Waves vibrations. Maxime Nicolas. To cite this version: HAL Id: cel Maxime Nicolas.

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1 Waves vibrations Maxime Nicolas To cite this version: Maxime Nicolas. Waves vibrations. Engineering school. France <cel > HAL Id: cel Submitted on 25 Jan 2017 HAL is a multi-disciplinary open access archive for the deposit and dissemination of scientific research documents, whether they are published or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers. L archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d enseignement et de recherche français ou étrangers, des laboratoires publics ou privés. Distributed under a Creative Commons Attribution - NonCommercial - NoDerivatives 4.0 International License

2 Waves & vibrations Maxime Nicolas Département génie civil november 2016 january 2017 november 2016 january / 31

3 Course 1 outline 1 Preamble Course schedule Online Course outline 2 Introduction 3 The harmonic oscillator november 2016 january / 31

4 Preamble PREAMBLE november 2016 january / 31

5 Course syllabus Preamble Course schedule Bâtiment Fermi, bureau 212 Schedule: 5lectures 5 workshops for civil eng. students 3 workshops for mech. eng. students Final exam: January 25th, 2017 november 2016 january / 31

6 Online Preamble Online This course is available on ENT/AmeTice : Sciences & technologies Polytech Cours communs [16] - S5 - JGC52D + JME51C - Ondes et vibrations (Maxime Nicolas) with lecture slides workshops texts past exams november 2016 january / 31

7 Abook Preamble Online This course is included in a book (paper and pdf versions available): november 2016 january / 31

8 Preamble Course outline Course outline 1 Introduction & harmonic oscillator 2 The wave equation and its solutions 3 1D transverse and longitudinal waves 4 2D waves: vibration of plates 5 beam vibration november 2016 january / 31

9 Introduction Introduction november 2016 january / 31

10 Introduction Where to find waves and vibrations? Short answer: everywhere : waves in fluids acoustics sound waves in fluids ripples, waves and tsunamis waves in solids compression waves vibrations of structures planes, cars, and many more electromagnetic waves light, radio, X-rays, chemical oscillators population dynamics health issues due to vibrations -rays november 2016 january / 31

11 Afewdefinitions Introduction wave: propagation of an oscillation or a vibration vibration: motion around an equilibrium state oscillation: motion of a body around an equilibrium point november 2016 january / 31

12 Equilibrium Introduction november 2016 january / 31

13 t-periodical functions Introduction F (t + ) = F (t), t f = 1,! = 2 f = 2 november 2016 january / 31

14 x-periodical functions Introduction F (x + ) = F (x), x k = 2 november 2016 january / 31

15 The harmonic oscillator The harmonic oscillator november 2016 january / 31

16 Example 1 The harmonic oscillator The mass-spring system: m d 2 x dt 2 + kx = 0 november 2016 january / 31

17 Example 2 The harmonic oscillator ml d 2 + mg sin = 0 dt2 november 2016 january / 31

18 The harmonic oscillator Examples comparison spring-mass pendulum m d 2 x dt 2 + kx = 0 ml d 2 dt 2 + mg sin = 0 d 2 x +! 2 dt 2 0 x = 0 d 2 +! 2 dt 2 0 sin = 0! 0 = km! 0 = gl linear equation for x non linear equation for november 2016 january / 31

19 Linearization The harmonic oscillator For any (continuous and derivable) function F near x 0 : F (x) x x0 = F (x 0 ) + n=1 1 n n! d f dx n (x 0)(x x 0 ) n for F ( ) = sin, x 0: F ( ) = november 2016 january / 31

20 The harmonic oscillator Examples comparison spring-mass pendulum m d 2 x dt 2 + kx = 0 ml d 2 dt 2 + mg sin = 0 d 2 x +! 2 dt 2 0 x = 0 d 2 +! 2 dt 2 0 = 0! 0 = km! 0 = gl linear equation for x linear equation for 1 november 2016 january / 31

