ex1_sol - Massachusetts Institute of Technology Physics...

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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 anti-node in p . At the open end, there is always an anti-node 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...
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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.

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ex1_sol - Massachusetts Institute of Technology Physics...

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