Formula Sheet Final Exam

Formula Sheet Final Exam Springs and masses: d2 d m 2 x(t) + b x(t) + kx(t) = F (t) dt dt More general differential equation with harmonic driving force: d2 F0 d cos (ωd t) x(t) + Γ x(t) + ω02 x(t) = 2 dt dt m Steady state solutions: xs (t) = A cos (ωd t − δ) where A= p (ω02 − F0 m ωd2 )2 + ωd2 Γ2 and tan δ = Γωd − ωd2 ω02 General solutions: For Γ = 0 (undamped system): x(t) = R cos (ω0 t + θ) + xs (t) where R and θ are unknown coefficients. For Γ < 2ω0 (under damped system): x(t) = Re −Γ t 2 r Γ2 ω02 − t+θ 4 cos ! + xs (t) where R and θ are unknown coefficients. For Γ = 2ω0 (critically damped system): Γ x(t) = (R1 + R2 t)e− 2 t + xs (t) where R1 and R2 are unknown coefficients. For Γ > 2ω0 (over damped system):  − x(t) = R1 e q Γ + 2 Γ2 −ω02 4  t + R2 e where R1 and R2 are unknown coefficients. 1   q Γ2 2 t − Γ − −ω 0 2 4 + xs (t) Coupled oscillators Fj = − n X Kjk xk k=1 Examples for n = 2   x1 (t) X (t) = x2 (t)   K11 K12 K= K21 K22   m1 0 M= 0 m2 Matrix equation of motion, matrices M, K, I are n × n, vectors X , Z are n × 1. d2 X (t) = −M−1 KX (t) dt2 Z(t) = Ae−iωt (M−1 K − ω 2 I)A = 0 To obtain the frequencies of normal modes solve: det(M−1 K − ω 2 I) = 0 For n = 2   M11 M12 = M11 M22 − M12 M21 det M21 M22 If the system is driven by force one can find the response amplitudes C(ωd ) F(t) = F0 e−iωd t W(t) = C(ωd )e−iωd t   c1 (ωd ) C(ωd ) = c2 (ωd ) (M−1 K − ωd2 I)C(ωd ) = F0 2 solving the equation above one can find the response amplitudies for the first (c1 (ωd )) and second (c2 (ωd )) objects in the system. Reflection symmetry matrix:   0 −1 S= −1 0 Eigenvalues (β) and  eigenvectors (A) of this 2 × 2 S matrix: 1 (1) β = −1, A =  1 1 (2) β = 1, A = −1 1D infinite coupled system which satisfy space translation symmetry: Given a eigenvalue β, the corresponding eigenvector is Aj = β j A0 where Aj (A0 ) is the normal amplitude of jth(0th) object in the system. Consider an one dimentional system which consists infinite number of masses coupled by springs, β can be written as β = eika where k is the wave number and a is the distance between the masses. Kirchoff’s Laws (be careful about the signs!) X X Node : Ii = 0 Loop : ∆Vi = 0 i Capacitors : ∆V = Q C i Inductors : ∆V = −L dI dt Current : I = Trigonometric equalities: sin(a ± b) = sin(a) cos(b) ± cos(a) sin(b) cos(a ± b) = cos(a) cos(b) ∓ sin(a) sin(b)     a−b a+b sin(a) + sin(b) = 2 sin cos 2 2     a+b a−b sin(a) − sin(b) = 2 cos sin 2 2     a+b a−b cos(a) + cos(b) = 2 cos cos 2 2 3 dQ dt  cos(a) − cos(b) = −2 sin a+b 2   sin a−b 2  eiθ = cos θ + i sin θ Some useful integrals involving sin and cos: ( 1, if n = m. sin sin dx = L L 0, otherwise. 