21 The harmonic oscillator The harmonic oscillator equation The equation for a physical quantity A(t) (x or ) is d 2 A dt 2 +!2 0A = 0 this is a 2nd order di erential equation. on the blackboard november 2016 january / 31

22 The harmonic oscillator The harmonic oscillator equation General solution: d 2 A dt 2 +!2 0A = 0 A(t) = A 1 e i! 0t + A 2 e i! 0t = B 1 cos(! 0 t) + B 2 sin(! 0 t) With the initial conditions (A 0,Ȧ 0 ) A(t) = A 0 cos(! 0 t) + Ȧ0! 0 sin(! 0 t) november 2016 january / 31

23 View of the solution The harmonic oscillator november 2016 january / 31

24 The harmonic oscillator Energy This equation describes a conservative system (no loss of energy): Back to the mass-spring example: d 2 A dt 2 +!2 0A = 0 m d 2 x dt 2 + kx = 0... november 2016 january / 31

25 The harmonic oscillator Harmonic oscillator with damping Introducing a fluid damping force (prop. to velocity): d 2 A dt 2 + da dt +!2 0A = 0 november 2016 january / 31

26 Solution with damping The harmonic oscillator Testing function A = e rt Characteristic equation: r 2 + r +! 0 2 = 0 with 2 solutions r 1,2 = ± 1 2 4! = 2 ± and A(t) = e t2 A 1 e t + A 2 e t november 2016 january / 31

27 The harmonic oscillator Weak damping solution 2 4! 2 0 < 0, = i! 1,! 1 = 1 2 4! A(t) = e t2 A 0 cos(! 1 t) + 1! 1 A Ȧ 0 sin(! 1 t) november 2016 january / 31

28 The harmonic oscillator Energy loss with damping 1 2Ȧ !2 0 = Ȧ2 dt E(t) = de dt = Ȧ2 Ȧ2 dt november 2016 january / 31

29 The harmonic oscillator Oscillator with energy input d 2 A dt 2 + da dt +!2 0A = A F cos(!t) with A F the forcing amplitude, and! the forcing angular frequency. proposed long time solution: A(t) = A 1 cos(!t + ') Now find A 1 and '... november 2016 january / 31

30 Solution The harmonic oscillator A(t) = A 1 cos(!t + ') A 1 1 = A F (! 2 0! 2 ) 2 +! 2 2 ' = arctan!! 2 0!2 november 2016 january / 31

31 View The harmonic oscillator november 2016 january / 31

32 View The harmonic oscillator november 2016 january / 31

33 Waves & vibrations Maxime Nicolas Département génie civil november 2016 january 2017 november 2016 january / 22

34 Course 2 outline 1 Coupled oscillators 2 1D infinite chain of oscillators 3 The simple wave equation november 2016 january / 22

35 Coupled oscillators Coupled oscillators november 2016 january / 22

36 Coupled oscillators Coupled oscillators 2 identical oscillators + coupling spring november 2016 january / 22

37 Coupled oscillators Coupled equations Reminder: x i variables are perturbations out of equilibrium. mẍ 1 (t) = kx 1 k c (x 1 x 2 ) mẍ 2 (t) = kx 2 k c (x 2 x 1 ) or ẍ 1 (t) =! 0x 2 1! c(x 2 1 x 2 ) ẍ 2 (t) =! 0x 2 2! c(x 2 2 x 1 ) with oscillating solutions: x 1 = X 1 exp(i!t), x 2 = X 2 exp(i!t) november 2016 january / 22

38 Coupled oscillators Coupled equations as a linear problem! 2 c!2 0!2 +! c 2! c 2! 0 2!2 +! c 2 X 1 X 2 = 0 trivial solution: X 1 = X 2 = 0, static equilibrium non trivial solution X 1 0, X 2 0 Two natural frequencies: det(m) = 0! g =! 0! u =! !2 c november 2016 january / 22

