(a) Explanation Theorem used: Hooke’s law: “The strain in a solid is proportional to the applied stress within the elastic limit of that solid.” Formula used: If m 1 = − λ + λ 2 − ω 2 and m 2 = − λ − λ 2 − ω 2 are the roots of the auxiliary equation m 2 + 2 λ m + ω 2 = 0 , then there are three different cases for which the solution exists. If λ 2 − ω 2 > 0 ,then x ( t ) = c 1 e m 1 t + c 2 e m 2 t .(Overdamped) If λ 2 − ω 2 = 0 ,then x ( t ) = e − λ t ( c 1 + c 2 t ) .(Critically damped) If λ 2 − ω 2 < 0 , m 1 = − λ + ω 2 − λ 2 i and m 1 = − λ − ω 2 − λ 2 i ,then x ( t ) = e − λ t ( c 1 cos ω 2 − λ 2 t + c 2 sin ω 2 − λ 2 t ) .(underdamped) Calculation: The mass of 1 slug is attached to a spring whose constant is 5 lb/ft . This implies m = 1 and k = 5 . The differential equation of a free damping motion is given as, d 2 x d t 2 + 2 λ d x d t + ω 2 x = 0 where, 2 λ = β m and ω 2 = k m ( or ) m d 2 x d t 2 + β d x d t + k x = 0 Here, β is the positive damping constant and the negative sign indicates the damping force acts in a direction opposite to the motion. Since the mass is released 1 foot below the equilibrium position with the downward velocity of 5 ft/s and the damping force is equal to 2 times the instantaneous velocity, the value of β = 2 and x ( 0 ) = 1 , x ′ ( 0 ) = 5 . Also, it is given that the mass is driven by an external force which is equal to f ( t ) = 12 cos 2 t + 3 sin 2 t Now, substitute all the values in the differential equation. 1 d 2 x d t 2 + 2 d x d t + 5 x = 12 cos 2 t + 3 sin 2 t . Thus, the differential equation with initial conditions are as follows. x ″ + 2 x ′ + 5 x = 0 , with x ( 0 ) = 1 , x ′ ( 0 ) = 5 . The auxiliary equation is, m 2 + 2 m + 5 = 0 . Here, 2 λ = 2 , ω 2 = 5 . λ 2 − ω 2 = 1 − 5 = − 4 < 0 This is said to be underdamped and the general solution is as follows. x ( t ) = e − t ( c 1 cos 5 − 1 t + c 2 sin 5 − 1 t ) = e − t ( c 1 cos 4 t + c 2 sin 4 t ) = e − t ( c 1 cos 2 t + c 2 sin 2 t ) The value of c 1 and c 2 is calculated using the initial conditions x ( 0 ) = 1 , x ′ ( 0 ) = 5 (b) To determine To sketch: The graph of the transient and steady-state solutions on the same coordinate axes. (c) To determine To sketch: The graph of the equation of the motion.

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Question

Chapter 3.8, Problem 34E

(a)

To determine

The equation of the motion.

(b)

To determine

To sketch: The graph of the transient and steady-state solutions on the same coordinate axes.

(c)

To determine

To sketch: The graph of the equation of the motion.

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Chapter 3.8, Problem 33E

Chapter 3.8, Problem 35E

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Chapter 3 Solutions

Advanced Engineering Mathematics

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The equation of the motion.

Solution Summary: The author explains Hooke's law: "The strain in a solid is proportional to the applied stress within the elastic limit of that solid."

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The equation of the motion.

To sketch: The graph of the transient and steady-state solutions on the same coordinate axes.

To sketch: The graph of the equation of the motion.

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Chapter 3 Solutions