonance We have found maximum of A(ω) at ω = ωr, but how ‘sharp’ is the resonance – i.e., what size of frequency ‘mismatch’ between drive frequency, and resonance, ω – ωr, can we tolerate to still get a considerable response from the oscillator? To answer this quantitatively, we find the ‘Full Width Half Maximum’ (FWHM) of the resonance function, A(ω). We define the half- width frequencie(s), ωFWHM, via A(ωFWHM) = 1/2 A(ωr). So, by definition, at ω1/2 (there will be two such frequencies), the amplitude response of the driven oscillator is half as high as the resonance amplitude. Groan, two more ω’s with yet another meaning. Exercise: In the limit of weak damping (ωr ≈ ω0), find ωFWHM ‘s from the defining equation. There are two solutions:

(eq.2.25)

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ωFWHM =ω0 ± 3γ Hence, the FWHM of the resonance peak is given by

(eq.2.26)

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ΔωFWHM = 2 3γor

ΔωFWHMωr

=3Q

(Always remember, for weak damping, ω0 ≈ ωd ≈ ωr): With Q the quality of the oscillator. Note that resonance curves sometimes are plotted as intensity or energy (rather than amplitude) against frequency, intensity being proportional to the square of amplitude. The FWHM frequencies of a plot of intensity or energy against frequency relate to Q slightly differently, namely,

(eq. 2.27)

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ΔωFWHM ,Intensityωr

=1Q

Eq. 2.26 or 2.27 give a third way of extracting Q directly from measured data: Finding the FWHM from a graph of A(ω) is easy, without having to measure m, c, k separately, and again, we are not restricted to mechanical oscillators. Q is a universal and dimensionless measure that allows us to compare the relative merit of different types of oscillators without (mentally or really) breaking them down into springs, masses, dashpots, or whatever. In fact, people commonly talk about Q in the context of atomic emission spectra. Plot the light emission intensity against frequency, find maximum and FWHM, calculate Q. (Note: Light intensity is proportional to the square of the amplitude, hence eq. 2.27 should be used). This is light coming from an atom: No springs, masses and dashpots, but Q remains a useful concept. Exercise: Read up what a Q- switch is in the context of laser physics. If you had an intuitive tummy ache when we mathematically forced the driven oscillator to respond with the drive frequency ω, rather than e.g with its ‘own’ frequency, ωd, you can relax now. Postulating response with ω is correct, otherwise the maths wouldn’t have worked out- after all, we have successfully solved the inhomogeneous differential eqn. But the oscillator gets its own back: Many oscillators have Q larger than 1000, which means they won’t respond with substantial amplitude unless driven with a frequency that is (roughly) within a fraction of 1/1000th or less of ωd.

If you try to drive the oscillator with a frequency that is substantially different from ωd, it will respond with very small amplitude. Mathematically, you force the oscillator to respond with ‘your’ frequency ω, not with the oscillator’s own ωd or ωr. Practically, however, the oscillator forces you to drive it at or near ωr, otherwise it just won’t oscillate much. Eq. 2.27 tells us that we have to match resonance frequency the better the higher Q gets. A table of a few typical Qs:

Table 2: Quality for typical oscillators

Oscillator Q Piano string 3000

Microwave cavity 104 Electron shell of atom 107 Nuclear γ- transition 1012…1013

The extremely high Q (extremely narrow spectral lines) of nuclear γ- transitions are the basis of physics’ most sensitive spectroscopy, Mössbauer spectroscopy (Nobel price 1961). Because Q is so high, Mössbauer spectroscopy can measure the tiny frequency shift from the energy loss of a γ- photon flying ‘uphill’ in Earth’s gravity, thus confirming one of the predictions of general relativity. It can also measure the very small energy shifts that occur in the energy levels inside γ- active atomic nuclei resulting from different chemical bonds the respective atom may be engaged in. Prominent example: Oxidation states of iron (Fe). Phase angle We briefly discuss the behaviour of the phase angle as a function of drive frequency. Since tanϕ =2γω/(ω02-ω2), we see that in the limit ω > ω0, tanϕ approaches zero from negative values, hence ϕ π: At high frequencies, oscillator response is 180 deg. out of phase with drive (like sine / -sine) At ω = ω0 ≈ ωr, tanϕ infinity, hence, ϕ = 90 deg. At resonance, drive and response are out of phase by 90 deg., e.g. like sine / cosine. Exercise: Show that the power, P, the driver delivers to the oscillator has a maximum at resonance. Use P = Fv (power = force x velocity) to do so. Reasoning will do, no calculation required to answer this! Feedback In a feedback loop, an oscillator is driven by a driver, as before, only now, the drive frequency is not set externally, but the driver’s frequency itself is set by the oscillator, so it is synchronised, and the system stabilises at the oscillator’s ωr and a constant amplitude, so the oscillation does not decay until the driver runs out of power. Frequency stabilisation is not perfect, relative tolerance is proportional to the width of the resonance curve, i.e. 1/Q. Typical application is in clocks. A very accessible example of feedback is the so- called Accutron, an electromechanical clock popular in the 1960s and 1970s. A tuning fork that resonates at 360 Hz drives a mechanical clockwork that turns the hands of the clock. The fork is made of magnetic material and is itself driven by the AC magnetic field of two induction coils, which periodically receive a current pulse from a drive transistor. Feedback is facilitated by a third coil that acts as pickup. The vibrating fork induces an AC current in the pickup coil, which is driven into the transistor’s base, that is its ‘control’ input. In this way, the oscillator synchronises its driver. The Accutron was the first