39 View of the solutions Coupled oscillators! g =! 0! u =! !2 c november 2016 january / 22

40 Coupled oscillators Complete solution The complete solution of the coupled oscillators problem is x 1 (t) = A g cos(! g t + ' g ) + A u cos(! u t + ' u ) x 2 (t) = A g cos(! g t + ' g ) A u cos(! u t + ' u ) The 4 constants A g, A u, ' g and ' u are to be determined by 4 initial conditions: x 1 (t = 0) and x 2 (t = 0) ẋ 1 (t = 0) and ẋ 2 (t = 0) november 2016 january / 22

41 N coupled oscillators Coupled oscillators For a set of N coupled oscillators, we can write a set of N equations with harmonic solutions The problem writes ẍ i =! 2 0x i N j i! 2 c(x i x j ) x i = X i exp(i!t) M X = 0 with M a N N matrix and X = (X 1,...,X N ) an amplitude vector. november 2016 january / 22

42 N coupled oscillators Coupled oscillators trivial solution: X = 0, static equilibrium non trivial solution X i 0 det(m) = 0 which leads to N natural frequencies! i,...,! N and the complete solution is a linear combination of these N individual solutions : x i = A i cos(! j t + ' i ) j the 2N constants A i and ' i are to be determined with 2N initial conditions. november 2016 january / 22

43 Partial conclusion Coupled oscillators N = 2easy N = 3 less easy but possible N 4 computer help is needed november 2016 january / 22

44 1D infinite chain of oscillators 1D infinite chain of oscillators november 2016 january / 22

45 N 1D infinite chain of oscillators For a very large number of oscillators, the previous method is too expensive! N equations or m d 2 A n dt 2 = k(a n A n+1 ) k(a n A n 1 ) d 2 A n dt 2 =! 0(A 2 n A n+1 )! 0(A 2 n A n 1 ) =! 0(A 2 n+1 2A n + A n 1 ) november 2016 january / 22

46 1D infinite chain of oscillators Continuous description A n (t) = A(z,t) november 2016 january / 22

47 1D infinite chain of oscillators Continuous description Two neighboring oscillator have a very close behavior A n+1 = A(z + z) = A(z) + ( A A n 1 = A(z z) = + ( A summing these 2 equations gives A n+1 + A n 1 = 2A(z) + ( z) 2 then A n+1 2A n + A n 1 = ( z) 2 november 2016 january / 22

48 Motion equation 1D infinite chain of oscillators Back to the motion equation with the continuous approach d 2 A n dt 2 =! 2 0(A n+1 2A n + A n 1 ) d 2 A n dt 2 and the motion equation is now A n+1 2A n + A n 1 = ( z) A c 2 = 0, c =! 0 z november 2016 january / 22

49 The simple wave equation The simple wave equation november 2016 january / 22

50 Wave equation The simple wave equation It is a 1D wave A c 2 = 0 with a constant velocity c. If a 3D propagation is needed, the wave equation writes with the laplacian operator A 1 c november 2016 january / 22

51 The simple wave equation The wave equation general solution The wave eq. can be written as (remember a 2 b 2 = (a b)(a + A = 0 The solution of the wave eq. is the sum of two functions: F and G are arbitrary functions. A(z,t) = F (z ct) + G(z + ct) Proof on the board... november 2016 january / 22

52 Example The simple wave equation november 2016 january / 22

53 Exercise The simple wave equation Let F a function such that F = 1if 1 < z < 1 F = 0elsewhere With c = 3, draw F for t=0 t=1 t=2 november 2016 january / 22

54 The simple wave equation The complete wave solution The complete solution needs: the initial condition A(z,0) the boundary conditions but since the WE is linear, its solutions can be written as a sum of elementary solutions november 2016 january / 22

55 Waves & vibrations Maxime Nicolas Département génie civil november 2016 january 2017 november 2016 january / 24

56 Lecture 3 outline 1 Compression waves 2 Vibration of a tensioned string Static equilibrium of a string Vibration of a string november 2016 january / 24

