Figure 1: Surface waves

Transcription

1 4 Surface Waves on Liquids 1 4 Surface Waves on Liquids 4.1 Introduction We consider waves on the surface of liquids, e.g. waves on the sea or a lake or a river. These can be generated by the wind, by a moving boat and in many other ways. One key factor is that if the surface is displaced from its equilibrium position z = 0 to z = η(x, y, t), gravity will tend to restore the surface to its equilibrium position. z y z=0 g x z= (x,y,t) Figure 1: Surface waves In practice we can assume that the disturbance is small (i.e. the amplitude, e.g. sup η, is much less than the wavelength) linear theory.

2 4 Surface Waves on Liquids 2 A further simplification is that for most liquids the equation of continuity Eq. (3.8) can be considerably simplified because liquids are difficult to compress volume of small piece of liquid is unchanged as it moves density is unchanged (since mass = density volume and mass is conserved). t t+ t O O Figure 2: Incompressible liquids. OO = u(x, t)δt Consider a small volume of liquid of density ρ. Suppose it is at x at time t ; in a small interval of time δt, the volume will have moved from x x + u(x, t)δt, so the density will have changed from ρ(x, t) ρ(x + uδt, t + δt). By hypothesis, these are the same. But...

3 4 Surface Waves on Liquids 3 ρ(x + uδt, t + δt) = ρ(x + uδt, y + vδt, z + wδt, t + δt) Thus ρ(x, y, z, t) +(u ρ x + v ρ y + w ρ z + ρ t )δt. Dρ Dt = ρ t + u ρ x + v ρ y + w ρ z = 0 (1) where D/Dt is the operator defined by ( D Dt = t + u x + v y + w ) z (2) D/Dt applied to any function of (x, t) measures rate of change when moving with the liquid (or fluid). From Eq. (3.8) we have ( ρ t + u ρ ) ( u x + v ρ y + w ρ +ρ z x + v y + w ) z = 0,

4 4 Surface Waves on Liquids 4 so Eq. (1) u x + v y + w z = 0 (3) which is the equation of continuity for an incompressible fluid or liquid. The reasoning applied to ρ(x, t) above can also be applied to u(x, t) (i.e. velocity). The rate of change of the velocity of the piece of fluid, i.e. its acceleration, is D Dt (u, v, w) = (u, v, w) t ( + u x + v y + w ) (u, v, w). z But our assumption that the disturbance is small second term is small ( u acceleration t, v t, w ). t (This has already been used in Eqs (3.3) and (3.10).)

5 4 Surface Waves on Liquids The governing equations for 1D water waves z z=0 z=-h x z= (x,t) Figure 3: 1-dimensional free surface We consider only cases where the disturbance of the free surface is independent of y z = η(x, t) is the disturbance. We therefore assume that u = u(x, z, t)i + w(x, z, t)k (4) Then Eq. (3) u x + w z = 0. (5)

6 4 Surface Waves on Liquids 6 Recalling N2 ρ u t = p x, ρ w t = p z ρg (6) where ρ can be regarded as constant, and the term ρg represents the weight ρg δv acting vertically downwards. This is an extension of Eq. (1) - we assume ρ is an absolute constant, independent of both x and t. In the ocean, ρ does vary (slightly) with height, but not enough to affect the analysis of surface waves. As in (3.3), it can be shown that, in most circumstances, there is a velocity potential, φ such that Eq. (3.13) holds 1. In the present case φ = φ(x, z, t) and u = φ x, w = φ z (7) 1 To show this is beyond the scope of this course. In brief we require the effects of viscosity to be small. See AMA 305.

7 4 Surface Waves on Liquids 7 Then Eq. (5) becomes 2 φ x + 2 φ 2 z = 0 (8) 2 This is the 2D form of Laplace s equation and is the PDE that must be solved. NB Surface waves are not governed by the wave equation! The special features of surface waves arise because of the boundary conditions. There will be three in the problems we consider: 1. w = 0 on z = h, where h is constant (see Fig 3). 2. the vertical velocity given by φ z at the free surface must equal the vertical velocity given by z = η(x, t). 3. the pressure at the free surface must be continuous and since the density of air is much less than that of water, we can assume the air pressure is constant p 0. (1): φ z = 0, z = h (9)

8 4 Surface Waves on Liquids 8 (2): D {z η(x, t)} = 0 at z = η Dt w η t u η }{{} x small = 0 at z = η. Since we are linearising, this condition can be applied at z = 0 η t = φ z at z = 0. (10) (3) : Eq. (6) { p p0 x ρ } + φ = z { p p0 ρ } + φ + gz = 0 p p 0 + φ + gz depends only on t. ρ t However we can incorporate this function of t by adding it to φ.

