4.19, 4.22, 4.35

# 4.19, 4.22, 4.35 - PHYS2626 Introductory Classical...

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Unformatted text preview: PHYS2626 Introductory Classical Mechanics Suggested Solutions for Assignment 2 October 2009 1. Problem 4.19 Method I: Initially, x ( t ) = d cos q 2 k/m t + φ , ˙ x ( t ) =- d q 2 k/m sin q 2 k/m t + φ . At t = 0, when x (0) = d/ 2, cos q 2 k/m t + φ = 1 2 , sin q 2 k/m t + φ = ± √ 3 2 . | ˙ x (0) | = d q 3 k/ 2 m. After the spring is removed, x ( t ) = C cos q k/m t + D sin q k/m t , ˙ x ( t ) =- C q k/m sin q k/m t + D q k/m cos q k/m t . Substituting the initial conditions x (0) = d/ 2 and | ˙ x (0) | = d q 3 k/ 2 m , we get x ( t ) = ( d/ 2) cos q k/m t + d q 3 / 2 sin q k/m t . The amplitude is A = √ C 2 + D 2 = d √ 7 / 2. Method II: (Energy approach) At the moment just before the spring is removed, the kinetic energy of the mass is 1 2 m ˙ x 2 = 1 2 (2 k ) h d 2- ( d/ 2) 2 i = kd 2 (3 / 4) . 1 Right after the spring is removed, 1 2 m ˙ x 2 = 1 2 k h A 2- ( d/ 2) 2 i kd 2 (3 / 4) = 1 8 k (4 A 2- d 2 ) 4 A 2 = 7 d 2 A = d √ 7 / 2 . = ⇒ x ( t ) = ( d √ 7 / 2) cos q k/m t + φ ....
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4.19, 4.22, 4.35 - PHYS2626 Introductory Classical...

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