Vibrations and Waves Exam 1 with Formula Sheet | Fall 2016 MIT, Exams of Physics

Download Vibrations and Waves Exam 1 with Formula Sheet | Fall 2016 MIT and more Exams Physics in PDF only on Docsity! Formula Sheet The differential equation ẍ+ γẋ+ ω0 2 x = f cos(ωt + φ) (1) Has the general solutions; γ −( γ 2 )t 2 < ω0 : X(t) = A1e cos(ω 0t + β) + Xp(t) (2) γ −( γ = ω0 : X(t) = (A1 + A2t) e 2 ) t + Xp(t) (3) 2 γ −Γ+t −Γ−t> ω0 : X(t) = A1e + A2e + Xp(t) (4) 2 with Xp(t) = A(ω) cos(ωt − δ(ω) + φ) (5) and r r γ2 γ γ2 ω0 = ω2 − Γ± = ± − ω2 (6)0 04 2 4 q A(ω) = f/ (ω2 − ω2)2 + (γω)2 tan(δ(ω)) = γω/(ω02 − ω2) (7)0 Idealized relations for voltage/emf across circuit elements: Q 1. Capacitor: VC = C 2. Resistor: VR = IR dI 3. Self Inductance: VL = L dt ∂2Ψ 2 2ΨClassical Scalar Wave Equation in 3-D: = v r ∂t2 The plane wave solution is: Ψ(~r, t) = A cos (~k · ~r ± ωt + φ) cos (kr ± ωt + φ) The spherical wave solution is: Ψ(~r, t) = A r ∂2Ψ ∂2Ψ2Classical Wave Equation in 1-D: = v ∂t2 ∂x2 The standing wave solution is: Ψ(x, t) = A cos ( ω x + φx) cos (ωt + φt)v The progressive wave solution is: Ψ(x, t) = f(t ± x )v r T √ 1 F02 For String: v = Z = Tµ hP i = µ 2 Z r κ √ ∂P For Sound: v = Z = κρ κ = −V ρ ∂V r √k For Torsion: v = Z = kI I r 1 L 1 V 2 For Transmission Line: v = √ Z = hP i = 0 LC C 2 Z ω ∂ω vphase = = νλ vgroup = k ∂k 1 2Z1 Z1 − Z2 For a displacement wave on a string: T = R = Z1 + Z2 Z1 + Z2 2Z2 Z2 − Z1 For a voltage wave on a transmission line: T = R = Z1 + Z2 Z1 + Z2 ρ ~ ~ ~ ~ ~r · E = F = q(E + ~v × B) 0 ~ ~r · B = 0 UE = 1 0E 2 2 ∂B~ 1 ~ ~ B2r× E = − UB = ∂t 2µ0 ~ ~∂E~ E × B ~ ~ ~ ~r× B = 0µ0 + µ0J S = ∂t µ0 ∂2 ~ ~E 1 S2 ~Electromagnetic wave in vacuum: = r E Radiation Pressure: ∂t2 µ00 c |E| 1 For a progressive wave solution in vacuum: = c = √ |B| µ00 Radiation due to the acceleration of charge: ~q~a⊥(t 0) r̂ × E |r|~ ~E(~r, t) = − B = t0 = t − 4π0rc2 c c 2q a2(t0) Total radiated power from accelerated charge (Larmor formula): P (t) = 6π0c3 Boundary conditions at the surface of a perfect conductor (for time-varying fields): Ek = 0 and B⊥ = 0. √ √ c For most dielectrics (KM ≈ 1): n = KE KM ≈ KE vphase = n Snell’s law: n1 sin θ1 = n2 sin θ2 Reflection and transmission of electromagnetic waves at normal incidence: n1 − n2 Ereflected = Eincident n1 + n2 2n1 Etransmitted = Eincident n1 + n2 For interference from N slits where a separation d between two slits, sin2( Nπ d sin θ)λI(θ) = I0 sin2( π d sin θ)λ Diffraction intensity from a slit of width D, sin2 ( π D sin θ)λI(θ) = I0 ( π D sin θ)2 λ Rayleigh’s criterion for resolution: Diffraction peak of one images falls on the first minimum of the diffraction pattern of the second image. 2 MASSACHUSETTS INSTITUTE OF TECHNOLOGY PHYSICS DEPARTMENT Physics 8.03: Vibrations and Waves Exam 1 FAMILY (Last) Name GIVEN (First) Name Student ID Number Recitation Section: (check one)  R01 TR 10 Prof. Jarillo-Herrero  R02 TR 11 Prof. Jarillo-Herrero  R03 TR 1 Prof. Weinberg  R04 TR 2 Prof. Weinberg Instructions: 1. Do not remove any pages of the exam, except the formula sheet. 2. This is a closed book exam. 3. Do all SIX (6) problems. 4. SHOW ALL WORK. Print your name on each sheet. 