9 4 Surface Waves on Liquids 9 This has no effect on u by Eq. (7). Since p = p 0 at z = η, and we are linearising, we can apply this condition at z = 0 as far as φ is concerned. Thus from (3) φ t + gη = 0 at z = 0. (11) 4.3 Monochromatic surface waves Monochromatic single wave number k, single (angular) frequency ω. We assume the free surface is given by η = η 0 sin(kx ωt) = η 0 sin k(x ct) ω = kc (12) Eq. (12b) has already been used several times, e.g. Eq. (2.28). We could also work with the complex form Eq. (1.27), viz η = η0e i(kx ωt). In order to satisfy Eqs. (10) and (11) we must have φ = f(z) cos(kx ωt) (13) where Eq. (8) f = k 2 f. (14)

10 4 Surface Waves on Liquids 10 In view of the BC (9), it is convenient to write the GS of Eq. (14) in the form f = A cosh k(z + h) + B sinh k(z + h), when from Eq. (9) B = 0, so φ = A cosh k(z + h) cos(kx ωt) (15) [or : GS of Eq. (14) is, using exp functions, Eq. (9) f = γe kz + δe kz. Thus kγe kh kδe kh = 0 δ = γe 2kh. f = γe kz + γe 2kh e kz = γe kh e k(z+h) + γe kh e k(z+h) = 2γe kh cosh k(z + h) = A cosh k(z + h) with A = 2γe kh.]

11 4 Surface Waves on Liquids 11 There remain Eqs. (10) and (11). From Eq. (10) ωη 0 cos(kx ωt) = ka sinh kh cos(kx ωt) ωη 0 = ka sinh kh (A) From Eq. (11) ωa cosh kh sin(kx ωt) + gη 0 sin(kx ωt) = 0 gη 0 = ωa cosh kh (B) Then (A)/(B) ω g = k ω tanh kh

12 4 Surface Waves on Liquids 12 and ω 2 = gk tanh kh, c 2 = g k tanh kh (16) φ = gη 0 ω cosh k(z + h) cosh kh cos(kx ωt). (17) Thus waves with different wavelengths travel at different speeds c, where c = ω k (18) is the phase velocity (speed). This phenomenon is known as dispersion. We note two special cases: Deep water h, ω 2 = gk, c 2 = g k (19a) Shallow water kh 1, ω 2 gk 2 h c 2 = gh (19b) NB: Shallow water waves are not dispersive. This is a progressive wave, but standing waves can be dealt with similarly - see S4 Q3.

13 4 Surface Waves on Liquids Energy The PE relative to the the undisturbed position is ( η 0 ) ρgzdz δa = 1 2 ρgη2 δa. Thus the PE in a wavelength per unit width in the direction of 0y is V, where V = 1 2 ρgη2 o 2π/k 0 sin 2 k(x ct)dx using Eq. (12). This is 1 2 ρgη2 0 π k = 1 4 ρgη2 0λ, where λ = 2π k is the wavelength. Thus the potential energy density per unit area of water surface is V ρ = V /λ. V ρ = 1 4 ρgη2 0 (20a)

14 4 Surface Waves on Liquids 14 Likewise the KE in a wavelength per unit width in the direction of 0y is T, where T = 1 2 ρ 0 h dz 2π/k 0 [ ( φ ) 2 + x ( ) ] 2 φ dx z ( ) 2 φ = g2 η0k 2 2 x ω 2 cosh 2 kh cosh2 k(z + h) sin 2 k(x ct) ( ) 2 φ = g2 η0k 2 2 z ω 2 cosh 2 kh sinh2 k(z + h) cos 2 k(x ct) Since (as with V ) 2π/k 0 we find sin 2 k(x ct)dx = 2π/k 0 cos 2 k(x ct)dx = π k,

15 4 Surface Waves on Liquids 15 T = π ρg 2 η0k 2 2 ω 2 cosh 2 kh 0 h cosh 2k(z + h)dz, (since cosh 2 θ + sinh 2 θ = cosh 2θ). Thus T = π 4 ρg 2 η0 2 ω 2 cosh 2 kh sinh 2kh = π ρg 2 η0 2 2 ω 2 tanh kh (since sinh 2θ = 2 sinh θ cosh θ). from Eq. (16), T = 1 π 2 ρgη2 0 k = 1 4 ρgη2 0λ = V. the kinetic energy density per unit area of the water surface is T ρ = T /λ T ρ = 1 4 ρgη2 0. (20b)

16 4 Surface Waves on Liquids 16 z A 0 x P z A Figure 4: 1-dimensional distorted free surface We now calculate the rate at which work is being done by the fluid on the left of AA on the fluid on the right. The force per unit width in the direction of 0y is p δz so its rate of working P (for power) per unit width is given by P = 0 h pu dz = 0 h ( p 0 ρ φ ) φ t gz x dz, since p = p 0 ρ φ t gz from derivation of Eq. (11). It is sufficient for our purposes to calculate the mean of P over one period. Since the mean of sin k(x ct) is 0, and the mean of sin 2 k(x ct) = 1 2, we let P be the mean of P and find:

