Explanation Given: The given mass is 32 pounds and spring is stretched to 2 feet . Calculations: According to Hooke’s law, m = k x . From the given conditions, m = 32 and x = 2 . Substitute m = 32 and x = 2 as follows: 32 = k ( 2 ) k = 32 2 k = 16 That is, the value of k is 16 lb/ft . Substitute W = 32 and g = 32 in the formula W = m g . The mass is calculated as: m = W g = 32 32 = 1 Therefore, the mass is 1 slug . The general equation of simple harmonic function is, d 2 x d t 2 + ( k m ) x = 0 (1) Substitute m = 1 and k = 16 in the equation (1) as follows: d 2 x d t 2 + ( 16 1 ) x = 0 d 2 x d t 2 + 16 x = 0 Since the initial length of spring is 1 foot, substitute t = 0 . Thus, x ( 0 ) = − 1 . Thus, the upward velocity is 2 ft/sec . That is, after differentiation, the value is x ′ ( 0 ) = − 2 . Substitute x ( 0 ) = ( − 1 ) and x ′ ( 0 ) = − 2 in the differential equation as follows: d 2 x d t 2 + 16 x = 0 m 2 = − 16 m = ± 4 i Therefore, the general solution of the differential equation is x ( t ) = c 1 cos 4 t + c 2 sin 4 t . Substitute x ( 0 ) = ( − 1 ) in the differential equation as follows: x ( 0 ) = c 1 cos 0 + c 2 sin 0 − 1 = c 1 + 0 c 1 = − 1 Differentiate x ( t ) with respect to t to obtain x ′ ( t ) = − 4 c 1 sin 4 t + 4 c 2 cos 4 t

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ISBN: 9781337652476

Author: ZILL

Publisher: CENGAGE L

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Chapter 5.1, Problem 8E

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The amplitude, period of motion and the completed cycles at the end of 4π seconds.

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Chapter 5.1, Problem 7E

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FIRST COURSE IN DIFF.EQ.-WEBASSIGN

Ch. 5.1 - 5.1.1 Spring/Mass systems: Free Undamped Motion A...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A mass...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A mass...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A force...Ch. 5.1 - Prob. 7E Ch. 5.1 - Prob. 8E Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A mass...Ch. 5.1 - 5.1.1Spring/Mass Systems: Free Undamped Motion A...

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The amplitude, period of motion and the completed cycles at the end of 4 π seconds .

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