electronic clock, one of the first commercial products using a transistor, and went into space on early satellites, e.g Telstar. Modern electronic clocks use an oscillating quartz crystal instead of a tuning fork. Via a phenomenon known as piezoelectricity, the quartz couples mechanical oscillations to an electric oscillator circuit, and stabilises the electric oscillator at the resonant frequency of the mechanical oscillations of the quartz. In terms of electronics, the drive and pickup of a quartz oscillator is ‘capacitative’ (via electric fields), while for the Accutron, it is ‘inductive’ (via magnetic fields). Systems similar to the Accutron, albeit with capacitative pickup, are making a comeback in microelectromechanical systems (MEMS), which essentially are micrometer- sized tuning forks. Exercise: Why do we need to couple the electric circuit to a mechanical oscillator to make an accurate clock- why not just build a purely electric oscillator? Feedback is an extremely important phenomenon both in electrical engineering, and nature, and can get extremely unpredictable or ‘chaotic’ when feedback is time- delayed (e.g. population dynamics in predator/prey ecosystems). It is a challenge to traditional scientific thinking that works ‘linearly’, from input = cause to output = effect. In feedback, output is returned to input, and you have to start thinking in self- consistent loops instead. 2.4 Coupled oscillators Often, oscillators interact with each other, i.e. they are coupled. Oscillations are then communicated between them. A typical example are atoms in a crystal. Atoms have clearly defined equilibrium positions, but due to thermal motion, will not sit still at these positions, but oscillate around them. A first approximation to thermal oscillations might be that atoms are bound to their equilibrium positions by harmonic forces, and indeed an early theory of heat capacity in crystals assumed just that (Einstein theory of heat capacity). However, if you think about it, atoms will rather be bound by harmonic forces to their neighbours, not their equilibrium positions – the ’spring’ will be a chemical bond, and the bond is between atom and atom, not between atom and lattice site. Hence, their oscillations will be coupled. That led to a treatment of heat capacity based on coupled oscillations by Debye, which gives much better agreement with measured data. As a very simple example of coupled oscillators, let us discuss two harmonic oscillators with the same spring constant k, and mass m. Now, we introduce a third spring, k’, that is parallel to the first two, and links the two masses. k’ couples the two previously independent oscillators. We call x1, x2 the displacement of mass 1/2 from its equilibrium position (note we use different origins for x1/2 but call the same direction ‘positive’). We assume one- dimensional motion along the springs only, and neglect damping. The motion of the coupled oscillators is described by a pair of coupled differential equations:

(eq. 2.28)

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m x1..

+ kx1 − k' (x2 − x1) = 0

m x2..

+ kx2 + k' (x2 − x1) = 0

Both equations begin like the eq. of the independent oscillator, but add another force that is due to the coupling spring. k’ ‘mixes’ the two equations. x1/2 are no longer independent of each other, because they appear in each other’s equations. Note the equations, while coupled, are still linear, and damping has been neglected. Let’s try to solve them again with an ‘Ansatz’, x1/2 = A1/2 exp(iωt). That means, we assume oscillations will be harmonic. Amplitudes may be different, and there may be a phase shift between the two oscillations – mathematically, this would be indicated by complex A1/2. What frequency ω the coupled oscillation will have, we have to work out. So, enter the ‘Ansatz’ into eq. 2.28, recalling d2x1/2/dt2 = -ω2x1/2.