57 Compression waves Compression waves november 2016 january / 24

58 Linear elasticity Compression waves Hooke s law: L L = 1 E F S = E november 2016 january / 24

59 1D compression waves Compression waves For u(x) the displacement of a section located at x: hence: u(x) u(x + dx) dx F (x) = F SE november 2016 january / 24

60 Compression waves 1D compression waves Newton s equation for the elementary mass dm u = F (x) F (x + u x which writes as a wave u E c 2 = 0, c = november 2016 january / 24

61 Compression waves Compression waves velocities Examples and order of magnitude of compression velocity: november 2016 january / 24

62 Vibration of a tensioned string Vibration of a tensioned string november 2016 january / 24

63 Vibration of a tensioned string Static equilibrium of a string Static equilibrium of a string november 2016 january / 24

64 Modeling Vibration of a tensioned string Static equilibrium of a string 1D system: L x L y, L x L z, L y L z without rigidity (for the moment, see lecture # 5) tension force: F = T 0 = L y L z november 2016 january / 24

65 Vibration of a tensioned string Shape of a string at equilibrium Static equilibrium of a string dm g + T (x) + T (x + dx) = 0 november 2016 january / 24

66 Vibration of a tensioned string Static equilibrium of a string Shape of a string at equilibrium Projection on the x- andy-axis T (x) cos (x) + T (x + dx) cos (x + dx) = 0 dm g T (x) sin (x) + T (x + dx) sin (x + dx) = 0 november 2016 january / 24

67 Vibration of a tensioned string Shape of a string at equilibrium The x-axis equation means Combining with the y-axis eq. leads or Writing gives Static equilibrium of a string T (x) cos (x) = T 0 = constant dm g + T 0 [tan (x + dx) tan (x)] = 0 mg + d mg tan = + d d tan = 0 LT 0 dx LT 0 dx d d 2 y dx 2 = mg tan = dy dx LT 0 2 dy 1 + dx november 2016 january / 24

68 Vibration of a tensioned string Shape of a string at equilibrium Static equilibrium of a string Using a variable u = dydx, one can finally find y(x) = L c cosh x cosh d, L c = LT 0 L c 2L c mg the shape of the string of length L, massm, attachedbetweentwofixed points with a d distance. Maximum bending at the center of the string: y m = L c cosh d 2L c 1 november 2016 january / 24

69 Vibration of a tensioned string Tension of a string at equilibrium Static equilibrium of a string The tensile force is T (x) = T 0 cos = T 0 cosh x L c and the tensile force variation along the rope is T T 0 = y m L mg T 0 This variation is thus negligible when the string is under tension (y m L) november 2016 january / 24

70 Vibration of a tensioned string Vibration of a string Vibration of a string november 2016 january / 24

71 Motion equation Vibration of a tensioned string Vibration of a string The local motion equation is y = T (x) sin (x) + T (x + dx) sin (x + dx) dm For a string under strong tension, the weight is negligible and the tension is constant: m L 2 = T 0 [sin (x + dx) sin (x)] Assuming only a weak deviation from the equilibrium (y = 0and dydx = 0) we y c 2 = 0, c = T0, L L = m L november 2016 january / 24

72 Boundary conditions Vibration of a tensioned string Vibration of a string The string is attached in two points: y(x = 0) = 0 y(x = L) = 0 Meaning that the wave propagation is confined between 0 and L. This gives a steady wave y(x,t) = A(x) cos(!t + ') introducing an amplitude function A(x). november 2016 january / 24

73 Amplitude equation Vibration of a tensioned string Vibration of a string Its general solution is d 2 A dx 2 +!2 c 2 A = 0 A = A 1 cos(kx) + A 2 sin(kx) and the constants A 1 and A 2 are to be determined with BC A 1 = 0, A 2 0 and sin(kl) = 0 november 2016 january / 24

