Explanation Given: The weight of the mass is 24 pounds attached to a spring stretches by 4 inches with a downward velocity of 2 ft / sec . Calculation: Convert the length 4 inches to feet as follows. Note that, 1 feet = 12 inches . Thus, 4 inches = 4 ( 1 12 ) feet = 1 3 feet Thus, the length of spring stretched is, x = 1 3 feet . Obtain the weight as follows. Given that, 24 pounds is attached to a spring. Thus, W = m g 24 = m ( 32 ) ( ∵ g = 32 ft/s 2 ) m = 24 32 m = 3 4 Now, using Hook’s law, W = k x 24 = k ( 1 3 ) ( ∵ W = 24 pounds and x = 1 3 feet ) k = 24 × 3 k = 72 So, when the mass is displaced from its equilibrium position by an amount x ( t ) , the differential equation will be, m d 2 x d t 2 + k x = 0 3 4 d 2 x d t 2 + ( 72 ) x = 0 d 2 x d t 2 + 96 x = 0 Thus, the differential equation is, d 2 x d t 2 + 96 x = 0 (1) At the initial conditions, the mass is released from the equilibrium with a downward velocity of 2 ft / sec . Thus, x ( 0 ) = 0 and the velocity is x ′ ( 0 ) = 2 . The auxiliary equation of (1) is m 2 + 96 = 0

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 3 Solutions

Advanced Engineering Mathematics

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