Striking a Beat. Ashley Martin PHY 495. Spring Striking a Beat. Ashley Martin PHY 495. Introduction. Outline. Cartesian Coordinates

Transcription

1 Spring 2012 Polar

2 Where is it optimal to strike a circular drum? Polar

3 Daniel Bernoulli ( ) - introduced concept of Bessel functions Leonhard Euler ( ) - used Bessel funtions of both zero and integral orders Friedrich Bessel ( ) - generalized the Bessel function Polar

4 I wish to ascertain the difference one hears between striking a drum at the center and off center. Two Dimensions Polar vs. Striking Off Center Polar

5 Rectangular Membrane The wave equation in 2 ( u 2 ) t 2 = u c2 x u T y 2 ; c = ρ t > 0, 0 x L, 0 < y H Polar

6 Conditions The fixed boundary conditions are with initial conditions u(0, y, t) = 0, u(l, y, t) = 0 u(x, 0, t) = 0, u(x, H, t) = 0 u(x, y, 0) = f (x, y) u t (x, y, 0) = g(x, y) Polar

7 Separation of Variables Let u(x, y, t) = F (x)g(y)t (t), then F (x)g(y)t (t) = c 2 (F (x)g(y)t (t) + F (x)g (y)t (t)) which leads to thus obtaining two equations 1 T c 2 T = F F + G G = λ T + c 2 λt = 0. (1) Polar F F + G G = λ (2)

8 Time Dependence T (t) = A cos ωt + B sin ωt, ω = c λ. Frequency of oscillations for the harmonics are ν = ω 2π = c λ. 2π Polar

9 Spatial Equations F F = G G This leads to two equations λ = µ, µ < 0. F + µf = 0 (3) Polar G + (λ µ)g = 0. (4)

10 Apply BCs Similarly, F (0) = 0 = A = 0, F (L) = 0 = B = 0 or sin λx L = 0 F n (x) = B n sin nπx ( nπ ) 2 L, λ =, n = 1, 2,.... L G(0) = 0, G(H) = 0 Polar G m (y) = D m sin mπy H ( mπ ) 2, λ µ =, m = 1, 2,.... H

11 Yields the product solutions u nm (x, y, t) = (A nm cos ω nm t + B nm sin ω nm t) sin nπx L where, ( nπ λ nm = L ω nm = c (nπ Thus the general solutions is u(x, y, t) = n=1 m=1 L ) 2 ( mπ ) 2 + H ) 2 ( mπ ) 2. + H (A nm cos ω nm t+b nm sin ω nm t) sin nπx L mπy sin H ; sin mπy H. Polar

12 Fourier Coefficients Initial displacement is u(x, y, 0) = f (x, y), thus f (x, y) = n=1 m=1 Rewriting as a single sum gives A nm sin nπx L mπy sin H. f (x, y) = A n (y) sin nπx L, where n=1 A n (y) = A nm sin mπy H. m=1 Polar

13 Fourier Coefficients The coefficients of Fourier sine series A n (y) = 2 L A nm = 2 H L 0 H 0 f (x, y) sin nπx L dx, A n (y) sin mπy H dy. Results in a double Fourier sine series, A nm = 4 LH H L 0 0 f (x, y) sin nπx L mπy sin H dxdy. Polar

14 Fourier Coefficients Initial velocity, u t (x, y, 0) = g(x, y), B nm = g(x, y) = n=1 m=1 4 H ω nm LH 0 L 0 1 B nm sin nπx ω nm L g(x, y) sin nπx L mπy sin H. mπy sin H dxdy. Polar

15 This completes the full solution of the vibrating rectangular membrane problem: u(x, y, t) = n=1 m=1 (A nm cos ω nm t+b nm sin ω nm t) sin nπx L A nm = 4 H LH 0 B nm = 4 ω nm LH L 0 H L 0 ω nm = c f (x, y) sin nπx L mπy sin H dxdy g(x, y) sin nπx mπy sin 0 L H dxdy (nπ ) 2 ( mπ ) 2 + L H mπy sin H, Polar

16 ϕ nm (x, y) = sin nπx L Nodal lines occur when ϕ nm (x, y) = 0. mπy sin H Polar

17 sin πx L sin πy H n = m = 1 = 0, x = 0, L; = 0, y = 0, H [3] Polar

18 [3] Polar n = 3, m = 2 sin 3πx L = 0, x = 0, L 3, 2L 3, L; sin 2πy H = 0, y = 0, H 2, H

19 Circular Membrane The wave equation in polar coordinates 2 [ u 2 t 2 = u c2 r u r r ] u r 2 θ 2 ; t > 0, 0 r a, 0 θ 2π Polar [3]

20 Conditions The fixed boundary conditions are The periodic conditions are with initial conditions u(a, θ, t) = 0 t > 0, 0 θ 2π. u(r, θ, t) = u(r, θ + 2π, t) u θ (r, θ, t) = u θ (r, θ + 2π, t), Polar u(r, θ, 0) = f (r, θ) u t (r, θ, 0) = g(r, θ)

21 Separation of Variables Let u(r, θ, t) = R(r)Θ(θ)T (t), Θ + n 2 Θ = 0 (5) T + c 2 λ 2 T = 0 (6) r 2 R + rr + (r 2 λ 2 n 2 )R = 0. (7) Polar

22 Simple Equations The solutions to (5) and (6) are Θ n (θ) = A n cos nθ + B n sin nθ; T (t) = C nm cos ω nm t + D nm sin ω nm t, ω nm = c λ nm. Polar

23 Radial Equation The solutions to r 2 R + rr + (r 2 λ 2 n 2 )R = 0, are Bessel functions. Polar Figure: A few Bessel Functions with their zeros, z nm [7].