74 Vibration modes Vibration of a tensioned string Vibration of a string The vibration is thus possible for a discrete set of k and! values: k = n L, and! = n L c and finally the vibration may be represented as y(x,t) = n with an integer n = 1,2,3... C n sin n x cos nc L L t + ' n The C n and ' n constants are to be determined by the initial conditions (IC). november 2016 january / 24

75 Vibration modes Vibration of a tensioned string Vibration of a string november 2016 january / 24

76 Afewdefinitions Vibration of a tensioned string Vibration of a string A nodal point (node): antinodal points (antinodes): A = 0, 0 < x < L, t da dx = 0, 0 < x < L, t november 2016 january / 24

77 Afewproperties Vibration of a tensioned string Vibration of a string The mode with the lowest frequency is the fundamental mode mode n has n antinodes and n 1nodes for a 1D system, nodes are points (0D objects) for a ND system, nodes are objects of (N-1)D dimension november 2016 january / 24

78 Movies Vibration of a tensioned string Vibration of a string View movies: mode 1 mode 3 modes november 2016 january / 24

79 Waves & vibrations Maxime Nicolas Département génie civil november 2016 january 2017 november 2016 january / 26

80 Lecture 4 outline 1 Static of a beam 2 Vibration of a beam november 2016 january / 26

81 Static of a beam Static of a beam november 2016 january / 26

82 Beam characteristics Static of a beam geometry: L x > L y L z cross-section S quadratic moment of inertia I material: E, november 2016 january / 26

83 Beam setup Static of a beam november 2016 january / 26

84 Force balance Static of a beam F F 2 + F 3 2 = 0 T + Sdx g + (T + dt ) = 0 dt dx = S g (1) november 2016 january / 26

85 Torque balance Static of a beam C + (C + dc) + T dx 2 + (T + dt )dx 2 = 0 using (1) dc dx = T d 2 C = S g (2) dx2 november 2016 january / 26

86 Elasticity of the beam Static of a beam dx dx dx y M M R november 2016 january / 26

87 Elasticity of the beam Static of a beam M dx R dx dx for the stretched part: y M dx dx = 1 df E ds dx y = dx R dx dx = y R df = y R EdS november 2016 january / 26

88 Torque Static of a beam C = S ydf= E R S y 2 ds C = E R S y 2 ds = E R I (3) The curvature radius is defined as 1 R = d 2y dx dy dx 2 32 d 2 y dx 2 (4) november 2016 january / 26

89 Static of a beam Quadratic moment of inertia Calculate I for a rectangular cross-section: I = WH3 12 november 2016 january / 26

90 Static of a beam Quadratic moment of inertia november 2016 january / 26

91 Static shape of a beam Static of a beam Combining equations (1)-(4) gives d 4 y dx 4 = S g IE Easily integrated as a polynomial function with 4 integration constants. = a 4BCneeded! november 2016 january / 26

92 Static shape of a beam Static of a beam BC: y = 0and dy dx = 0atx = 0 T = 0andC = 0atx = L november 2016 january / 26

93 Static shape of a beam Static of a beam y(x) = a 2 x 2 x 2 The maximum deviation is at x = L 12 Lx 3 + L2 2 y max = al4 8 november 2016 january / 26

94 Vibration of a beam Vibration of a beam november 2016 january / 26

95 Vibration equation Vibration of a beam Out of static equilibrium, the motion equation is d 2 C dx 2 = S 2 Other geometric equations remain unchanged 1 R = d 2 y dx 2, C = IE S 2 g = 0 november 2016 january / 26

96 Hypothesis Vibration of a beam We assume that so that the vibration equation is y r 2 c 2 = 0 with c = E I, r = S november 2016 january / 26

97 Gyration radius Vibration of a beam The gyration radius r is defined by or I = (2r)4 64 r = 4I 14 november 2016 january / 26

