Angular Velocity - Vibrations and Waves - Solved Past Exam, Exams of Physics

Download Angular Velocity - Vibrations and Waves - Solved Past Exam and more Exams Physics in PDF only on Docsity! Massachusetts Institute of Technology Physics 8.03 Exam 1 Solutions Thursday, October 14, 2004 Problem 1 Without slipping: when the center of the wheel moves a distance x, the wheel has rotated an angle 0 = 52920 = @ radians. Angular velocity w = @ = 4, thus the velocity of the center of the wheel v=wR. gq A A mF qm | A (a) The total energy at x point is Evot(x) = ake + 5M + 5 = she + 5M? + iures = ake +Mé? (1) (b) Time derivative dE/dt = 0 gives IME + kx =0 (2) (c) The angular frequency is wa (3) Problem 2 (a) v & 340 m/sec is the speed of sound in air (measured during lecture on Oct. 5). (b) The sound wave in this pipe is shown in the following figure. Where € represents the position of the air molecules, and p « a is the pressure over and above 1 atm. At the closed end there is always a node in € and an anti-node in p. At the open end, there is always an anti-node in € and a node in p. Thus, in p space we only have the cos term because B=0. The boundary conditions are Pz =0,t=0)=po=A (4) and at any time f, pz = L,t) = 0 = po cos kink (5) Therefore (2n − 1)π kn = (6)2L NOTICE: at any time, p(z = L, t) = 0 thus 0 = p0 cos knL + B sin knL ⇒ B = −p0 tan(knL) tan(knL) = ±∞ for any n. This is consistent with B = 0. And (2n − 1)πv ω = vkn = (7)2L This can be found by substituting p(z, t) = p0 cos kz cos ωt into the wave equation. ∂p = −p0k sin kz cos ωt (8) ∂z ∂2p = −p0k2 cos kz cos ωt (9) ∂z2 ∂p = −p0ω cos kz sin ωt (10) ∂t ∂2p = −p0ω2 cos kz cos ωt (11) ∂t2 Thus 1 −p0k2 cos kz cos ωt = − 2 p0ω2 cos kz cos ωt (12) v which leads to ω2 k2 = ⇒ ω = vk (13) 2v 2π(c) L = 0.5 cm, k1 = π ⇒ λ1 = k1 = 2 m (λ1 = 4L). 340λ1 = f v 1 ⇒ f1 = 2π = = 170 Hz. ω1 2 2π 340 k2 = 3π ⇒ λ2 = k2 = 2 m ⇒ f2 = 2/3 = 510 Hz = 3f1.3 2