Lecture 6: Differential Equations Describing Vibrations

Transcription

1 Lecture 6: Differential Equations Describing Vibrations In Chapter 3 of the Benson textbook, we will look at how various types of musical instruments produce sound, focusing on issues like how the construction of the instrument affects the spectrum of the musical tones it produces. The link between the physics and the music is mathematical, given by a differential equation that describes how something vibrates a guitar string, the skin of a drum, or the air column inside a wind instrument. In this lecture I want to introduce the kinds of ordinary and partial differential equations which we will use in analyzing certain musical instruments. Kinds of Differential Equations A differential equation (DE) is an equation whose solution is a function, and which involves derivatives of that unknown function. For example, the equation dy dt = 5y specifies a function y(t) whose derivative is 5 times itself. A solution of this equation is y = Ce 5t for any constant C. (In fact, all solutions of the equation are of this form; a formula which gives all solutions for a DE is called the general solution.) The order of a DE is the order of the highest derivative of the unknown function involved. For example, you ve already encountered the equation describing simple harmonic motion, dt 2 = k m y, where m is the mass of an object attached to a spring, and k is the spring constant. This is a second order equation because the highest derivative 1

2 of y(t) involved is the second derivative. We saw earlier that the general solution for this equation is of the form ( ) ( ) k (1) y = c 1 cos m t k + c 2 sin m t, with two arbitrary constants c 1, c 2. (For a DE of order n, we expect the general solution to contain n arbitrary constants.) This is also an example of a linear combination of solutions, since each of the sine and cosine terms is itself a solution of the DE. The kind differential equations we ll be using, called linear equations, will usually have the property that we can mix solutions in this way. More formally, we say that a differential equation has the superposition property if, whenever y 1, y 2, y 3,... are solutions, then for any constants c 1, c 2, c 3,... the linear combination y = c 1 y 1 + c 2 y 2 + c 3 y is also a solution. Examples dy 5y = 0 1st order, linear homogeneous dt dt + k y = 0 2nd order, linear homogeneous 2 m dt + 2 sin2 (t)y = 0 2nd order, linear homogeneous (Hill s equation) dt + k y = cos t 2nd order, linear non-homogeneous 2 m + sin y = 0 2nd order, non-linear (pendulum eqn.) dt2 d 4 y dt 4 k4 y = 0 4th order, linear homogeneous (vibrating beam eqn.) Among the above equations, the ones that are linear homogeneous have the superposition property. 2

3 Simple Harmonic Motion with Damping The DE for simple harmonic motion my + ky = 0 models the oscillations of a mass m attached to a spring which exerts a force ky in response to a stretch by y units. The solutions of this equation, given by (1), oscillate forever with the same period and amplitude. But in the real world, such vibrations die down after a while. To get more realistic behaviour, we add a term to DE to take friction into account: (2) my + µy + ky = 0, where the frictional force is proportion to velocity y, with constant of proportionality µ > 0. (The frictional force is also known as damping.) Solutions of this equation can be determined by solving a related algebraic equation, mr 2 + µr + k = 0, known as the characteristic equation. (This equation comes from substituting a guess y = e rt, for some constant r, into the equation.) This is a quadratic equation for r, and its solutions are given by r = µ ± 2m, = µ2 4km, where is the discriminant of the quadratic. If there is strong damping (i.e., µ is large) then is positive and both possible values of r are negative real numbers. Then the motion of the mass on the spring dies down to zero quite quickly. Example y + 5y + 4y = 0 has = 9, giving r = 1 and r = 4; the general solution is y = c 1 e t + c 2 e 4t. If there is weak damping (µ is small) then is negative and the possible values of r are complex numbers r = α ± iβ. In this case, the general 3

4 solution is y = e αt (c 1 cos(βt) + c 2 sin(βt)), α = µ 2m, β = 2m. Because α is negative, the solutions look like a sine wave with amplitude dying down to zero as time goes on. Example Take m =.05kg or 50g, k = 5N/m and µ = 0.6N/(m/s). Then =.64, α = 6, β = 8 and the general solution is y = e 6t (c 1 cos(8t) + c 2 sin(8t)). Note that the periodic factor in this solution has period 2π/β.785 seconds. Resonance We now want to fill in some detail in our heuristic explanation for how our ears transform sound waves into nerve impulses to the brain namely, that a sine wave of frequency f causes a specific part of the basilar membrane to vibrate at the same frequency. If we visualize this part of the membrane as a underdamped spring that vibrates at a certain natural frequency (determined by β), then the sound waves act on that spring by adding a periodic external force, which we can model by adding a cosine term to the equation (2) (3) my + µy + ky = R cos(ωt), making it non-homogeneous. Here, the right-hand side represents the force due to the cosine wave entering the inner ear (with amplitude R and period 2π/ω), while k, m, µ are determined by which part of the basilar membrane we are watching. We want to see what (if any) vibrations of the same frequency are set up in the membrane. (Note that the frequency of the incoming wave is actually ω/(2π), but I will refer to ω as the frequency.) If we plug in a guess of the form y = A cos(ωt) + B sin(ωt) 4