98 Amplitude equation Vibration of a beam We seek solution written as leading to an amplitude equation The amplitude A(x) is y(x,t) = A(x)e i!t d 4 A dx 4!2 c 2 r 2 A = 0 A(x) = B 1 cosh( x) + B 2 sinh( x) + B 3 cos( x) + B 4 sin( x) with the wave number. =!cr november 2016 january / 26

99 BC Vibration of a beam The B i are determined through the BC : A = 0atx = 0 dadx = 0atx = 0 d 2 Adx 2 = 0atx = L d 3 Adx 3 = 0atx = L november 2016 january / 26

100 BC Vibration of a beam B 1 = B 3 B 2 = B 4 cosh( L) + cos( L) sinh( L) + sin( L) sinh( L) sin( L) cosh( L) + cos( L) B 1 = 0 B 2 The equation for is cosh( L) cos( L) + 1 = 0 or 1 cos( L) = cosh( L) november 2016 january / 26

101 Graphical solution Vibration of a beam november 2016 january / 26

102 Vibration of a beam Vibration modes for the beam Fundamental mode: Other (higher) modes: L,! 0 = IE 4L 2 S n1 = (2n + 1) 2L,! n1 (2n + 1) 2 2 4L 2 IE S november 2016 january / 26

103 Back to the hypothesis Vibration of a 2 =!A2 y m! 2 y = g y m! 2 = 128 (2n + 1) 4 2 n gy m! november 2016 january / 26

104 Concluding remarks Vibration of a beam The shape of the beam in the fundamental mode (n = 0) is very close to the shape of the static beam under gravity. The shape of the beam of mode n 1hasn nodes november 2016 january / 26

105 Waves & vibrations Maxime Nicolas Département génie civil november 2016 january 2017 november 2016 january / 31

106 Lecture 5 outline 1 Equation of vibration of membranes 2 Solving method 3 Vibration of rectangular membranes 4 Vibration of a square membrane 5 Circular membranes 6 Vibration of a plate november 2016 january / 31

107 Equation of vibration of membranes Equation of vibration of membranes november 2016 january / 31

108 Membrane setup Equation of vibration of membranes 2D system: L x L y, L x L z, L y L z without rigidity november 2016 january / 31

109 Equation of vibration of membranes Tension of a membrane F = L y L z = T 1 L y T 1 is a tension per unit length (in N m 1 ) november 2016 january / 31

110 Equation of vibration of membranes Evidence of the internal tension of a membrane The force needed to close the fracture is F = T 1 L fracture november 2016 january / 31

111 Equation of vibration of membranes Motion equation Hypothesis: gravity force is negligible compared to tension force tension is uniform and isotropic We derive the motion equation from the dynamic balance of a small element of the membrane: along the z-axis: dm d 2 r dt 2 = C T 1 dc dm d 2 z dt 2 = C T 1 dcz = T 1 cos dc = T 1 dz dn dc november 2016 january / 31

112 Equation of vibration of membranes Motion equation Using the Green s theorem: dz dn dc 2 zds= we c 2 2 = 0 which is a 2D wave equation with a velocity c = T1 S with S = MS the surface density of the membrane. november 2016 january / 31

113 Equation of vibration of membranes About wave velocities compression waves (see lecture #3): transversal waves of a string: c comp = E c string = transversal waves of a membrane: c membrane = T0 L T1 S november 2016 january / 31

114 Equation of vibration of membranes About wave velocities compression waves (see lecture #3): E c comp = transversal waves of a string (see lecture #3): with T 0 = L y L z and L = ML x = L y L z c string = transversal waves of a membrane: with T 1 = L z and S = M(L x L y ) = L z c membrane = november 2016 january / 31

115 Solving method Solving method november 2016 january / 31

116 What we want to know Solving method From the vibration c 2 2 = 0 we want to know: the natural frequencies of the membrane the shape of the deformed membrane under vibration nodes and anti-nodes november 2016 january / 31

