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4 Oct 2023: S(t) = cos 2ωtS (1 cos 2ωt) B̂ (B̂ S) sin 2ωt B̂S. ... H = γB(t) S where B(t) = B0ẑ b (x̂ cos ω1t ŷ sin ω1t).

1 Nov 2023: 5. Diagonalise the quadratic form in R2 defined by. F(x, y) = (a cos2 θ b sin2 θ) x2 2(a b)(sin θ cos θ) xy (a sin2 θ b

16 Jan 2024: Find the time-averaged energy flux vector 〈 I 〉 for (i) a plane harmonic S-wave with u =. B cos(k x ωt), (ii) a plane harmonic P-wave with u = Ak̂ ... An SV-wave with displacement. u = B(cosθ,0, sinθ)eik(x sin θz cos θ)iωt.

23 Oct 2023:   cos φ sin φ 0sin φ cos φ 00 0 1. ... ψ̇ φ̇ cos θ. . (b) [optional extra] Calculate the matrix B = R>Ṙ and verify that it is antisymmetric.

6 Oct 2023: I0(x) =1. π. π. 0. ex cos θ dθ. Oscillatory Integrals and Stationary Phase. ... a). 1. 0. cos(. x tp). dt , with p > 1 , real ,. (b). π. 2. 0. (. 1 (. 2θ. π. ))γ. cos(.

2 Feb 2024: Let there be a magnetic field B, sothat each molecule has a potential energy mB cos θ (θ being the angle between theaxis of the molecule and the magnetic field). ... N where. zrot =. [. 2I. 2mBβ2. ]. sinh(mβB). Evaluate the total magnetisation, M =

15 Jan 2024: What. motivated your choice? [Suggestions: You could try ψ = cos(πx/2α) or ψ = (α2 x2) for |x| < α and vanishing. ... b. A defect is introduced into the crystal. As a result, the amplitude for jumping.

1 Nov 2023: 10. The linear map R3 R3 defined by. x 7 x′ = cos θ x (x n) (1 cos θ) n sin θ (x n) (). is a rotation by angle θ in ... Hence show that. Rii = 2 cos θ 1 , εijkRjk = 2ni sin θ.

23 Oct 2023: q1 = Q1 cos λ P2 sin λ, q2 = Q2 cos λ P1 sin λ,. ... p1 = Q2 sin λ P1 cos λ, p2 = Q1 sin λ P2 cos λ.

8 Mar 2024: dt=. 1. Ra, where R2 = k1. Derive the. equationsa(η) = B(1 cos η) , t(η) = BR(η sin η) ,. where B is a constant, and hence show that the universe ... ds2 = sin θ (dφ2 dθ2) 2 cos θ dθ dφ ,. where 0 θ 2π and 0 φ 2π. Show that, for a null

18 Oct 2012: y =. . . . . . 0 , x < 0 ,. 1 cos x , 0 < x < ǫ ,. cos(x ǫ) cos x , x > ǫ. ... 1 cos x. x. )2. dx = π. (b) Use Parseval’s theorem and the result of question 7b to evaluate the integral.

26 Jan 2024: Show that the resulting state is. |ψαβ⟩ = 12 (cos(α β) |00⟩ sin(α β) |01⟩ sin(α β) |10⟩ cos(α β) |11⟩). ... B) to be independent of the choiceof angle chosen and used remotely at B (resp.

31 Oct 2012: x = ax′ , y = by′ , z = cz′ ,. x′ = r′ sin θ′ cos φ′ , y′ = r′ sin θ′ sin φ′ , z′ = r′ cos θ′ ,. where a, b and c are positive constants. Calculate ... If an extra transverse force F sin(πx/L) cos(πct/L) per unit length

25 Jan 2007: es = cos φ i sin φ j, eφ = sin φ i cos φ j, ez = k. ... For (iii), from a decent diagram of spherical polars, find k = er cos θ eθ sin θ. ].

16 Oct 2021: Γ. θ. r. θ. =. Γ. r. θ. θ. =. 1. r. Γ. ϕ. ϕ. θ. =. -. sin. θ. cos. ... θ. ϕ. =. cos. θ. sin. θ. (1.11). There are some important lessons here.

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)].