24 Apply BCs u(a, θ, t) = 0 for t > 0 and 0 θ 2π = R(a) = 0. Since we expect solutions to be finite at the center, therefore R(r) = CJ n ( λr). Polar

25 Apply BCs Since, R(a) = 0, J n ( λa) = 0. Listing a few of the mth zeros of J n z nm n = 0 n = 1 n = 2 n = 3 m = m = m = m = Polar Table: Approximate location of the zeros of Bessel functions of the first kind, J n (z nm ) = 0 [7].

26 Denoting that z nm is the mth zero of J n (x), then, J n ( λa) = 0, tells us that λa = znm λ nm = ( znm Substituting this into R(r) gives a ) 2. R(r) = J n ( znm a r ). Polar

27 General Solution u(r, θ, t) = J n (λ nm r) [A nm cos nθ + B nm sin nθ] cos ω nm t n=0 m=1 +J n (λ nm r) [C nm cos nθ + D nm sin nθ] sin ω nm t. Polar

28 The circular membrane harmonics are given by ( ) z nm r Φ(r, θ) = cos(nθ)j n. a Nodal curves occur when cos nθ = 0 or J n ( z nmr a ) = 0. Polar Figure: Vibrational Modes of a Circular Membrane with dashed nodal curves [3].

29 Nodal Circles We wish to find values of r such that z nmr a Bessel function. Thus, z nm r a = z nj, j m r = z nj z nm a, r a. is a zero of the Polar

30 [3] For m = 2, we have two circles, r = a and r = z n1 z n2 for each n. We will need to calculate r for each n = 0, 1, 2 by using the table. For n = 0, For n = 1, and for n = 2, r = z n1 z n2 = a a. r = a a, r = a a. Polar

31 General Solution u(r, θ, t) = J n (λ nm r) [A nm cos nθ + B nm sin nθ] cos ω nm t n=0 m=1 +J n (λ nm r) [C nm cos nθ + D nm sin nθ] sin ω nm t Polar

32 Initial displacement, u(r, θ, 0) = f (r, θ), u(r, θ, 0) = J nm (λ nm r) [A nm cos nθ + B nm sin nθ]. n=0 m=1 Let a n (r) = A nm J n (λ nm r) and b n (r) = B nm J n (λ nm r). m=1 m=1 Rewriting the equation gives u(r, θ, 0) = [a n (r) cos nθ + b n (r) sin nθ], where n=0 a n (r) = 1 π 2π 0 f (r, θ) cos nθdθ, Polar b n (r) = 1 π 2π 0 f (r, θ) sin nθdθ.

33 Since a n (r) = Similarly, A nm = B nm = m=1 A nm J n (λ nm r), 2 a 2 [J n+1 (z n )] 2 2 a 2 [J n+1 (z n )] 2 a 0 a 0 a n (r)j n (λ nm r)dr. b n (r)j n (λ nm r)dr Polar

34 Intitial velocity, u t (r, θ, 0) = g(r, θ), C nm = D nm = 2 ω nm a 2 [J n+1 (z n )] 2 2 ω nm a 2 [J n+1 (z n )] 2 a 0 a 0 c n (r)j n (λ nm r)dr, d n (r)j n (λ nm r)dr, Polar

35 Axisymmetric Solution The solution independent of θ, u(r, t) = m=1 ( z0m r ) J 0m [α m cos ω 0m t + β m sin ω 0m t] a. Notice for n 1, each vibrational mode has nodal curves that pass through the center so none of these modes can be excited. Polar

36 Conclusion Where is it optimal to strike a drum? Off Center Polar

37 Acknowledgements Polar Dr. Russel Herman, Allison Martin, Navid Sharifian, and my friends and family!

38 A. Bruder. Two mathematical models for the tympanic membrane. pages , Models_for_the_Tympanic_Membrane.pdf. R. Herman. An introduction to fourier and complex analysis with applications to the spectral analysis of signals. FCABook/Book2010/FCA_Main.pdf, March R. Herman. An introduction to mathematical physics via oscillations. Polar MathPhysBook/MathPhys_Main.pdf, March M. Kac. Can one hear the shape of a drum? The American Mathematical Monthly, 73:1 23, 1966.

39 22/Ford/MarkKac.pdf. O. Shipilova. Differential equations lecture vi, February bm20a2101/luennot/lecture_4.pdf. A. Wazwaz. Partial Differential Equation and Solitary Wave Theory. Higher Education, Beijing, D. Young. An introduction to partial differential equations in the undergraduate curriculum. // [4] [6] [3] [2] [1] [5] [7] Polar