Explanation Given information: The weight is pulled 1 2 foot below the equilibrium and released. The 64-pounds weight stretches a spring 4 3 foot from its natural length, so by Hooke’s Law, 64 = k ( 4 3 ) 64 × 3 4 = k 48 = k As the weight w is given by m g , we have m = w g = 64 32 = 2 So, the resulting differential equation for the undamped motion is d 2 y d t 2 + 48 2 y = 0 d 2 y d t 2 + 24 y = 0 The characteristic equation for the differential equation d 2 y d t 2 + 24 y = 0 is m 2 + 24 = 0 . The roots of the characteristic equation m 2 + 24 = 0 are: m 2 + 24 = 0 m 2 = − 24 m = − 24 m = ± 2 6 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 + 24 y = 0 is y = C 1 cos 2 6 t + C 2 sin 2 6 t

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Chapter 16, Problem 35RE

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The formula for the position of weight as a function of time when a 64-pounds weight stretches a spring 43 foot from its natural length. The weight is pulled 12 foot below the equilibrium and released.

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Chapter 16, Problem 34RE

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The formula for the position of weight as a function of time when a 64-pounds weight stretches a spring 4 3 foot from its natural length. The weight is pulled 1 2 foot below the equilibrium and released.

Solution Summary: The author explains how the 64-pounds weight stretches a spring 43 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 64-pounds weight stretches a spring 43 foot from its natural length. The weight is pulled 12 foot below the equilibrium and released.

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