27 Mar 2022: We write. 𝐞. a. ′′. =. R. 1. (. θ. ). a. b. 𝐞. b. ′. with. R. 1. (. θ. ). =. (. 1. 0. 0. 0. cos. θ. sin. θ. 0. -. ... V. eff. (. θ. ). =. I. 1. (. b. -. a. cos. θ. ). 2. 2. sin. 2. θ. M. g.

1 Nov 2023: For b > 0 and x R, we have. cos(x ib) = 12(ei(xib) ei(xib)) = 1. ... 2(ebeix ebeix). and thus Re (cos(x ib)) = cosh(b) cos(x) and Im (cos(x ib)) = sinh(b) sin(x).

9 Jan 2024: given boundary conditions (λ a constant):. (i) r < R with ψ(R,θ) = λ cos θ,. ... ii) r > R with ψ(R,θ) = λ cos θ and ψ(r,θ) 0 as r ,. (iii) a < r < b with n ψ(a,θ) = 0 and ψ(b,θ) =

10 Mar 2020: ȧ. a= ḃ. b= S sin θ cos θ, θ̇ =. S(b2 cos2 θ a2 sin2 θ)a2 b2. Combine these effects to obtain. ... ȧ. a= ḃ. b= S sin θ cos θ,. θ̇ =S(b2 cos2 θ a2 sin2 θ).

27 Feb 2020: 2 S)zByBz. Bx. Bz. K cos(Kz). sin(Kz+) 1. z+. ,. vy = v2az2. ... BxBz. K cos(Kz). sin(Kz+). The meridional (x and z) components of B are called the poloidal magnetic field, while the azimuthal(y) component is called the toroidal magnetic field.

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.

18 Oct 2021: The wavefunction (2.12) can be viewed as a sum of sine and cos functions and describes a complex-valued wave of wavelength. ... 2. x. The most general solution has the form. x. (. t. ). =. A. cos. (. ω. (. t. -. t.

11 Feb 2014: α =. C. A dx. α =. C. A dx =. B dS. Aϕ =g. 4πr. 1 cos θsin θ. B =g. ... 1 cos θsin θ. B =g. 4πr2̂r. References. [1] Gibbons and Rychenkova, 9608085.

9 Oct 2023: F8. (a) i. (b) 1. (c) 12(1 i). (d) e(cos 1 isin 1). ... F13. (a)sin 6θ sin 5θ sinθ. 2(1 cosθ). (b). cos 6θ cos 5θ cosθ 12(1 cosθ).

12 Jun 2024: x(, ) = ( cos , sin , ) (4.18). with a   b and 0  < 2. ... equation tells us that. µ0I =. Z. C. B dx =. Z 2. 0. d b() =) B(x) =µ0I. 2( sin , cos , 0).

8 Nov 2019:   cos φ sin φ 0sin φ cos φ 00 0 1. ... ψ̇ φ̇ cos θ. . (b) [optional extra] Calculate the matrix B = R>Ṙ and verify that it is antisymmetric.

5 Dec 2008: So general solution becomes. f = A(cos 2φ 1) B(sin 2φ 2φ). ... The boundary condition at φ = α gives. A(cos 2α 1) B(sin 2α 2α) = 12.

28 Nov 2008: B. B3 = B cos θ. B2 = B sin θ cos ψ. ... H = 2. 2m2 12 B σ cos θ. 12B. 2 sin2 θ 12.

12 Nov 2023: b) 2(cos θ θ sin θ)/ cos2 θ. (c) t(1 2 ln t). ... a tan1 y. a). H14. (b) J = 1b(1 eax cos bx) a.

3 Oct 2023: a)d. dx(sin x) (b). d. dθ(cos θ). (c)d. dt(tan t) (d). ... x2 4)dx (b) 20. (xa)2dx. (c) π0eiθdθ (d). π0. cos(x)dx. (e)π/4π/4.

8 Jul 2021: ψ. x. (x(1)s ,y(1)s ). = b cos(x t ), (18). ... ψ. x. (x(2)s ,y(2)s ). = b cos(x t ), (20).

14 Oct 2014: obtain:. Φ(ϕ) = A sin(λϕ) B cos(λϕ),. with positve constants A, B. ... Aλrλ Bλr. λ) sin(λϕ) (Cλrλ Dλr. λ) cos(λϕ). E0 F0 ln(r).