(eq.2.29)

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−mω2A1+ kA1+ k'A1 − k

'A2 = 0

−mω2A2 + kA2 + k'A2 − k

'A1 = 0or

−mω2 + k + k ' −k '

−k ' −mω2 + k + k '

A1A2

=

00

Again, all the ‘oscillatory’ exp(iωt) parts can be cancelled from the equation, and we are left with an algebraic rather than a differential equation. Only, as we had coupled differential equations, we end up with a ‘coupled’ algebraic equation- that is, two equations with three unknowns, A1, A2, ω. The alternative version of eq. 2.29 re- writes the coupled equations in matrix form. Exercise: Recall the rules of multiplying a matrix and a vector. Show that the two ways of writing eq. 2.29 are equivalent. We use the matrix form, because there is a well- rehearsed standard procedure how to solve it, which basically is a souped- up version of the rule that a product is zero when one of its factors is zero. So, we want the product of a matrix and a vector to equal the zero- vector. One way of making the product zero is to make the A- vector zero (i.e., A1 = A2 = 0). That means, the oscillator has zero amplitude, i.e. it is at rest. That is the so- called trivial solution- not quite what we had in mind. The other possibility is that the matrix somehow is ‘zero’. Only, we can’t make all elements of the matrix 0- the off- diagonal elements, for example, are –k’≠ 0. Instead, we need the determinant of the matrix to equal 0. In a 2x2 matrix, the determinant is the product of the diagonal elements minus the product of the off- diagonal elements:

(eq. 2.30)

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(−mω2 + k + k ' )2 − k '2 = 0 eq. 2.30 is a characteristic equation again, namely a quadratic equation for the only unknown, ω. (strictly, it’s a quadratic eqn. for ω2, but as ω has to be positive, just solve for ω2 and only consider the positive roots). Exercise: Solve eq. 2.30. Eq. 2.30 has two solutions for ω, namely:

(eq.2.31)

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ωS =km

ωA =k +2k '

m

The first solution, ωS, is the same as for the uncoupled oscillator. We can substitute this solution back into eqn. 2.29 to mathematically work out a relationship between A1 and A2, but I hope as physicists, we can do better, simply by interpreting the result. ωS does not contain k’. The only way how the coupling spring k’ can possibly not matter is that it never gets stretched! Both masses oscillate in phase with the same amplitude, hence A1=A2. Exercise: Show that eq. 2.29 is satisfied when ω = ωs, and A1=A2.

We will call this solution the symmetric solution, hence index S. The other solution, ωA, contains k’ (in fact, weighs it double compared to k), so the coupling spring will get stretched. We can substitute ωA into eq. 2.29 to find the relation between A1 and A2 mathematically. Exercise: Find the relation between A1 and A2 for ω = ωA. But we can again use our intuition… coupling spring k’ is weighed double, compared to k, in ωA. The obvious reason is that it is stretched or compressed twice as far as k. That means, the two masses will oscillate 180deg out- of- phase’ in other words, A1 = -A2. We call this the antisymmetric solution, hence index A. The frequency of the antisymmetric solution is higher, because now, the coupling spring contributes to the restoring force. E.g., if we assume k’ = k, then ωA = √3ωS. Together, the two solutions we have found for the coupled oscillator are known as normal modes. Off course, we recall that superpositions of solutions of linear diff. eqn’s again are solutions. Hence, the coupled oscillator can undergo a motion that is neither purely symmetric, nor purely antisymmetric. However, every motion of the oscillator can be broken down into its (in this case, two) normal modes. The normal modes are a bit like the ‘unit vectors’ of a vector space- not every vector is a unit vector, but every vector can be expressed as a linear combination of the unit vectors. Just like different unit vectors are ‘linearly independent’, the normal modes never mix. If a motion of the coupled oscillator can be broken down at any point in time into, say, 70% symmetric and 30% antisymmetric, then it will always remain like that. In other words, no energy is exchanged between normal modes. Energy may be exchanged between the masses though. Applications of the ‘coupled oscillator’ concept include the vibrations of molecules, and 3- dimensional crystal, consisting of many coupled oscillators (but with a small number of different masses, and springs). Of course, strictly this requires quantum mechanical treatment. However, the conclusions about breaking down oscillations into normal modes carries through to the quantum mechanical treatment, only that resulting normal modes are quantised, i.e. they can only carry integer multiples of a basic unit of energy. Quantised normal modes in a crystal are known as Phonons. In thermodynamics, normal modes are also known as degrees of freedom. Like in our very simple example, normal modes in molecules or crystals will either be symmetric, or antisymmetric. The symmetry of the mode has important consequences: A symmetric mode does never have an oscillating dipole moment associated with it, while an antisymmetric mode may have. Therefore, antisymmetric modes can couple to electromagnetic radiation (i.e, they can absorb or emit radiation – typically in the infrared (IR)) – symmetric modes can’t. Similarly, the symmetric / antisymmetric phonon modes in crystals are called ‘acoustic’ and ‘optical’ modes, respectively. Infrared spectroscopy is an important tool in analytical chemistry, as the precise location of resonances in the IR spectrum reveals the presence, and concentration, of certain chemical groups- but IR spectroscopy is ‘blind’ to symmetric modes. Highly symmetric, small molecules do not have an antisymmetric mode, hence don’t absorb or emit IR. Examples: N2, O2, the most common molecules in our atmosphere. However, CO2 does have an antisymmetric mode, and does absorb in the IR: Global warming! The ‘equipartition theorem’ of thermodynamics says that in thermodynamic equilibrium, all degrees of freedom on average have the same energy. This can only be true if there is a way of distributing energy between the degrees of freedom = normal modes. But we concluded earlier that normal modes don’t mix… how does that go together? Normal modes strictly don’t mix as long as the forces are precisely harmonic, F = -kx. Practically, that is never true. In particular at high x, every real ‘spring’ will have anharmonic contributions to the restoring force, i.e. contributions that are not exactly proportional to displacement. Anharmonic contributions to the force law do allow the ‘mixing’ of normal modes, that is the transfer of energy between them. Still, coupling will be

