Lecture 1: Simple Harmonic Oscillators

Matthew Schwartz Lecture 1: Simple Harmonic Oscillators 1 Introduction The simplest thing that can happen in the physical universe is nothing. The next simplest thing, which doesn’t get too far away from nothing, is an oscillation about nothing. This course studies those oscillations. When many oscillators are put together, you get waves. Almost all physical processes can be explained by breaking them down into simple building blocks and putting those blocks together. As we will see in this course, oscillators are the building blocks of a tremendous diversity of physical phenomena and technologies, including musical instruments, antennas, patriot missiles, x-ray crystallography, holography, quantum mechanics, 3D movies, cell phones, atomic clocks, ocean waves, gravitational waves, sonar, rain- bows, color perception, prisms, soap films, sunglasses, information theory, solar sails, cell phone communication, molecular spectroscopy, acoustics and lots more. Many of these topics will be covered first in lab where you will explore and uncover principles of physics on your own. The key mathematical technique to be mastered through this course is the Fourier transform. Fourier transforms, and Fourier series, play an absolutely crucial role in almost all areas of modern physics. I cannot emphasize enough how important Fourier transforms are in physics. The first couple of weeks of the course build on what you’ve covered in 15a (or 16 or 11a or AP50) – balls and springs and simple oscillators. These are described by the differential equation for the damped, driven oscillator: d2x(t) dx(t) F (t) + γ + ω2 x(t)= (1) dt2 dt 0 m Here x(t) is the displacement of the oscillator from equilibrium, ω0 is the natural angular frequency of the oscillator, γ is a damping coefficient, and F (t) is a driving force. We’ll start with γ =0 and F =0, in which case it’s a simple harmonic oscillator (Section 2). Then we’ll add γ, to get a damped harmonic oscillator (Section 4). Then add F (t) (Lecture 2). The damped, driven oscillator is governed by a linear differential equation (Section 5). Linear equations have the nice property that you can add two solutions to get a new solution. We will see how to solve them using complex exponentials, eiα and e−iα, which are linear combi- nations of sines and cosines (Section 6). A review of complex numbers is given in Section 7. Studying multiple coupled oscillators will lead to the concept of normal modes, which lead naturally to the wave equation, the Fourier series, and the Fourier transform (future lectures). 2 Why waves? Why oscillators? Recall Hooke’s law: if your displace a spring a distance x from its equilibrium position, the restoring force will be F = kx for some constant k. You probably had this law told to you in high school or 15a or wherever.− Maybe it’s an emperical fact, deduced from measuring springs, maybe it was just stated as true. Why is it true? Why does Hooke’s law hold? To derive Hooke’s law, you might image you need a microscopic description of a spring (what is it made out of, how does it bend, how are the atoms arranged, etc.). Indeed, if you hope to compute k, yes, absolutely, you need all of this. In fact you need so much detail that generally it’s impossible to compute k in any real spring. But also generally, we don’t care to compute k, we just measure it. That’s not the point. We don’t want to compute k. What we want to know is why is the force is proportional to displacement. Why is Hooke’s law true? 1 2 Section 2 First of all, it is true. Hooke’s law applies not just to springs, but to just about everything: Figure 1. The restoring force for pretty much anything (bending trees, swings, balls, tires, etc) is linear close to equilibrium. You can move any of these systems, or pretty much anything else around you, a little bit away from its equilibrium and it will want to come back. The more you move it, the stronger the restoring force will be. And often to an excellent approximation, the distance and force are directly proportional. To derive Hooke’s law, we just need a little bit of calculus. Let’s say we dispace some system, a spring or a tire or whatever a distance x from its equilibrium and measure the function F (x). We define x =0 as the equibrium point, so by definition, F (0)=0. Then, we can use Taylor’s theorem ′ 1 ′′ F (x)= F (0)+ xF (0)+ x2F (0)+ (2) 2 ··· Now F (0) = 0 and F ′(0) and F ′′(0) etc are just fixed numbers. So no matter what these numbers are, we can always find an x small enough so that F ′ 1 xF ′′ . Then we can neglect (0) 2 (0) ′′ ′ ≫ the 1 x2F (0) term compared to the xF (0) term. Similarly, we can always take x small enough 2 that all of the higher derivative terms are as small as we want. And therefore, F (x)= kx (3) − with k = F ′(0). We have just derived Hooke’s law! Close enough to equilibrium, the restoring force for anything− will be proportional to the displacement. Since y = kx is the equation for a line, we say systems obeying Hooke’s law are linear. Thus, everything− is linear close to equilibrium. More about linearity in Section 5. You might also ask, why does F depend only on x? Well, what else could it depend on? It could, for example, depend on velocity. Wind resistance is an example of a velocity-dependent force. However, sinec we are assuming that the object is close to equilibrium, its speed must be small (or else our assumption would quickly be violated). So x˙ is small. Thus we can Taylor exapnd in x˙ as well ∂F (x,x˙) ∂F (x,x˙) F (x,x˙)= x + x˙ + (4) ∂x ∂x˙ ··· x=x=0˙ x=x˙ =0 where the terms are higher order in x or x˙, so they are subleading close to equilbrium. ··· Writing ∂F (x,x˙) = mγ we then have ∂x˙ x=x˙ =0 − F (x)= kx mγx˙ (5) − − Then F = ma with a = x¨ gives d2x(t) dx(t) + γ + ω2 x(t)=0 (6) dt2 dt 0 k as in Eq. (1) with ω0 = . γ is called a damping coefficient, since the velocity dependence q m tends to slow the system down (as we will see). The other piece of Eq. (1), labeled F (t), is the driving force. It represents the action of something external to the system, like a woman pushing the swing with the girl on it, or the car tire being compressed by the car. Simple harmonic motion 3 3 Simple harmonic motion We have seen that Eq. (1) describes universally any system close to equilibrium. Now let’s solve it. First, take γ =0. Then Eq. (1) becomes d2 x(t)+ ω2 x(t)=0 (7) dt2 0 ω k ω g For a spring, 0 = m , for a pendulum 0 = L . Other systems have different expressions for q q ω0 in terms of the relevant physical parameters. We can solve this equation by hand, by plugging into Mathematica, or just by guessing. Guessing is often the easiest. So, we want to guess a function whose second derivative is proportional to itself. You know at least two functions with this property: sine and cosine. So let us write as an ansatz (ansatz is a sciency word for “educated guess”): x(t)= A sin(ωt)+ B cos(ωt) (8) This solution has 3 free parameters A, B and ω. Plugging in to Eq. (7) gives ω2[A sin(ωt)+ B cos(ωt)] + ω2 [A sin(ωt)+ B cos(ωt)]=0 (9) − 0 Thus, ω = ω0 (10) ω ω k That is, the angular frequency of the solution must be the paramter 0 = m in the differential equation. We get no constraint on A and B. q ω is called the angular frequency. It has units of radians per second. The frequency is ω ν = (11) 2π units of 1/sec. The solution x(t) we found goes back to itself after t t + T where → 1 2π T = = (12) ν ω is the period. T has units of seconds. The function x(t)= A sin(ωt)+ B cos(ωt) satisfies x(t)= x(t + nT ) for any integer n. In other words, the solutions oscillate! A and B are the amplitudes of the oscillation. They can be fixed by boundary conditions. For example, you specify the position and velocity at any given time, you can determine A and B x m x′ m . To be concrete, suppose we start with (0)=1 and (0)=2 s . Then, 1m = x(0)= A sin(ω0)+ B cos(ω0) = B (13) m 2 = x′(0)= ωAcos(ω0) ωB sin(ω0)= ωA (14) s − A 2 m B m So we find = ω s and =1 . Keep in mind that the angular frequency ω is not fixed by boundary conditions. It is deter- mined by the physical problem: ω = k where k = F ′(0) and m is the mass of the thing oscil- q m − lating. That is why if you start a pendulum from any height and give it any sort of initial kick, it will oscillate with the same frequency. Another representation of the general solution x(t) = A sin(ωt) + B cos(ωt) is often conve- nient. Instead of using A and B we can write x(t)= C sin(ωt + φ) (15) using trig identities, we find C sin(ωt + φ)= C cos(φ)sin(ωt)+ C sin(φ)cos(ωt) (16) and so A = C cos(φ) B = C sin(φ) (17) 4 Section 4 Thus we can swap the amplitudes A and B for the sine and cosine components for a single amplitude C and a phase φ.

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