18 Nov 2019: q1 = Q1 cos λ P2 sin λ, q2 = Q2 cos λ P1 sin λ,. ... p1 = Q2 sin λ P1 cos λ, p2 = Q1 sin λ P2 cos λ.

30 Oct 2008: k=0. (1)k[2ω(n m)]2k2 [f. (2k1)(b) cos 2ω(n m)b. f(2k1)(a) cos 2ω(n m)a]}. ... m. ). {. k=0. (1)k[ω(2n 2m 1)]2k1. [f(2k)(b) cos ω(2n 2m 1)b f(2k)(a) cos ω(2n 2m 1)a].

23 Jan 2004: B(t) = v(t). [. ω1 sin ωt cos ωt. ]. [ cos ωt ω1 sin ωt ].(2.3). ... ω3,. b(2)2 (ω) =. 4(1 cos ω). ω2+. 8 sin ω.

23 Dec 2020: vr = rν cos φ[c(ν 1)Pν1 b(ν 1) cos θ Pν. ], ... 2.12a). vθ = rν cos φ[cP′ν1 2a csc θ Pν b.

16 Oct 2023: λk(A) =λk(C). λk(B)=. 12µsin2 12πkh1 2µsin2 12πkh. |λk(A)| 1, k = 1.M. ... iγ where γ = 2 cos(πh) for = 1,.

8 Oct 2015: x = A cos ωt B sin ωt a cos(ωt ϵ) Peiωt Qeiωt, (2). ... The solution isx = A cos ωt B sin ωt. Initially, x = a and ẋ = 0, so.

13 Mar 2019: What. motivated your choice? [Suggestions: You could try ψ = cos(πx/2α) or ψ = (α2x2) for |x| < α and vanishingoutside this interval.]. ... 8. Consider the scaled Schrödinger equation. d2ψ. dx2+ λ cos 2xψ = Eψ.

19 Oct 2013: exponential counterpart, cos(µ) = 12 (e. ... Then φ̂ = φ̂(x, z) exp(kz) cos(kx),equivalent to a single pair of progressive deep-water waves travelling in the x directions.

26 Jun 2024: or,. (x) = B cos kx |x| < a (2.34). where B is again a normalisation constant and k is, like , related to the energy, now. ... automatically patched at x = a as well. We have. (x) cts at x = a ) B cos ka = Aea.

1 Dec 2023: sin α) 2π(2 cos φ, sin φ)/ log Bi1 (see (3.2)). ... Ṽs =U cos(φΦ), ω̃=URH. sin(φΦ). (4.5a,b). Dow. nloa. ded. from. htt.

24 Jul 2023: Ω sed = mg16πa2η. ln(a. b. )(κ̃x sin θ κ̃y cos θ. ... sin θ, (3.40). Usedx = mg. 8πaηln(a. b. )cos θ sin θ, (3.41).

10 Oct 2017: 1 q cos p. ). 6. Show that the following transformation is canonical, for any constant, λ :. q1 = Q1 cos λ P2 sin λ , q2 = Q2 cos λ P1 sin λ ,. p1 = ... Q2 sin λ P1 cos λ , p2 = Q1 sin λ P2 cos λ.

3 Jul 2024: We then define the usual. polar coordinates. x = r cos , y = r sin. – ... l2(4.13). with A and 0 integration constants. (You might be tempted instead to write u =. A cos B sin k/l2 with A and B as integration constants.

13 Mar 2008: 4 A circular helix is given by. x = (a cos t, a sin t, ct). ... Its mass density is ρ0 cos θ. By evaluating a volume integral find the mass ofthe cone.

19 Oct 2013: expected. (ii) all A’s, B’s zero except A1:. φ = A1r cos θ = A1z , u = φ = A1ez. ... z. (iii) all A’s, B’s zero except B1:. φ =B1 cos θ.

1 Nov 2017: 3ρ̄r. (b) Consider linear perturbations about the homogeneous solution. Show that the density. ... t = B (1 cos θ). where you should determine A, B and θ.

1 Dec 2023: the boundary condition to be imposed is. u=cos θ and τrθ =%Bi sgn(y) at r =1 (2.7a,b). ... balance. But FEAE′ =2FAE, where FAE =rEI Bi[2(πβ2) cos β2 sin β2] is the force on the section AE (see appendix B.1).