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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.

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).

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.

2 Jul 2024: 45 –. by a sign, since this can be compensated by the cos function. ... Insisting that the two. cos functions agree, up to sign, gives us the condition.

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.

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).

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.

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.

3 Jul 2024: This too can be. resolved as. T = T cos r̂ T sin ̂. ... ei!t sin A cos. rg. lt B sin. rg. lt.

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.

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.

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.

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).

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.

29 Aug 2014: cosh ξ cos η ,. z =c sinh ξ. cosh ξ cos η , (B.1). ... UndVndη. , (B.17). ψ̂. ξ= 3. 2sinh ξ. (cosh ξ cos η)5/2.

1 Dec 2023: 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.

2 Jan 2024: cos u (1 b sin u) þ O ln Rðh RÞ. : ... u1. hvg. x. y. DD. (a) (b). =. =. cos(θ)2. –cos(θ).

23 May 2006: Solution. h1 = cos θ A(T ) cos(θ τ ) B(T ) sin(θ τ ). n>1. [An cos n(θ τ ) Bn sin n(θ τ )] eαn2(n21)τ. ... Secularity condition – resonant sin(θ τ ) and cos(θ τ ). d. dT. (. A. B. ). =. (. a b. b a. )(.

18 Oct 2021: cos. (. ω. t. -. 𝐤. 𝐱. ). B. (. 𝐤. ). sin. (. ω. t. -. 𝐤. 𝐱. ). (3.143). ... This form of the solution is more useful simply because it’s quicker to write exponentials rather than cos and sin.

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.

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,.

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.

2 Jul 2024: along the a = (sin, 0, cos) axis or along the b = (sin, 0, cos) axis. ... hEPR|Az B |EPRi = sin2. 2 cos2. 2= cos. This means that if we take the four operators A2, B1, A1 and B2 to be spin operators,.

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.

5 Dec 2008: 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 ... 1 e cos φ,. where (ρ, φ) are plane polar coordinates and e is a

13 Sep 2023: and b =pI1. (3.66). Then we can write. ̇ =b a cos. ... sin2 (3.67). and. ̇ =I1a. I3. (b a cos) cos. sin2 (3.68).