(a) Explanation Results used: Exercise 50: The general equation of the motion of the buoyant cylinder is y ′ ′ + ω 0 2 y = 0 with the general solution y ( t ) = c 1 cos ω 0 t + c 2 sin ω 0 t . Calculation: The value of ω 0 is ω 0 = ρ w g ρ L . Calculate the value of ω 0 in the given case as follows. ω 0 = 0.7 ( 9.8 ) 2 = 49 = 7 Therefore, using the above result the general equation is y ′ ′ + 7 y = 0 . Similarly the general solution is y ( t ) = c 1 cos 7 t + c 2 sin 7 t . It is given that θ ( 0 ) = 0.5 . Substitute t = 0 in y ( t ) = c 1 cos 7 t + c 2 sin 7 t . y ( 0 ) = c 1 cos 7 ( 0 ) + c 2 sin 7 ( 0 ) 0 (b) To determine To express: The solution obtained in part (a) in amplitude phase form. (c) To determine The period of the oscillation of buoyant cylinder.

2nd Edition

ISBN: 9781323178522

Author: Briggs

Publisher: PEARSON

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Chapter D2.4, Problem 52E

(a)

To determine

To solve: The initial value problem with L=2 m,ρρw=0.7,y(0)=0.5,y′(0)=0 .

(b)

To determine

To express: The solution obtained in part (a) in amplitude phase form.

(c)

To determine

The period of the oscillation of buoyant cylinder.

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Chapter D2.4, Problem 51E

Chapter D2.4, Problem 53E

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Chapter D2 Solutions

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D2 - Prob. 6RE Ch. D2 - Prob. 7RE Ch. D2 - Prob. 8RE Ch. D2 - Prob. 9RE Ch. D2 - Prob. 10RE Ch. D2 - Prob. 11RE Ch. D2 - Prob. 12RE Ch. D2 - Prob. 13RE Ch. D2 - Prob. 14RE Ch. D2 - Prob. 15RE Ch. D2 - Prob. 16RE Ch. D2 - Prob. 17RE Ch. D2 - Prob. 18RE Ch. D2 - Prob. 19RE Ch. D2 - Prob. 20RE Ch. D2 - Prob. 21RE Ch. D2 - Forced undamped oscillations A 4-kg block hangs on...Ch. D2 - Free damped oscillations A 0.2-kg block hangs on a...

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The initial value problem with L = 2 m, ρ ρ w = 0.7 ,y ( 0 ) = 0.5 , y ′ ( 0 ) = 0 .

Solution Summary: The author explains how to solve the initial value problem with L=2m,rho

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To solve: The initial value problem with L=2 m,ρρw=0.7,y(0)=0.5,y′(0)=0 .

To express: The solution obtained in part (a) in amplitude phase form.

The period of the oscillation of buoyant cylinder.

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Chapter D2 Solutions