Chapter Nine - Vibrating Membranes

Transcription

1 Chapter Nine - Vibrating Membranes The vibrating membrane problem is simply the two-dimensional version of the vibrating string problems. Specifically, we are given a plane region R and we want to find ux,y,t so that 2 u 1 k 2 u tt 0, forx,y R, t 0 ux,y,t 0 forx,y boundary of R; ux,y,0 fx,y and u t x,y,0 gx,y. Here k is a constant which depends on the physical properties of the membranedensity and tension. We begin with the case in which R is a disc of radius c and centered at the origin. It should come as no surprise that we use polar coordinates. Fortunately, we know all about the associated eigenvalue problem 2 2 c, 0. Recall we have eigenvalues nm z nm /c, n 0,1,2,, m 1, 2,3,, where z nm is the m th zero of the Bessel function J n. The corresponding eigenfunctions nm are 0m r, J 0 m r, m 1,2,3, nm r, J n nm rcosn J n nm rsinn. We thus let ur,,t m1 0m tj 0 m r nm tcosn nm tsinnj n nm r. Hence, 1

2 2 u 1 k 2 u tt m1 2 0m 0m t 1 k 2 0m t J 0 m r 2 nm nm t 1 k 2 0m tosn 2 nm nm t 1 k 2 0m t sinn J n nm r We now have a bunch of ordinary differential equations of the form t 2 k 2 t 0. Thus, nm t a nm cos nm ktb nm sin nm kt. nm t c nm cos nm ktd nm sin nm kt, We substitute these vales back into our expression for ur,,t and when the dust settles, we see ur,,ta 0m cos 0m ktb 0m sin 0m ktj 0 m r m1 a nm cos nm ktb nm sin nm ktj n nm rcosn c nm cos nm ktd nm sin nm ktj n nm rsinn All the constants are determined from the initial conditions. First, ur,,0 fr, gives us Hence, ur,,0 fr, a 0m J 0 m r m1 a nm J n nm rcosn c nm J n nm rsinn 2

3 a nm a 0m c nm rfr,j 0 m rdrd rj 0 m r 2 drd rfr,j n nm rcosndrd rj 0 m rcosn 2 drd rfr,j n nm rsinndrd rj n nm rsinn 2 drd,, and. Exercises 1. Find expressions for the constants b nm and d nm in the preceding discussion. Consider a solution u 0m in which all series terms except those for 0m are zero. Thus, ur,,t a 0m cos 0m ktb 0m sin 0m ktj 0 m r Acosm ktj 0 m r. Ignoring the phase shift, we have that u is a constant multiple of u 0m : u 0m osm ktj 0 m r. Let s see what this looks like. Notice first, that the solution does not depend on. Thus anywhere you take a cross-sectional slice of the membrane, you see the same curve. Initially, we see u 0m r,,0 J 0 m r. For m 1, we have u 01 J 0 1 r. This looks like 3

4 Now, as t increases, we see this same shape, but with an amplitude of cos1 kt. Here is a picture for a few values of t: The membrane thus oscillates up and down with a frequency of 01 k 2.405k. For m 2, the membrane oscillates with a frequency of 02 k 5.520k, and the corresponding pictures of the cross-section look like Observe that here there is a so-called nodal curvea set of points that do not move. It is a circle of radius1 / 02c c. Looking down on the membrane, we see I hope it is clear that the corresponding pictures for an oscillation frequency 1/ 03 k look like 4

5 Exercises 2. Find the radii of the nodal circles for the solution u 03 os3 ktj 0 3 r. 3. Describe and draw pictures of the oscillation having frequency 05 k. Next, consider the solution in which all the terms save the ones for 11 are zero. Here ur,,t a 11 cos 11 ktb 11 sin 11 ktj 1 11 rcos c 11 cos 11 ktd 11 sin 11 ktj1 11 rsin. Look at the first of the two terms: ur,,t a11 cos 11 ktb 11 sin 11 ktj 1 11 rcos Acos 11 ktj 1 11 rcos. As before, look at u 11 r,, t cos 11 ktj 1 11 rcos. In this case, a cross-sectional slice through the membrane does indeed depend on.for 0, we see cos 11 ktj 1 11 r and for, we see cos 11 ktj 1 11 r.thus, for various values of t, this slice looks like 5

6 As we take slices for increasing values of,the picture looks the same, but with decreasing amplitude until /2, at which the amplitude is zero; i.e., there is a nodal line. Look at the picture r Note that we get nothing new from the cos 11 ktj 1 11 rsin term; it is just this picture turned ninety degrees. It should be clear how to see what the remaining vibration modes look like. For the solution corresponding to nm, the nodal lines are the solutions to the equation J n nm rcosn. For instance, forn.m 3,2, the nodal lines look like 6

7 A normal mode is a solution in which each point of the membrane oscillates about equilibrium with the same frequency. Thus the solutions just discussed are normal modes. Exercises 4. The picture shows the nodal lines of a vibrating membrane (same membrane). Which is vibrating with the higher frequency? Explain. Now let s consider the case of a square membrane: 2 u 1 k 2 u tt 0, 0 x,y L, t 0 ux,0 ul,y ux,l u0,y 0, and ux,y,0 fx,y, u t x,y,0 gx,y.. From our previous work on eigenvalue problems, etc., we know to let ux,y,t nm tsin m L xsin n L y, which gives us the solution 7

8 ux,y,t a nm cos nm ktb nm sin nm kt sin m L xsin n L y, where nm n m, L L and the constants a nm and b nm are determined from the initial conditions. Exercise 5. Find expressions for the constants a nm and b nm. The normal modes of oscillation are more exciting in this case. Now if we assume the membrane vibrates with a fundamental frequency nm, there is a significant complication compared to the situation with a circular membrane. In the circular case, the integers n and m determined exactly one term of our series as the solution. Here this is not the case. The normal modes are not simply the terms u nm a nm cos nm ktb nm sin nm ktsin n L Acos nm ktsin m L xsin n L y xsin m L y because the frequency nm k may appear in more that one such term. Let s illustrate with an example. Suppose k 1, and L. Then nm n 2 m 2. Now look at the natural frequency Then all solutions of the form cost 10asinxsin3y bsin3xsiny are normal modes. Look at the case where b 0. Then each point x,y moves sinusoidally up and down with the frequency 10 with an amplitude given by a constant times sx,y sinxsin3y The nodal lines are simply places at which sx,y 0. 8

9 But with this same frequency, we also have lots of other normal modes. Let s take a look at one of them: The nodal lines: cost 10sinxsin3y 2sin3xsiny As you can imagine, the possibilities are almost endless. Exercises 6. Draw some more graphs of nodal lines for normal modes of frequecy 10. 9