Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
The Vibrating String Kent Merryfield Assumptions: 1. A thin, elastic string is stretched between two fixed supports. It is held under a tension (that is, a force that acts lengthwise along the string) of size T. 2. The string is of uniform composition, with a constant mass per unit length of size . In what follows, let “horizontal” mean the same direction as the “home” or “rest” position of the string, and let “vertical” be a direction perpendicular to that. The motions of the string are small in the following senses: 3. When the string moves, the horizontal displacement of any particle of the string is insignificantly small compared to its vertical displacement. 4. The angle that the string makes with the horizontal is so small that for all practical purposes we may assume that sin = tan = . 5. The change in the length of the string is so small that for all practical purposes the tension T remains unchanged. Under these assumptions, define u(x, t) to be the “vertical” displacement of the string as a function of x, which represents the position along the length of the string and of t, which is time. ∂u At any point x along the string, the string has a slope ∂x . ∂u The angle with the horizontal is ∂x = tan sin . The horizontal component of the tension is T cos T. ∂u The vertical component of the tension is T sin T ∂x . Consider a small piece of the string of length dx. The horizontal component of tension is essentially the same at each end. ∂u At the left end, the slope is ∂x and the vertical component of ∂u tension is about –T ∂x . ∂u ∂2u At the right end, the slope is ∂x + 2 dx . ∂x ∂u ∂2u Thus the vertical component of tension about T ∂x + 2 dx . ∂x ∂2u The net vertical force on this piece of string is T 2 dx. ∂x Newton’s law of motion is F = ma. ∂2 u The mass is m = dx. The (vertical) acceleration is 2 . ∂t ∂2u ∂2u T 2 dx = 2 dx ∂x ∂t Divide by dx to get the wave equation: ∂2u T ∂2u ∂2u ∂2u 2 = or = c ∂x 2 ∂t 2 ∂t 2 ∂x 2 T . If we measure in kg/m and T in N = kgm/s2, then c 2 is measured in m2/s2, so c is in m/s and is a velocity. where c 2 = If ƒ and g are completely arbitrary functions, the function u(x, t) = ƒ(x – ct) + g(x – ct) is a solution of the wave equation at least if you don’t think about the ends of the string. In our musical instruments, the string is stretched over two fixed supports that are a distance L apart, leaving a string of length L. We can use this to say our equation has boundary values: u(0, t) = 0 for all t, and u(L, t) = 0 for all t. Separation of Variables ∂2 u ∂2u 2 Our problem is: 2 = c ; u(0, t) = u(L, t) = 0. ∂t ∂x 2 This is a linear homogeneous partial differential equation. Because it is linear and homogeneous, if you have a set of functions that are solutions, then any linear combination of those functions is a solution. We try to find some such basis set of solutions. Assume there is a solution of the form u(x, t) = H(x)V(t). H(x) V(t) Then H(x)V(t) = c 2H(x)V(t), or H(x) = c 2 V(t) = . H(x) = H(x); H(0) = H(L) = 0 is an eigenvalue problem. We can’t have ≥ 0 as we would only get H = 0. So assume that = –k 2. We then get H(x) = C sin kx, and sin kL = 0. This forces kL = nπ for some natural number n, so nπ n 2 π2 k = L and = – 2 . L n 2 π2 c 2 Then H(t) = – H(t). This has solutions L2 nπct nπct Hn(t) = An cos L + Bn sin L . nπct nπct The function Hn(t) = An cos L + Bn sin L is a simple nc T harmonic motion of frequency 2L . Since c 2 = , the frequency n T (in time) of this simple harmonic motion is = 2L . The most general solution to the partial differential equation is u(x, t) = ∞ n=1 nπx sin L Hn(t) , where Hn(t) is as above. The most important term of this sum is the n = 1 term, in which the full string vibrates at the fundamental frequency. The terms with n ≥ 2 create overtones. 1 The fundamental frequency is = 2L T . This depends on , L, and T. If , L, and are known, then T = 4L 22. For the violin or viola A string, we may assume = 440 Hz. The other strings are tuned above or below that in “fifths”. A “pure” fifth is a frequency ratio of 3:2 or 1.5:1, but an “equally tempered” fifth is a frequency ratio of 27/12:1 or 1.4983:1. Calculations of string tensions: Violin Strings String (Hz) E 659 A 440 D 294 G 196 L = 32.4 cm T (lbs) (g/cm) 16 0.0066 12 0.0112 9 0.0273 10 c (mph) 956 638 426 284 Viola Strings String (Hz) A 440 D 294 G 196 C 131 L = 37.1 cm (g/cm) ? 0.0104 0.0225 0.0534 c (mph) ? 487 325 217 T (lbs) ? 11 11 11