17 4 Surface Waves on Liquids 17 P = ρg2 η 2 0k 2ω cosh 2 kh 0 h cosh 2 k(z + h)dz after some algebra (exercise for student). Since c = ω/k and cosh 2 θ = 1 2 (1 + cosh 2θ), we find P = ρg 2 η0 2 4c cosh 2 kh = ρg2 η 2 0 8kc [ h + [ 1 + 2kh sinh 2kh ] sinh 2kh 2k ] 2 tanh kh since sinh 2θ = 2 sinh θ cosh θ. Thus, using Eq. (16), P = 1 [ 4 ρgη2 0c 1 + 2kh ]. (21) sinh 2kh There is an interesting and important consequence of Eqs. (20a), (20b) and (21) which can be extended to many sorts of waves leading to the concept of group velocity.

18 4 Surface Waves on Liquids 18 As a consequence of the passage of the waves, energy is being transmitted from left to right with a (mean) speed U to be determined. z V + T U Figure 5: Concept of group velocity In a time τ, this results in new energy per unit width equal to (V ρ + T ρ )Uτ, and this must be equal to P τ, the work done. U = P/(V ρ + T ρ ) = P/( 1 2 ρgη2 0), i.e. U = c g = 1 [ 2 c 1 + 2kh ] (22) sinh 2kh where c g is known as the group velocity for reasons that will be discussed later.

19 4 Surface Waves on Liquids 19 From the first of Eq. (16), we have 2ω dω dk = g tanh kh + gkh cosh 2 kh (since d dθ (tanh θ) = sech 2 θ = 1 ). Thus cosh 2 θ dω dk g tanh kh = 2ω = kc2 2ω [ 1 + [ 1 + 2kh sinh 2kh kh tanh kh cosh 2 kh ], ] i.e. see Eqs. (18) and (22) c g = dω dk. (23) Eq. (23) is the general definition of group velocity. [Note that ω = kc so that when c is independent of k, as for waves on strings and sound waves, i.e. when the waves are non-dispersive, Eq. (23) gives c g = c (24) i.e. the group velocity c g is equal to c, the phase velocity.]

20 4 Surface Waves on Liquids 20 Finally, we record the results for the two special cases considered in Eqs. (19a)-(19b) Deep water h ω 2 = gk, c 2 = g k, c g = 1 2 g k = 1 2 c (25a) Shallow water kh 1 ω 2 gk 2 h, c 2 = gh, c g = gh = c (25b)

21 4 Surface Waves on Liquids Group velocity Energy is (often and at an average over time) transported at the group velocity c g. This applies to many sorts of wave. There are other important properties of c g. Consider the superposition of two waves like Eq. (12) in the case when the amplitudes are equal but the waves numbers and frequencies are slightly different. We have η = η 0 sin(kx ωt) + η 0 sin [(k + δk)x (ω + δω)t] = 2η 0 sin [(k + 12 δk)x (ω + 12 ] δω)t [ ] 1 δω cos δk(x 2 δk t) η 2η 0 cos [ ] 1 2 δk(x c gt) sin[kx ωt] ( c g δω δk (26) The combined displacement can be thought of as the original wave but with an amplitude that gradually changes between ±2η 0 over a distance π/( 1 2δk) = 2π/(δk). )

22 4 Surface Waves on Liquids 22 The surface will be a series of groups of waves, separated by essentially smooth water where cos[ 1 2 δk(x c gt)] 0. The groups are travelling at speed c g, whereas the individual waves within each group are travelling at speed c. See top sketch in handout. NB In passing, suppose η is density or velocity potential in sound waves, where c g = c. Then Eq. (26) becomes η 2η 0 cos [ ] 1 (δkx δωt) sin [kx ωt], 2 so that the wave has a fluctuating intensity known as beats ; the beat frequency is δω. This phenomenon can be used to determine unknown frequencies, by determining the beat frequency between a standard tuning fork and the unknown.

23 4 Surface Waves on Liquids 23 We can develop the above analysis to consider a wave packet. As noted in (2.5i), we can generalise to consider the disturbance η(x, t), where η(x, t) = A(k)e i(kx ωt) dk, and the real part of this is eventually to be taken. Here we shall consider the special case when A(k) = Ae d2 (k k 0 ) 2, where A, d, k 0 are constants. This gives the Gaussian wave packet η(x, t) = A e d2 (k k 0 ) 2 e i(kx ωt) dk, (27) The dominant contribution comes from values of k near k 0 because of the nature of e d2 (k k 0 ) 2. We write ω = ω(k 0 )+ dω dk (k k 0)+... = ω 0 +c g (k k 0 )+...

