This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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 θ = x 2 πR 2 π = x R radians. Angular velocity ω = ˙ θ = ˙ x R , thus the velocity of the center of the wheel v = ωR . k R M x = 0 (a) The total energy at x point is E tot ( x ) = 1 2 kx 2 + 1 2 Mv 2 + 1 2 Iω 2 = 1 2 kx 2 + 1 2 M ˙ x 2 + 1 2 MR 2 ˙ x 2 R 2 = 1 2 kx 2 + M ˙ x 2 (1) (b) Time derivative dE/dt = 0 gives 2 M ¨ x + kx = 0 (2) (c) The angular frequency is ω = k 2 M (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 ξ represents the position of the air molecules, and p ∝ ∂ξ ∂z is the pressure over and above 1 atm. At the closed end there is always a node in ξ and an antinode in p . At the open end, there is always an antinode in ξ and a node in p . Thus, in p space we only have the cos term because B = 0. The boundary conditions are p ( z = 0 , t = 0) = p = A (4) and at any time t , p ( z = L, t ) = 0 = p cos k n L (5) 1 Therefore (2 n − 1) π k n = (6) 2 L NOTICE: at any time, p ( z = L, t ) = 0 thus 0 = p 0 cos k n L + B sin k n L ⇒ B = − p 0 tan( k...
View
Full
Document
This note was uploaded on 04/22/2011 for the course PHYS 1441 taught by Professor White during the Spring '08 term at UT Arlington.
 Spring '08
 WHITE
 Physics, Mass

Click to edit the document details