weak, and the exchange of energy will be ‘slow’- which means, it takes much longer than the period of the oscillation. Exercise: If binding forces in crystals where precisely (not just approximately) harmonic, crystals would have zero thermal expansion. Explain that! Parametric oscillator Another important aspect is the parametric oscillator. A mechanical example is a playground swing. Did you ever wonder how you can get yourself to swing high on a playground swing when you sit on it, feet off the ground? The trick is that you periodically change the ‘parameters’ of the oscillator: By moving your centre of gravity, you effectively change the length of the swing’s ropes- hence, ω0 and ωr! In this way, you can amplify the amplitude of pre- existing oscillations (and a small amplitude oscillation always pre- exists, e.g. from when you sat down, or the wind blows you a little). The key ‘trick’ is that you have to change the ‘parameter’ with twice the frequency of the average ω0. Think how you swing your legs forward and back during every single swing cycle. It takes a while to learn that, that’s why young children want pushing. The optical parametric oscillator is a key tool in modern laser physics, enabling frequency mixing and photon entanglement. 3 Waves Definitions and Terms Assume a medium of infinite size that is at rest and in equilibrium. For simplicities’ sake, we will assume a one- dimensional medium, and treat waves in 3 dimensions ‘anectodatally’ rather than precisely. The medium can be ‘deflected’ or ‘distorted’ somehow away from its equilibrium state. Initial deflection is described by a function y = f(x,t=0). (Note: Often it is useful to write f(x,t=0) = Ag(x,t=0), wherein A is a constant physical quantity (or a vector) with units appropriate of the respective deflection, and g a dimensionless mathematical function). There will be an energy associated with this distortion, typically scaling with A2. In many cases, the initial deflection will not be stationary, nor will it simply return to equilibrium. Instead, it will travel along the medium. Such a travelling distortion is a wave. Waves are of course closely related to oscillations, the difference is that oscillations don’t travel. There is, however, a hybrid between the two, the so- called ‘standing wave’, which we will discuss briefly later. If the deflection is parallel to the direction the wave propagates, we speak of a longitudinal wave. If the deflection is orthogonal to the direction the wave travels, the wave is called transversal. Note there are two directions orthogonal to any given direction of propagation, therefore, there is two possible, mutually perpendicular directions for the distortion. Which of these two possible directions the distortion has is known as the polarisation of the wave. (Careful- polarisation is a term that even within physics has more than one meaning- not to be confused with polarisation of a dielectric). Let’s illustrate a travelling distortion. At first, we assume it does not change in shape as it travels, only in location. The deflection that was peaked at x=0 for t=0 peaks at Δx the little time, Δt, later, while its shape has not changed. From this simple observation we can reach an important conclusion: The function f(x,t) that describes the deflection must be of either of the two forms:

(eq.3.1)

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y = f (x − ct)ory = f (x+ ct)

