Equations of Motion - Vibrations and Waves - Solved Past Exam, Exams of Physics

Download Equations of Motion - Vibrations and Waves - Solved Past Exam and more Exams Physics in PDF only on Docsity! Midterm 2: Example Solutions (1a) In the limit where the oscillations are small, we can make the following approximations: (i) Tension in strings is the same as in equilibrium (2Mg and Mg for upper and lower strings respectively) (ii) Motions are nearly horizontal (iii) For any angle θ to the vertical, sinθ ~ tanθ ~ θ è Force exerted by upper string on mass A = – 2MgΨA/L Force exerted by lower string on mass A = – Mg(ΨA – ΨB )/L Force exerted by lower string on mass B = + Mg(ΨA – ΨB )/L Equations of motion: d2ΨA/dt2 = (g/L)(ΨB – 3ΨA) d2ΨB/dt2 = (g/L)(ΨA – ΨB ) (b) Normal coordinates are of the form ΨΝ = ΨA + rΨB for which d2ΨN/dt2 = d2ΨA/dt2 + r d2ΨB/dt2 = (g/L)(ΨB – 3ΨA + rΨA – rΨB) = (g/L)([r – 3] ΨA + [1– r] ΨB) The right-hand-side is a multiple of ΨΝ (in fact, [g/L][r – 3] times ΨΝ) provided (1 – r)/(r – 3) = r. This requires 1 – r = r2 – 3r <==> r2 – 2r –1 = 0 <==> r = 1 ± √2 Normal coordinates are Ψ1 = ΨA + (1 + √2) ΨB Ψ2 = ΨA + (1 – √2) ΨB Equation of motion is then d2ΨN/dt2 = (g/L)[–2 ±√2] ΨN Frequencies are √[(g/L)(2 +√2)] (c) Normal modes: Ψ1 = 0 ==> ΨA = – (1 + √2) ΨB Antisymmetric mode in which masses move in opposite directions. Amplitude of mass A ~ 2.4 times amplitude of mass B Frequency = √[(g/L)(2 + √2)] Ψ2 = 0 ==> ΨA = (√2 – 1) ΨB Symmetric mode in which masses move in same directions Amplitude of mass A ~ 0.4 times amplitude of mass B Frequency = √[(g/L)(2 –√2)] (2a) Equation of motion: d2Ψ/dt2 + Γ dΨ/dt + ω02 Ψ = F0 cos ω0t where ω0 = √(K/M) (b) The transient solutions to the equation die away on a timescale 1/Γ, leaving us with the steady-state solution for time, t > few x 1/Γ (c) Try solution of form Ψ = Ψ0 sin ω0t (for which dΨ/dt = ω0Ψ0 cos ω0t and d2Ψ/dt2 = –ω02 Ψ0 sin ω0t ) Substituting into equation of motion, we find that the first and third terms on the left-hand-side cancel, leaving us with Γω0Ψ0 cos ω0t = F0 cos ω0t, which is satisfied for all t if Ψ0 = F0/(Γω0)