Problem B.2 Each of the ODES shown below is second order in y, with y1 as a solution. Reduce the ODE from being second order in y to being first order in w, with w being the only response variable appearing in the ODE. Combine like terms. Show your work. B.2.a. xy" –y' +y = 0 Y1 = x* В.2.b. y" + 9у %3D 0 sin(3t) B.2.c. x²y" +y = 0 Y1 = x²/3
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- Solve for question 3 The θ integration fails if α is non-constant. There is also a type of situation where the step before the integration fails, involving ω. Question 1: derive ω = αt + ω0. Start with the derivatives I began with, and derive this equation. Question 2: derive θ = 1/2αt2 + ω0t + θ0. You may use prior derived results. Question 3: derive θ = θ0 + (ω + ω0) t/2. You probably need to use prior derived results.PART A k = 2.000 N/m M=0.125kg %3D A 0.125kg mass is attached to a horizontal spring of spring constant k = 2 N/m and set into oscillation. Assume frictionless floor. What is the angular frequency o of the vibration? o = How about the vibrating frequency f in Hz? f = The position of the mass is given as x = A cos (ot + o). With your calculated value of w, fill in the following table (don't forget to set your calculator to "radian mode"). Each row represents a different oscillation with the same spring, pay attention to sign: А (m) x (m) v (m/s) a (m/s?) t (s) фо (rad) F (N) 0.03 2.5 0.3 0.04 0.02 0.2 0.06 0.1 0.4 0.04 0.3 -0.2 0.5 0.04 0.1 0.1Part B: Error propagation Consider that you measured an angle 0 to be 40 degrees. The uncertainty of your measurement is + 1 degree. Using the rules for error propagations, determine the uncertainty of sin(0). Clearly show your work. Hint: The uncertainty of any function of a single variable, z = f(x), is the derivative of the function calculated at that point multiplied by the uncertainty of the variable. A(z) df Ax dx
- Solve d,e,f B) : v = -R w(sin(wt)î-cos(wt)j) C): v =( (sin(wt))^2 (cos(wt))^2)^1/2 For C I raised the entire equation to 1/2 because my keyboard doesn’t allow me to do square root but it is basically the Pythagorean theoremSelf-test 1B.1 Derive the expression for (v²) in eqn 1B.7 by evaluating the integral in eqn 1B.6 with n = 2.Exercise b. Hard spin glasses.Physicists use a model called an Ising spin glass to study magnetic materials. We havea graph G = (V, E) and a function J : E → Z, where J(u, v) is the interaction strengthof edge {u, v} ∈ E. A state is a function s : V → {−1, +1}.1 The energy of a states is H(s) = −P{u,v}∈EJ(u, v) · s(u) · s(v).2 An edge {u, v} is called ferromagnetic ifJ(u, v) > 0, and antiferromagnetic if J(u, v) < 0. Ferromagnetic edges “pressure” uand v to be the same spin, and antiferromagnetic edges “pressure” them to be different.We wish to find the ground state, i.e., values of s(u) for each u ∈ V that minimize theenergy. Phrased as a decision problem, we ask whether a state exists below a certainenergy threshold:Spin-GlassInput: a graph G = (V, E), interaction strengths J : E → Z, and a threshold hˆ ∈ ZQuestion: is there a state s : V → {−1, +1} with H(s) ≤ hˆ?Show that Spin-Glass is NP-complete.1Each s(u) is called the spin of u; if u is a magnetic domain, think of s(u) as…