0 ( Z  mπx   nπx  1, if n = m. 2 L cos dx = cos L 0 L L 0, otherwise. Z  nπx   mπx  2 L cos sin dx = 0 L 0 L L Z x sin(x)dx = sin(x) − x cos(x) + C Z x cos(x)dx = cos(x) + x sin(x) + C 2 L Z L  nπx   mπx  4 Maxwell Equations in vacuum ∂Ey ∂Ex ∂Bz ∂Ez ∂Ey ∂Bx ∂Ex ∂Ez ∂By − =− ; − =− ; − =− ∂x ∂y ∂t ∂y ∂z ∂t ∂z ∂x ∂t ∂By ∂Bx ∂Ez ∂Bz ∂By ∂Ex ∂Bx ∂Bz ∂Ey − = µ0 0 ; − = µ0 0 ; − = µ0 0 ∂x ∂y ∂t ∂y ∂z ∂t ∂z ∂x ∂t ∂Ex ∂Ey ∂Ez ∂Bx ∂By ∂Bz + + =0; + + =0 ∂x ∂y ∂z ∂x ∂y ∂z Lorentz force ~ + ~v × B) ~ F~ = q(E Wave equation for EM fields in vacuum ∂ 2 Ei ∂ 2 Ei ∂ 2 Ei 1 ∂ 2 Ei + + = where i = x, y, z ∂x2 ∂y 2 ∂z 2 c2 ∂t2 ∂ 2 Bi ∂ 2 Bi ∂ 2 Bi 1 ∂ 2 Bi + + = where i = x, y, z ∂x2 ∂y 2 ∂z 2 c2 ∂t2 For EM plane waves in vacuum: ~ r, t) ~ r, t) = 1 k̂ × E(~ B(~ c ~ r, t) = cB(~ ~ r, t) × k̂ E(~ Linear energy density in a string with tension T and mass density ρL  2  2 dK 1 dU 1 ∂y ∂y = ρL = T dx 2 ∂t dx 2 ∂x EM energy per unit volume and Poynting vector: 1 ~2 UE = 0 E 2 UB = 1 ~2 B 2µ0 ~= 1E ~ ×B ~ S µ0 Transmission and reflection R= Z1 − Z2 2Z1 , T = Z1 + Z2 Z1 + Z2 Phase velocity and impedance: 5 s p T , Z = T ρL (string) ρL r r 1 L v= , Z= (transmission line) LC C v= Snell’s law n1 sin θ1 = n2 sin θ2 Fourier transform Z ∞ dωC(ω)e−iωt f (t) = −∞ 1 C(ω) = 2π Delta function 1 2π Z ∞ Z ∞ dtf (t)eiωt −∞ 0 ei(ω−ω )t dt = δ(ω − ω 0 ) −∞ Z ∞ δ(x)dx = 1 −∞ Z ∞ δ(x − a)f (x)dx = f (a) −∞ Electric and magnetic field from an accelerated charge: ~ r, t) = − q a~⊥ (t − |r|/c) E(~ 4π0 rc2 ~ ~ r, t) = r̂ × B E(~ c Total power emitted by the accelerated charge: q 2 a2 (t − r/c) 6π0 c3 Interference of two sources with amplitudes A1 and A2 with a relative phase difference δ: P (t) = < I >∝ (A21 + A22 + 2A1 A2 cos δ) Interference of N fields of equal amplitude with phases δm+1 − δm = δ: 6  sin(N δ/2) < I >=< I0 > sin(δ/2) 2 Single slit diffraction where β is the phase difference between rays coming from edges and the center of the slit:  2 sin(β) < I >=< I0 > β Rayleigh’s criterion for resolution: Diffraction peak of one image falls on te first minimum of the diffraction pattern of the second image. Electric field transmission and reflection ratios, magnitude and sign, for radiation incident normally on an interface between lossless dielectrics with indices of refraction n1 and n2 . Er n1 − n2 = Ei n1 + n2 2n1 Et = Ei n1 + n2 Schrodinger’s Equation   ∂ ~2 ∂ 2 i~ ψ(x, t) = − + V (x, t) ψ(x, t) ∂t 2m ∂x2 where V is the potential energy, m is the mass of the particle and ψ is the wave function. 7