Wherein the ‘-‘ describes a wave travelling from left to right, and the ‘+’ describes a wave that travels from right to left. In either case, f is a function of x and t only in their combination x-ct or x+ct, not of x and t separately or in any other combination. Exercise: Only one of the following three describes a wave: f1=A/(x+ct2), f2=B/(x+ct)2, f3= Cexp(-at)sin(x+ct). Which? Mathematically, f is a function of only one variable (x+/-ct), not two (x and t). Physically, this ‘lumping’ of two variables into one precisely captures the most important aspect of a wave, namely, that it travels in space and time. To observe a wave, you can take a snapshot at one fixed time of the wave in all space. Since x ranges over all space (-∞ to ∞), you get the full range of x +/- ct, and hence the ‘full picture’ of the wave. Or, you can observe at a fixed point in space, and let the wave wash over you for a long time (strictly speaking, forever). You then again get the full range of the combined variable x +/- ct, hence the full picture of the wave. Waves travel, so you don’t have to. The description of a wave as f(x+/-ct) leaves a loose end to tie up: The combined variable, (x+/-ct), has a unit. Exercise: What are the dimensions of x + ct ? It is not generally convenient to feed a variable with units into a mathematical function. We know sin(π), but what is sin(π meters)? exp(seconds)? In other words, mathematical functions like sine or log or exponential assume a dimensionless argument. We therefore introduce a constant, k, with dimension L-1, to multiply with x+/-ct before feeding it into the wave function. Surely, a function of k(x+/-ct) is a function of x+/-ct, since k is a constant. Further, we introduce the definition ck = ω . So, we name the product of our constant k with c, and call it ω. Note, ω = ck cannot be ‘derived’ somehow. It is a definition of ω, hence always true, and shall be memorised! Exercise: What are the dimensions of ω? It is of course no coincidence that ω gets the same symbol as the angular frequency of an oscillator. We will later see that c usually is a function of k, and therefore, ω will be as well: ω(k)=c(k)k. But still, ω = ck. Now, we can write our wave function in the form f(kx-ωt), or f(kx+ωt), with the dimensionless group kx+/-ωt. The group kx+/-ωt that combines time and space into a single variable is so important to the description of waves that we give it a name: kx+/-ωt is known as the phase (ϕ) of the wave. Points in space that have equal phase are known as wave fronts. The wave function has the same value along a wave front, because by definition, it has the same argument. (Note, wave fronts are a sensible concept in 2 or 3 spatial dimensions, not in 1). To interpret the meaning of k, ω, and c, we now assume one specific form for the wave function, namely, (eq. 3.2)

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f (kx −ωt) = A sin(kx −ωt) Note it is a big misconception to believe that waves necessarily must be sinusoidal. Our initial sketch was a finite- size ‘hump’ with a single peak, not even periodic! Nevertheless, it is instructive to discuss waves in terms of sine functions. The deeper reason why waves are commonly treated as sinusoidal waves is Fourier’s theorem. According to Fourier, every function can be decomposed into a superposition of sine (and cosine) waves of different ω (this is known as Fourier analysis). So, at least in principle, the problem of wave propagation can be considered solved if we can describe the propagation of sine (and cosine) waves: Take whatever wave function we have (e.g., a ‘hump’), decompose it into its sine/cosine components, let these components propagate, and in the end, add them up again (this is known as Fourier synthesis). This may be awkward and indeed,

issues arise when waves of different ω propagate with different c – this is known as dispersion, and we will discuss it in some detail later. However, at least conceptually, the complete problem is solved if it is solved for sinusoidal waves, that is why ‘sine’ and ‘wave’ go together so well. We know that mathematically, the sine is periodic with 2π. On the other hand, physically, sine waves are periodic in space with a wavelength, λ, and periodic in time with a period, T. Let’s equate the mathematical and physical periodicity. First, periodicity in space:

(eq. 3.3)

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k(x+λ)−ωt = kx −ωt+2π

That is, if we move on in space by wavelength λ, the phase has to move on by 2π. We can cancel most of the terms in eq. 3.3 to arrive at the important equation:

(eq. 3.4)

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k = 2πλ

Which gives meaning to the constant k we had previously introduced only to fix the unit problem. k is confusingly called wave number although it has a unit. For three- dimensional waves, k becomes a vector, the so- called wave vector. The modulus of wave vector k will again equal 2π/λ, but k also has a direction, namely, the direction the wave propagates into. In the phase of the wave, the product kx is replaced by the scalar product kx (or, kr) between wave vector and location vector. A completely analogous reasoning can be applied to periodicity with time, T, resulting in eqn. 3.5. Exercise: Apply the above reasoning to periodicity with time, T, to derive eq. 3.5.

