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19 Jan 2013: Find the time-averaged energy flux vector 〈I〉 for (i) a plane harmonic S-wave with u =B cos(kx ωt), (ii) a plane harmonic P-wave with u = Ak̂ ... An SV-wave with displacement. u = B(cos θ, 0,sin θ)eik(x sin θz cos θ)iωt.

26 Jul 2013: cosh cos ,. it is found that. θ. (a) (b) (c). ... cos cos cosh cos. 4 sin2. cosh cos 2 sin sin.

26 Jul 2013: b) Two dimers. We now consider the case of two dimersa and b. ... 482 2d3, 3a. V = ̄ab2 b1b1b2. ̄ba2 a1. a1a2 f a f b.

26 Jul 2013: evaluated at =. In Fourier notation and given that t= cos t, they become. ... 0. = 0. 2. Pressure and stress field. We have. ̃20 =. 2De1 De21 iDe1. B. 2 cos 2 sin 22, 43. as well as zero pressure p2 = 0, so that the average

26 Jul 2013: With this in mind we let g2 = g1(z φ) with g1(z) = cos(z), andwe set equal for both sheets the bending stiffness B and the forcing amplitude A. ... B(576 B2. ) C, (3.7a)η2(z) = A. 12B [cos z cos(z φ)] [288 sin z (288 B2) sin(z φ)]B(576 B2).

4 Oct 2013: cos A cos B =1. 2[cos(AB) cos(A B)]. sin A sin B =1. ... 2[cos(AB) cos(A B)]. sin A cos B =1. 2[sin(A B) sin(AB)].

26 Jul 2013: 1(a). Our new result is illustrated in Fig. 1(c), where we showthe dynamics of swimmers undergoing reciprocal motionwith velocity Uswim UðtÞe and UðtÞ U cos!t withU 5 ... a) No swimming, (b) steady swimming at speed U 5 m=s, (c) reciprocalswimming at

4 Oct 2013: TA sin θATA cos θA. µgδxT. (3). Now. TB sin θBTB cos θB. ... T(t) = cos jmnct and sin jmnct. Putting all this together i.e.

28 Nov 2013: X = 0 is given byF′ G = c0, (cos X σ0) G sin XF = 0, (4.4a,b). ... Θ = Ae(1κ2)τ cos κX Z, W = Ae(1κ2)τ cos κX, (5.1a,b)for any amplitude A.

19 Oct 2013: be similar to that of the spherical harmonicP4(cos ), and so we adopt the estimate L 4:5. ... Then equations(1) and (2) imply T=T r2c 2i =gL and juj 2g T=2N2T , where we have takensin cos 21=2.