5. CALCULATORS, BOOKS, COMPUTERS and CELL PHONE are NOT ALLOWED. Points: Problem Maximum Score Grader Problem 1: 16 Problem 2: 16 Problem 3: 16 Problem 4: 16 Problem 5: 18 Problem 6: 18 5 Problem 1: 16 Points Figure 1: A perfectly conducting waveguide. The electric field for a TE mode in an infinitely long (in the x direction) perfectly conducting rectangular waveguide (a < b) is given by; ~E(x, y, z, t) = E0 cos(ky y + φy) cos(kxx − ωt)ẑ (8) (1.a) (4pts) Find ky and φy that satisfy the boundary conditions. (1.b) (4pts) Write down the dispersion relation for this mode of the waveguide. (1.c) (4pts) What is the lowest frequency that will propagate in this mode? (1.d) ~(4pts) What is the magnetic field B(x, y, z, t) associated with the electric field of this mode? 6 Problem 2: 16 Points Figure 2: A system of coupled oscillators. The figure above shows a system of masses. The mass of 2m is connected to an immobile wall with a spring of constant 2k, while the mass of m is connected to an immobile wall with a spring of constant k. The masses are then coupled to each other with an elastic band of length L, under tension T = 2kL. The masses are constrained to move in the x direction only. At equilbrium the masses have the same x position and the springs are uncompressed. There is no friction or gravity. The displacements from equilibrium are small enough (x1, x2  L), so that the tension in the band stays constant. (2.a) (5pts) Write down the coupled differential equations describing the displacement of the masses from equilibrium {x1, x2}. (2.b) (7pts) Find the normal mode frequencies of the system. (2.c) (4pts) On the two figures included on the next page sketch the normal modes of the system, be sure to clearly indicate both the magnitude and direction of the motion of the masses. 7 Problem 4: 16 Points A monochromatic beam is incident on N slits, which results in a intensity pattern as a function of angle on a screen some distance away as shown in the figure below. Each slit has a width D and the distance between the centers of the slits is d. The distance between the screen and the slits is very large. Figure 5: Interference pattern due to N slits. From the pattern deduce the following: (4.a) (6pts) The number of slits N on which the beam is incident. Explain your reasoning. (4.b) (6pts) The ratio d/D. Explain your reasoning. (4.c) (4pts) Now suppose that the width of the slits is reduced to ∼ 0, while the intensity of the monochromatic beam is increased so that the intensity of the central maximum is unchanged. On top of the plot (showing the original intensity pattern in dashed lines) on the next page, draw the resulting intensity pattern. 10 Figure 6: Plot the resulting intensity pattern as D → 0. 11 Problem 5: 20 Points A string of length 2L with mass density µ is placed under tension T and is fixed at both ends. At time t = 0, the displacement of the string is zero everywhere but it is struck so that a transverse velocity is imparted to a section of the string. The intial conditions of the string are (a  L); y(x, t = 0) = 0 (10) ⎧ ⎨ v0 : L − a ≤ x < L ẏ(x, t = 0) = −v0 : L ≤ x < L + a (11)⎩ 0 : elsewhere Figure 7: The intial transverse velocity of the string at time t = 0. The initial displacement is zero everywhere. (5.a) (3pts) Using the plot (Figure #8) provided on the next page, sketch the first three normal modes of vibration for this string, regardless of whether or not they are excited. (5.b) (10pts) What is the amplitude of the n-th normal mode after the string is struck? What is the lowest unexcited mode? (Problem continues on the next page.) 12