(eq. 3.5)

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ω =2πT

ω is known as angular frequency, just as it was for oscillations. That is why it even gets the same letter, ω. From the previous definition, ω = ck, we now arrive at

(eq. 3.6)

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c = ωk

=λT

= λf c is known as the phase velocity of the wave, and f = 1/T is the frequency. (Not to be confused with the wave function f = f(kx+/-ωt)). You see that ω = ck is just another way of saying c = λf. Note that a phase is a mathematical, not a physical object. Phase is dimensionless, it has neither mass, momentum, nor energy. The phase velocity is therefore not the speed of any physical object. On the other hand, waves very much are physical objects, in particular, they do transport momentum and energy. Phase velocity can therefore not be the last word on how fast a wave travels. Phase velocity does describe how fast a sine wave travels- another good reason for using sine waves as examples when exploring the properties of waves. But for other waves, we need to think harder- prepare for a re- visit later. The wave equation We have already reached a rather advanced description of waves. To give this description a sounder footing in theory, we are again looking for a differential equation. Remember, differential equations are descriptions of functions, the solution is a function that answers to the description. We already have established what functions are solutions: All those that have as their argument time and space only in the combination of the phase ϕ , kx-ωt or kx+ωt. So, here we are looking for a differential equation that describes a specific argument, rather than a specific function.

The following partial differential equation fits the bill:

(eq.3.7)

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∂2 f∂t2

= c2 ∂2 f∂x2

Eq. 3.7 holds true for any function, f, as long as it is a function of the phase, ϕ. To show that, we need to apply the chain rule to f = f(ϕ), and ϕ = kx +/- ωt:

(eq.3.8)

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∂f∂t

=∂f∂ϕ

∂ϕ∂t

= ±ω∂f∂ϕ

and∂f∂x

=∂f∂ϕ

∂ϕ∂x

= k ∂f∂ϕ

hence : ∂2 f∂t 2

=ω 2∂ 2 f∂ϕ 2

and ∂2 f∂x 2

= k 2 ∂2 f

∂ϕ 2

Substitute the bottom line of eq.3.8 into 3.7, and you will see d2f/dφ2 will cancel out, whatever f is. Use c=ω/k to find eq. 3.7 is fulfilled. Eq. 3.7 prescribes the argument of the functions to be the phase, it does not prescribe any particular function. Exactly what we ordered! Eq. 3.7 is known as the wave equation, one of the key equations of physics. In three dimensions, the wave equation is extended to

(eq 3.9)

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∂2 f∂t2

= c2 ∂2 f∂x2

+∂2 f∂y2

+∂2 f∂z2

= c

2(∇)2 f = c2Δf

with the Laplace operator, Δ, which is defined as the square of the del operator. Note that even the 3- dimensional wave equation usually is an oversimplification- e.g. like oscillators, real waves are usually damped, hence travel only a finite distance. To describe that, we would again have to add first derivative terms to 3.7 or 3.9, but we won’t do that here. Also, not all waves are described by 3.7: For example, the namesake of the entire discipline- waves on the surface of water- are described by the much more difficult Korteweg- de Vries (KdV) equation. Sine waves are not solutions of the KdV eqn- ouch!- and can at best serve as an approximation for very low amplitudes. Still, the terms and concepts we have and will develop in this chapter (ω, k, phase velocity, group velocity, dispersion) can be applied to water (and other) waves. You may question if the exercise of finding a differential equation is worthwhile when we already know the solution. The solution is dead simple; any function that is a function of the phase is a wave- why do I need the differential equation? Why find a question when we already know the answer? The reason why we want eq. 3.7 is that it provides a bridge between the general and the specific. Given any specific medium- will it sustain waves? The answer is, whenever the possible ‘deflections’ of the medium (fields, pressures, strains,…) can be shown to satisfy an eqn. of the general form 3.7, that is there is a relation between the second derivative to space, and the second

derivative to time, then this will be a wave- sustaining medium. Off course, for any specific medium, we will not end up with an equation that says c2 between the two second derivatives, but some constant that depends on the properties of the specific medium. That means, we have found c for that medium! Exercise: According to Maxwell, in vacuum, magnetic induction B and electric field E are related by the following two equations (simplified to 1 dimension):

