(a) Explanation For the given spring, a mass weighing 10 pounds stretches 1 4 foot. The mass is removed and replaced with a mass of 1.6 slugs The mass is initially released from 1 3 a foot above the equilibrium position with a downward velocity of 5 4 ft/s . Consider Hooke’s law in different form, W = k s , where k is a spring constant and s is the displacement of spring from equilibrium position when weight W is put on the spring from one end. So, to obtain the value of spring constant from the initial data which states that 10 pounds weight stretches the spring 1 4 foot. W = − k s 10 = − k ( − 1 4 ) k = 40 lb/ft Therefore, spring constant for the given spring is 40 lb/ft . Consider the relationship between circular frequency and spring constant to obtain the circular frequency as follows. ω 2 = k m = 40 1.6 = 25 Therefore, the circular frequency is 5 rad / sec . From the situation in the problem, it is undammed spring motion as there is no damping force or driving force. Therefore, consider the second-order differential equation of undammed spring motion is d 2 x d t 2 + ω 2 x = 0 with a general solution x ( t ) = c 1 cos ω t + c 2 sin ω t . So, the general solution of the given problem with circular frequency 5 rad / sec is x ( t ) = c 1 cos 5 t + c 2 sin 5 t . Now, to obtain arbitrary constant use the initial conditions x ( 0 ) = − 1 3 and x ' ( 0 ) = 5 4 . x ( 0 ) = c 1 cos ( 5 ( 0 ) ) + c 2 sin ( 5 ( 0 ) ) − 1 3 = c 1 cos ( 0 ) + c 2 sin ( 0 ) − 1 3 = c 1 ( 1 ) + c 2 ( 0 ) c 1 = − 1 3 Now, differentiate x with respect to t as follows x ( t ) = c 1 cos 5 t + c 2 sin 5 t x ' ( t ) = − 5 c 1 sin 5 t + 5 c 2 cos 5 t Substitute the initial condition x ' ( 0 ) = 5 4 . 5 4 = − 5 c 1 sin ( 5 ( 0 ) ) + 5 c 2 cos ( 5 ( 0 ) ) 5 4 = − 5 c 1 ( 0 ) + 5 c 2 cos ( 0 ) 5 c 2 = 5 4 So, the value of c 2 is (b) To determine To express: The equation of motion in the form x ( t ) = A cos ( ω t − ϕ ) . (c) To determine To find: The times at which displacement below the equilibrium position equal to half of the amplitude of the motion.

6th Edition

ISBN: 9781284127003

Author: ZILL

Publisher: JONES+BARTLETT LEARNING,LLC (CC)

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Question

Chapter 3.8, Problem 10E

(a)

To determine

To express: The equation of motion in the form of x(t)=Asin(ωt+ϕ).

(b)

To determine

To express: The equation of motion in the form x(t)=Acos(ωt−ϕ).

(c)

To determine

To find: The times at which displacement below the equilibrium position equal to half of the amplitude of the motion.

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Chapter 3.8, Problem 9E

Chapter 3.8, Problem 11E

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Chapter 3 Solutions

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To express: The equation of motion in the form of x ( t ) = A sin ( ω t + ϕ ) .

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To express: The equation of motion in the form of x(t)=Asin(ωt+ϕ).

To express: The equation of motion in the form x(t)=Acos(ωt−ϕ).

To find: The times at which displacement below the equilibrium position equal to half of the amplitude of the motion.

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Chapter 3 Solutions