Explanation Given information: The weight is pulled 1 2 foot below the equilibrium and released. The motion takes place in a medium that furnishes a damping force of magnitude 1 8 | v | at all times. 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 motion is d 2 y d t 2 + p m d y d t + k m y = 0 d 2 y d t 2 + ( 1 8 ) 1 d y d t + 48 1 y = 0 d 2 y d t 2 + 1 8 d y d t + 48 y = 0 The characteristicequation for the differential equation d 2 y d t 2 + 1 8 d y d t + 48 y = 0 is m 2 + 1 8 m + 48 = 0 . The roots of the characteristic equation m 2 + 1 8 m + 48 = 0 are: m 2 + 1 8 m + 48 = 0 8 m 2 + m + 384 = 0 m = − 1 ± 1 − 4 × 8 × 384 2 × 8 m = − 1 ± 1 − 12288 16 m = − 1 ± − 12287 16 m = − 1 ± 12287 i 16 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 + 1 8 d y d t + 48 y = 0 is y = C 1 e − 1 16 t cos 12287 16 t + C 2 e − 1 16 t sin 12287 16 t . Now, when t = 0 , y = 1 2 , so y = C 1 e − 1 16 t cos 12287 16 t + C 2 e − 1 16 t sin 12287 16 t 1 2 = C 1 ( 1 ) cos ( 0 ) + C 2 ( 1 ) sin ( 0 ) 1 2 = C 1 ( 1 ) ( 1 ) 1 2 = C 1 Now, differentiate y = C 1 e − 1 16 t cos 12287 16 t + C 2 e − 1 16 t sin 12287 16 t with respect to 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 63E

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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 12 foot below the equilibrium and released. The motion takes place in a medium that furnishes a damping force of magnitude 18|v| at all times.

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Chapter 16.2, Problem 62E

Chapter 16.2, Problem 64E

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

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Solution Summary: The author explains how the 32-pounds weight stretches a spring 23 foot from its natural length.

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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 12 foot below the equilibrium and released. The motion takes place in a medium that furnishes a damping force of magnitude 18|v| at all times.

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