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∂E∂x

= −∂B∂t

∂B∂x

= −µ0ε0∂E∂t

Manipulate these two equations so that you find a relation between the second derivative to time and the second derivative to space for E. Show that it is a wave equation, and give the phase velocity of the corresponding waves. We will discuss examples of media that can sustain waves. But first, another lose end to tie up… Waves we have discussed so far are waves in infinite media. That means, eq. 3.7 has initial conditions (f(x, t=0)), but no boundary conditions- an infinite medium has no boundaries. Quite a different situation arises when we discuss a finite medium with boundaries, e.g. solid walls. This adds a boundary condition, f(x=boundary, t) = 0 to eq 3.7. A differential eqn. is only complete when initial and boundary conditions are specified, and indeed, the addition of a boundary condition completely changes the character of the solutions of eqn. 3.7. Eq. 3.7 plus boundary conditions leads to so- called standing waves. The term ‘standing wave’ is an oxymoron, the defining property of a wave is that it travels, but the term is commonly used. Exercise: A philosophical dispute: Your friend says you can see wild animals in the zoo, in fact, they have a lion. You say, no you can’t. Sure they have a lion, but… Base your reasoning on a definition of the term ‘wild’, and how the zoo’s boundary conditions affect the nature of the lion. To cope with the presence of boundaries in mathematical terms, you are often encouraged to solve the wave equation by separation of variables. That is, assume your wave function can be written as f(x,t)=T(t)X(x), as the product of one function T(t) of time only, and another function X(x) of location only. Lo and behold, separation of variables leads to solutions, but you see that this approach is a head- on collision with our previous demand that waves are described by functions that combine location and time into a single variable, the phase. Funny that separation of variables works at all, when eqn. 3.7 was specifically designed to give only solutions that are functions of the combined variable, phase. Physically, the apparent contradiction can be reconciled by describing the standing wave as the sum of two travelling waves (remember the superposition principle- it applies here, as well). The two travelling waves have the same amplitude, only, one travels from left to right (call this the incoming wave), and another from right to left (you can understand that as the reflection of the incoming wave at the boundary). That is, we superimpose one solution with phase in the form kx-ωt with another with phase kx+ωt. You can separate variables to solve eqn. 3.7, and it works- but you don’t have to: Exercise: Use the mathematical identity sin(kx-ωt) + sin(kx+ωt) = 2sin(kx)cos(ωt) to come to terms with the standing wave problem. How can apparently separated variables nevertheless be interpreted as phases? Let us now consider some examples of wave- carrying media.

Waves on strings A piece of string is held under tension, σ =F0/A (F0 force, A crosssection of string). The string’s material has density ρ. We pluck the string at time t = 0, giving it an initial deflection y(x, t=0) = f (x). In 1st year, you have already discussed that this leads to waves propagating on the string with phase velocity, c:

(eq. 3.10)

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c = σρ

=F0µ

Wherein µ = ρA is the mass/length of the string. Of course, on strings of finite length, we will get standing waves again, with wavelength given by the length of the string. Exercise: Verify that the units of √σ/ρ are m/s Exercise: Discuss musical string instruments using Eq. 3.10, c = λf, and λ=2L (wherein L is the length of a vibrating string). How do you tune the guitar with ‘tuning pegs’ (screws) that tighten strings? How do you play it by ‘fretting’ the strings, i.e. shortening L? Why does a bass guitar or double bass have thicker and longer strings than a lead guitar or violin? Pressure waves in gases and liquids (‘sound’) Sound waves are an important example of waves, we even have a sense that picks them up. In a fluid (gas or liquid), sound waves are always longitudinal waves, i.e. the fluid is compressed or expanded along the direction of the wave’s propagation. Assume the gas is contained in a linear tube of crosssection A, wherein it may move forwards or back with velocity, v. We look at a small volume of gas (or liquid), V = Al, between locations x, and x+l. At location, x, gas has pressure, p(x), and moves with velocity v(x) along the tube (i.e, longitudinally). At x+l, pressure is p(x+l), velocity is v(x+l). Since the pressure at x and x+l may be different, an overall force F acts on V: F = -AΔp ≈ -Aldp/dx =-Vdp/dx, assuming l to be small compared to the length scale of pressure change, l

Exercise: Combine eqn’s 3.11, 3.12 into a single equation. Now, take the derivative of 3.11 w.r.t. x, and of 3.12 w.r.t. t. The left- hand sides will then be equal, because the order of differentiation does not matter. Consequently, equate the right hand sides:

(eq.3.13)

