Explanation Given information: The weight is pulled 2 3 foot above the equilibrium and released. 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

11th Edition

ISBN: 9781337275378

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 16.2, Problem 60E

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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 pulled 23 foot above the equilibrium and released.

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Chapter 16.2, Problem 59E

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 2 3 foot from its natural length. The weight is pulled 2 3 foot above the equilibrium and released.

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 pulled 23 foot above the equilibrium and released.

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Chapter 16 Solutions