24 4 Surface Waves on Liquids 24 and neglect terms of higher order to obtain η(x, t) = Ae i(k 0x ω 0 t) e d2 (k k 0 ) 2 +i(k k 0 )(x c g t) dk Substitute d(k k 0 ) = ξ to obtain (after some algebra): η(x, t) = Ae i(k 0x ω 0 t e {ξ 2d i (x c gt)} 2 e (x c gt) 2 dξ 4d 2 d Now substitute ζ = ξ i 2d (x c gt) to get η(x, t) = A d e (x c gt) 2 /(4d 2) e i(k 0x ω 0 t) e ζ2 dζ Now e ζ2 dζ = π (28) Proof of Eq. (28) is via a clever trick!

25 4 Surface Waves on Liquids 25 Proof: y x { r r Figure 6: Infinitesimal element in polar coordinate system I = e x2 = e y2 I 2 = = 2π 0 = 2π dθ 0 0 e x2 e y2 dxdy e r2 rdr re r2 = 2π [ 1 2 e r2 ] 0 = π I = π

26 4 Surface Waves on Liquids 26 Thus η(x, t) = A π (x cgt) 2 d e 4d 2 e i(k 0x ω 0 t) (29) c = g d dk 0 c 0 = k 0 Figure 7: Wavepacket

27 4 Surface Waves on Liquids The Doppler effect It is convenient here 2 to consider another general phenomenon connected with waves, namely the changes in frequency of waves sent out by a moving source and perceived by a stationary observer. Consider sound waves for sound waves for definiteness. We shall work in terms of the actual frequency ν and the wavelength λ, where ν = ω 2π, λ = 2π, c = νλ (30) k As seen in the sketch, in a time t the source emits νt waves. For a stationary source these occupy a length νtλ, whereas, for a source moving with speed u towards the observer, the wavelength changes to λ and the νt waves occupy a distance νtλ. Thus νtλ = νtλ + ut λ = λ u ν See bottom sketch on Handout. λ = λ(1 u c ) (31) 2 But not logical since it is more relevant to sound waves and radio waves than to surface waves than to surface waves on water!

28 4 Surface Waves on Liquids 28 As a result the observer measures the frequency of the waves as ν where ν λ = c = νλ. Thus ν = νc c u (32) Example An observer at rest notices that the frequency of the sound waves from a car appears to drop from 281 Hz to 257 Hz as the car passes. Given that the speed of sound is 330 ms 1, estimate the speed of the car. From Eq. (32) 281 = ν 1 u, 257 = ν c 1 + u c = 1 + u c 1 u c u c = u 14.7 m s 1 (About 33 mph)

29 4 Surface Waves on Liquids Particle paths in surface waves Consider a particle whose equilibrium position is (x 0, z 0 ). Suppose its position at time t is (x 0 + X(t), z 0 + Z(t)), where the time means of X and Z will be chosen to be zero. Then dx dt = φ x (x0 +X,z 0 +Z) φ x (x0,z 0 ) = kgη 0 cosh k(z 0 + h) ω cosh kh sin(kx 0 ωt) dz dt = φ z (x0 +X,z 0 +Z) φ z (x0,z 0 ) (33) = kgη 0 sinh k(z 0 + h) ω cosh kh cos(kx 0 ωt) using Eq. (17). Thus, integrating and ensuring zero time means:

30 4 Surface Waves on Liquids 30 X = kgη 0 cosh k(z 0 + h) cos(kx ω 2 0 ωt) cosh kh = η 0 cosh k(z 0 + h) sinh kh cos(kx 0 ωt) Z = kgη 0 sinh k(z 0 + h) sin(kx ω 2 0 ωt) cosh kh sinh k(z 0 + h) = η 0 sin(kx 0 ωt) sinh kh (34) using the first of Eq. (16). It follows on eliminating cos(kx 0 ωt) and sin(kx 0 ωt) that X 2 a + Z2 2 b = 1 where 2 a = η 0 cosh k(z 0 +h) sinh kh, b = η 0 sinh k(z 0 +h) sinh kh (35) Thus the particle paths are ellipses. As z 0 h, b 0, a η 0 / sinh kh rectilinear motion in direction of 0x.

31 [ [ GROUP VELOCITY 1 cos [ 0 2 k(x-c t) g sin[kx- t] 1 cos 0 [ 2 k(x-c t) g x Figure 1: Sketch for Eq. (4.26). DOPPLER EFFECT t A B (a) source ut t observer A B (b) Figure 2: (a) Waves when source is stationary; (b) Waves when source is moving. Sketch for Eq. (4.31). R. von Fáy-Siebenbürgen November