Explanation Given information: The weight is pulled 2 3 foot above the equilibrium and released with an initial velocity of 1 2 foot per second. The 32-pounds weight stretches a spring 2 3 foot from its natural length, so by Hooke’s Law, 32 = k ( 2 3 ) 32 × 3 2 = k 48 = k As the weight w is given by m g , we have m = w g = 32 32 = 1 So, the resulting differential equation for the undamped motion is d 2 y d t 2 + 48 1 y = 0 d 2 y d t 2 + 48 y = 0 The characteristicequation for the differential equation d 2 y d t 2 + 48 y = 0 is m 2 + 48 = 0 . The roots of the characteristic equation m 2 + 48 = 0 are: m 2 + 48 = 0 m 2 = − 48 m = − 48 m = ± 4 3 i If m 1 = α + i β and m 2 = α − i β are complex zeros of the characteristics equation, then the general solution is y = C 1 e α t cos β t + C 2 e α t sin β t . Thus, the general solution of the equation d 2 y d t 2 + 48 y = 0 is y = C 1 cos 4 3 t + C 2 sin 4 3 t

Student Solutions Manual For Larson/edwards? Multivariable Calculus, 11th

11th Edition

ISBN: 9781337275392

Author: Larson, Ron; Edwards, Bruce H.

Publisher: Cengage Learning

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Chapter 16.2, Problem 61E

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The formula for the position of weight as a function of time when a 32-pounds weight stretchesa spring 23 foot from its natural length. The weight is raised 23 foot above the equilibrium and released with an initial downward velocity of 12 foot per second.

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Chapter 16.2, Problem 60E

Chapter 16.2, Problem 62E

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Chapter 16 Solutions

Student Solutions Manual For Larson/edwards? Multivariable Calculus, 11th

Ch. 16.1 - Exactness What does it mean for the...Ch. 16.1 - Integrating Factor When is it beneficial to use an...Ch. 16.1 - Testing for Exactness In Exercises 3-6, determine...Ch. 16.1 - Testing for Exactness In Exercises 3-6, determine...Ch. 16.1 - Prob. 5E Ch. 16.1 - Prob. 6E Ch. 16.1 - Prob. 7E Ch. 16.1 - Solving an Exact Differential Equation In...Ch. 16.1 - Prob. 9E Ch. 16.1 - Prob. 10E

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The formula for the position of weight as a function of time when a 32-pounds weight stretchesa spring 2 3 foot from its natural length. The weight is raised 2 3 foot above the equilibrium and released with an initial downward velocity of 1 2 foot per second.

Solution Summary: The author explains how the 32-pounds weight stretches a spring 23 foot from its natural length, so by Hooke's Law, the resulting differential equation for the unda

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The formula for the position of weight as a function of time when a 32-pounds weight stretchesa spring 23 foot from its natural length. The weight is raised 23 foot above the equilibrium and released with an initial downward velocity of 12 foot per second.

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