5 then we find that A, B are given by A = (k mω 2 )R (k mω 2 ) 2 + µ 2 ω 2, B = µωr (k mω 2 ) 2 + µ 2 ω 2. Notice that this motion of the membrane doesn t have a decreasing exponential factor; instead, it has constant amplitude (at least, as long as the incoming sound wave is present) given by A2 + B 2 R = (k mω2 ) 2 + µ 2 ω 2. The ratio of the amplitude R of the incoming signal with the amplitude of the response is known as the frequency response: 1 M(ω) = (k mω2 ) 2 + µ 2 ω 2. Remember, the constants k, m, µ are fixed by our focusing on one particular part of the membrane. We see that the magnitude of the response of that part to an incoming sine wave depends on the frequency ω of the incoming wave. For what frequency is the response the biggest? This is a question we can answer by applying first-year calculus to the function M(ω). (Actually, it s easier to do calculus on the expression M(ω) 2 = (k mω 2 ) 2 + µ 2 ω 2.) When µ 2 > 2km, there is no positive maximum for M(ω). However, if µ 2 < 2km, then the maximum response occurs at the frequency ω r = 2km µ 2 2m 2. The phenomenon that an underdamped system has maximum amplitude response at one certain frequency is called resonance, and ω r is the resonant frequency. (Note that when the damping is zero, the resonant frequency is the same as the natural frequency of the system.) When a pure cosine wave enters your inner ear, it will excite vibrations in every part of the basilar membrane, but the wave will resonate with only one part of the membrane. 5

6 The hair cells in that part detect the resonance, and signal your brain that that particular frequency has been detected. Of course, real musical sounds are composed of many pure sine/cosine waves put together, and so several parts of your basilar membrane will resonate at the same time. Partial Derivatives In real musical instruments, different parts of the object vibrate at different speeds; for example, the middle of a plucked string may move the most, while the parts near the clamped ends of the string only move a small amount. Therefore, we will want to describe the vibrations of a musical instrument as a function of time but also position along the instrument. (So, instead of one spring with a mass on it, imagine a row of springs, each oscillating up and down, but with the spring at position x connected to the other springs to the left and right.) For this reason, we will need to consider the calculus of functions u(x, t) that depend on a time variable t and a spatial coordinate x. The way we differentiate such functions is called partial differentiation, and it s done by taking an ordinary derivative with respect to one variable (say, t) while pretending that the other variable is a constant. This kind of differentiation is denoted by t and. For example, if x then u(x, t) = 4x + 5t + t 3 x 2 t + sin(xt) u x = 4 2xt + t cos(xt), u t = 5 + 3t2 x 2 + x cos(xt). 6

7 For higher partial derivatives we use notation like 2 / x 2, e.g. x = ( ) u = 2t t 2 sin(xt), 2 x x t = 6t 2 x2 sin(xt), = 2x + cos(xt) xt sin(xt). t x If a function depends on three or more variables, the rule for partial differentiation is the same: pretend all the other variables are constants while you take a derivative with respect to one of them. For example, if t, x, y are three independent variables, then ( ) x 2 y + e y2 t = y 2 e y2t, t ( ) x 2 y + e y2 t = 2xy, x ( ) x 2 y + e y2 t = x 2 + 2tye y2t. y Differential equations satisfied by functions of two (or more) variables are known as partial differential equations (PDE). The most prominent PDE we ll encounter are the wave equation t = u 2 c2 2 x 2, which describes the propagation of waves through a medium, where c is the speed of transmission; and the Laplace equation x + 2 u 2 y = 0, 2 which can describe the shape of standing waves on a surface with coordinates x and y. The wave equation and Laplace equation are second-order, linear homogeneous PDE. As such, they again have the superposition property: if u 1 7

8 and u 2 are two functions that satisfy one of these PDE, then any linear combination c 1 u 1 + c 2 u 2 is also a solution of that PDE. Assignment Read Benson sections 1.10, 1.11 Problems: (1) Write down the general solutions to the equations (i) y +5y +6y = 0 and (ii) y + 2y + 17y = 0. (2) Let m = 1, k = 17 and µ = 4 in equation (3). Apply first-year calculus (i.e., the first- and second-derivative tests) to show that M(ω) has a maximum at ω = 3. (3) Show that e kx, e kx, cos(kx) and sin(kx) are solutions of the vibrating beam equation. Use superposition to make a general solution y(x) (involving four arbitrary constants) and find which of those solutions satisfy the boundary conditions y(0) = y (0) = 0 (corresponding to clamping one end of the beam). 8