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∂2 p∂t2

=1κρ

∂2 p∂x2

Compare to 3.7- you see 3.13 is a wave equation for the pressure in the tube. Such waves are known as sound, or acoustic waves. The point of having a general wave eqn is illustrated nicely, also. Not only can we tell that fluids can sustain pressure waves, but also, we can immediately read their phase velocity as c = 1/√κρ . As an example, the adiabatic compressibility of water is κ = 5x10-10 m2/N, and the density of water is 1000 kg/m3. Consequently, sound in water has phase velocity 1.4 km/s. Exercise: You are assigned to measure the compressibility of a number of hydraulic oils. Given that compressibility of liquids is extremely low, this is far from easy to do directly. Propose a simple method that avoids use of high pressure rams and ultra- strong vessels. Exercise: Assuming that phase velocity c in fluids is controlled by compressibility and density, use dimensional analysis to show c = 1/√κρ. Interpret the dimensionless Mach number from aerodynamics, M = v√κρ, with v being the velocity of the moving object. Exercise: Show that for an ideal gas, c ∝ √T, with temperature T. (This is not easy. You need to calculate the adiabatic compressibility of an ideal gas, using the adiabatic equation, p ∝ V-γ). The root- mean- square velocity of molecules in gases also is proportional to √T. Interpret why the phase velocity of sound in a gas is coupled to rms velocity of gas molecules. It is different for solids- why? Mechanical waves in solids A similar reasoning as for pressure in fluids can be applied to the mechanical tension (σ, also called ‘stress’) in solids. We find that a solid can also sustain longitudinal mechanical waves, which are also called sound, or acoustic waves. The relevant mechanical property that replaces the inverse of compressibility is the elastic modulus E (also known as Young’s modulus) of the material. E is defined similar to a spring constant (but has different dimensions!). When a certain tension (stress) is applied to a solid of length, l, and, it will stretch by δl, called ‘strain’. How big δl is depends on E as follows:

(eq. 3.14)

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δll

=σE

3.14 serves as the defining equation for E. You see it can be re- arranged as σ = Εδl/l, which compares to Δp = -1/κ ΔV/V (apart from the sign). The phase velocity of sound in solids is given by

(eq 3.15)

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c = Eρ

Solids can also sustain another type of waves, known as shear waves. Shear waves are transversal rather than longitudinal waves, and therefore can be polarised. Shear is defined as a ‘sideways’ deformation of a body, rather than expansion or compression. A force F is applied along the surface, A, of a body, not normal to it. F/A = τ is known as shear stress. As a result of shear stress,

the body tilts sideways (‘shears’) by a little angle, α. The size of α relates to τ via a property of the solid called shear modulus, G:

(eq. 3.16)

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α =τG

τ, G have the same dimensions as σ, E, respectively, but the direction of the force involved is different. Shear waves propagate with phase velocity c given by:

(eq. 3.17)

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c = Gρ

Some materials may display anisotropic shear modulus, i.e. shear modulus is different for the two possible polarisations of shear waves. However, for almost all materials, shear modulus is smaller than elastic modulus, typically E/3 < G < E/2. Consequently, shear waves have lower phase velocity than longitudinal mechanical waves. Shear waves are unique to solids, fluids (liquids or gases) do have zero shear modulus and therefore, cannot sustain shear waves. Exercise: Discuss why liquids and gases have zero shear modulus. Discuss why G < E. Longitudinal and transversal waves in solids are collectively known as elastic waves. An important example are seismic waves - ‘earthquakes’. A seismic event in the Earth’s mantle generates both longitudinal and shear waves. In general, propagation of seismic waves is quite complicated, because modulus and density depend on depth below ground, the Earth’s mantle is solid, but the core is liquid, and waves can travel along the surface as well as through the bulk. But often, a quake hits you twice: The longitudinal ‘primary’ or ‘P’ wave first, then the secondary ‘S’ wave, which is a shear wave. The S- wave usually is more destructive (why?). When a quake hits your building, but does not destroy it, be prepared for a more destructive second hit coming in a few seconds. Get out immediately! Seismic waves hitting the sea bed from below can cause tsunamis. Seismic waves originating from a ‘shallow’ centre, having low frequency and with little surface wave contribution are the telltale of a nuclear test rather than a natural earthquake. Exercise: Is it the P- wave or the S- wave that causes a tsunami? Intensity of waves We will now study the very important behaviour of waves that encounter the boundary between 2 media. Waves transport energy and momentum. The density of the energy flux, that is energy transported through unit area in unit time, is called Intensity I of the wave. The question is, what fraction of intensity will be transmitted, what fraction will be reflected, at the boundary? Exercise: Give the SI units, or the dimensions, of intensity. To calculate intensity, we first define the energy density w, Energy/Volume, of a wave. Intensity is then given by how fast the energy density propagates. A harmonic wave, similar to an oscillator, can store energy in two ways: Either, as kinetic energy (e.g., a moving mass of gas, or piece of string), or, as potential energy (e.g., as a compressed gas, or a stretched piece of string). As the wave propagates, it continuously ‘swaps’ one form of energy for the other; however, at every point in space, the sum of the two is the same at all times. E.g, at the node of pressure, you will have an antinode of velocity (maximum velocity), and vice versa. The energy density w can therefore be calculated in two ways, either from the velocity amplitude (all energy = kinetic), or the pressure ampli