(a) Explanation Given: The given mass is 20 pounds and spring is stretched by 6 inches . Calculations: Convert the given mass as follows: m = 20 32 = 5 8 slug Therefore, the mass is 5 8 slug . Convert the stretch 6 inches into feet as follows. 1 feet = 12 inches 6 inches = ( 1 12 ) 6 feet Thus, the stretch is x = 6 12 feet . According to Hooke’s law, m = k x . Substitute m = 20 and x = 6 12 as follows: 20 = k ( 6 12 ) k = 20 × 12 6 k = 40 Thus, the value of k is 40 lb/ft . The general equation of simple harmonic function is, m d 2 x d t 2 = − k x (1) Substitute m = 5 8 and k = 40 in the equation (1) as follows: 5 8 d 2 x d t 2 = − 40 x d 2 x d t 2 = − 40 x × 8 5 d 2 x d t 2 = − 64 x d 2 x d t 2 + 64 x = 0 Since the initial length of spring is 6 inches, substitute t = 0 . x ( 0 ) = 6 12 = 1 2 ft Differentiate both sides with respect to t . x ′ ( 0 ) = 0 The frequency of harmonic function is, ω = k m = 40 5 × 8 = 320 5 = 8 Therefore, the frequency of harmonic function is 8. The solution of differential equation is, x ( t ) = c 1 cos 8 t + c 2 sin 8 t x ′ ( t ) = 8 [ − c 1 cos 8 t + c 2 sin 8 t ] Here, x ( 0 ) = c 1 . Therefore, the value of c 1 is 1 2 ft . Here, x ′ ( 0 ) = c 2 . Therefore, the value of c 2 is 0 . The amplitude of simple harmonic function is, A = c 1 2 + c 2 2 = 1 2 Thus, tan ϕ = c 1 c 2 = ∞ Therefore, the angle of amplitude ϕ is, ϕ = π 2 (b) To determine The velocity of the mass at t = 3 π 16 and the direction at this instant. (c) To determine The times at which the mass pass through the equilibrium position.

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Question

Chapter 3.8, Problem 5E

(a)

To determine

The position of the mass at the times t=π12,π8,π6,π4and9π32.

(b)

To determine

The velocity of the mass at t=3π16 and the direction at this instant.

(c)

To determine

The times at which the mass pass through the equilibrium position.

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Chapter 3.8, Problem 4E

Chapter 3.8, Problem 6E

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Chapter 3 Solutions

Advanced Engineering Mathematics

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The position of the mass at the times t = π 12 , π 8 , π 6 , π 4 and 9 π 32 .

Solution Summary: The author calculates the position of the mass at the required times: cm=2032.

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The position of the mass at the times t=π12,π8,π6,π4and9π32.

The velocity of the mass at t=3π16 and the direction at this instant.

The times at which the mass pass through the equilibrium position.

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