117 Method of the solution Solving method Wee look for the z(x,y,t) solution c 2 2 = 0 Separating space and time variables gives z(x,y,t) = A(x,y) cos(!t) and the motion equation writes as an 2 A(x,y) 2 c 2 A(x,y) = 0 november 2016 january / 31

118 Method of the solution Solving method This amplitude equation ( 2A + k 2 A = 0) can be (a priori) solved explicitly with the knowledge of the boundary conditions (BC) the initial conditions (IC) There is no analytical solution of the membrane vibration for an arbitrary shape! We know solutions for simple shapes: a rectangular membrane (or square) acircularmembrane november 2016 january / 31

119 Vibration of rectangular membranes Vibration of rectangular membranes november 2016 january / 31

120 Arectangularframe Vibration of rectangular membranes amplitude solution (extension from string amplitudes solution): A(x,y) = A mn sin m x L x sin n y L y november 2016 january / 31

121 Vibration of rectangular membranes Vibration freq. of a rectangular membrane Combining 2 A(x,y) 2 c 2 A(x,y) = 0 A(x,y) = A mn sin m x L x f mn =! mn 2 = c 2 m L x sin n y L y 2 + n 2 L y The fundamental frequency is for m = 1andn = 1andis f 11 = c 1 2 L x L y november 2016 january / 31

122 Vibration of rectangular membranes Viewing a few modes november 2016 january / 31

123 Complete solution Vibration of rectangular membranes The complete solution of the linear vibration equation for a rectangular membrane is with z(x,y,t) = A mn sin m x sin n y cos(! mn t + ' mn ) (m,n) L x L y! mn = c m L x and A mn, ' mn determined through IC. 2 + n 2, c = L y november 2016 january / 31

124 Vibration of rectangular membranes Nodes and anti-nodes A nodal line (or curve) is the set of points where A(x,y) = 0, 0 < x < L x, 0 < y < L y Amode(m,n) has m + n 2nodallines anti-nodes are points where the amplitude is extremal. november 2016 january / 31

125 Vibration of a square membrane Vibration of a square membrane november 2016 january / 31

126 Square membrane Vibration of a square membrane Taking L x = L y describes a square membrane, with natural frequencies f mn =! mn 2 = c m 2 + n 2 2L x important remark: for example f 21 = f 12. f mn = f nm DEF: degenerated frequency: when two (or more) modes have the same frequency. november 2016 january / 31

127 Square membrane Vibration of a square membrane november 2016 january / 31

128 Circular membranes Circular membranes november 2016 january / 31

129 Circular membranes Circular membranes The amplitude equation written in polar (r, ) coordinates is with BC Separating r and, theamplitudeis 2A + k 2 A = A r 2 +!2 c 2 = 0 A(r = R, ) = 0 A(r, ) = A n (r) cos(n + ' n ) n november 2016 january / 31

130 Circular membranes Circular membranes Step to solution: variable change x =!r c the amplitude equation writes now as a Bessel equation d 2 A n dx x + 1 n2 x 2 A n = 0 november 2016 january / 31

131 Circular membranes Bessel equations and Bessel functions with A n (x) = n J n (x) + n Y n (x) J n (x) = x n 2 ( x 2 4) m m=0 m! (n + m + 1) Y n (x) = J n(x) cos(n ) J n (x) sin(n ) (k) = 0 e t t k 1 dt november 2016 january / 31

132 Bessel functions Circular membranes for n = 0, the J 0 and Y 0 Bessel functions are november 2016 january / 31

133 Circular membranes Circular vibration modes z(r,,t) = cos(!t) n J n (kr)cos(n + ' n ) n mode (01) mode (13) november 2016 january / 31

134 Vibration of a plate Vibration of a plate november 2016 january / 31

135 Vibration of a plate Vibration of a rigid membrane Combining membrane + beam gives Eh 2 12 (1 2 ) 2 z 2 = 0. where h is the thickness of the plate november 2016 january / 31