(a) Explanation Given: For small displacement angle θ 1 ( t ) and θ 2 ( t ) , the system of linear differential equations of motion are as follows: θ 1 ′ ′ + g l θ 1 = − k m ( θ 1 − θ 2 ) θ 2 ′ ′ + g l θ 2 = k m ( θ 1 − θ 2 ) Definition used: The Laplace transformation of a linear differential equation with constant coefficient is expressed as follows: a n L { d n y d t n } = a n [ s n Y ( s ) − s ( n − 1 ) y ( 0 ) − ⋯ − y ( n − 1 ) ( 0 ) ] Calculation: Substitute ω 2 for g l and K for k m in equations θ 1 ′ ′ + g l θ 1 = − k m ( θ 1 − θ 2 ) and θ 2 ′ ′ + g l θ 2 = k m ( θ 1 − θ 2 ) . θ 1 ′ ′ + ω 2 θ 1 = − K ( θ 1 − θ 2 ) ...... ( 1 ) θ 2 ′ ′ + ω 2 θ 2 = K ( θ 1 − θ 2 ) ...... ( 2 ) Use definition (1) to obtain the Laplace transform of equation (1). s 2 L { θ 1 } − s θ 1 ( 0 ) − θ 1 ′ ( 0 ) + ω 2 L { θ 1 } = − K L { θ 1 } + K L { θ 2 } ( s 2 + ω 2 ) L { θ 1 } − s θ 0 = − K L { θ 1 } + K L { θ 2 } ( θ 1 ( 0 ) = θ 0 and θ 1 ′ ( 0 ) = 0 ) ( s 2 + ω 2 ) L { θ 1 } = − K L { θ 1 } + K L { θ 2 } + s θ 0 ...... ( 3 ) Use definition (1) to obtain the Laplace transform of equation (2). s 2 L { θ 2 } − s θ 2 ( 0 ) − θ 2 ′ ( 0 ) + ω 2 L { θ 2 } = K L { θ 1 } − K L { θ 2 } ( s 2 + ω 2 ) L { θ 2 } − s ψ 0 = K L { θ 1 } − K L { θ 2 } ( θ 2 ( 0 ) = ψ 0 and θ 2 ′ ( 0 ) = 0 ) ( s 2 + ω 2 ) L { θ 2 } = K L { θ 1 } − K L { θ 2 } + s ψ 0 ...... ( 4 ) Add equation (3) and equation (4). ( s 2 + ω 2 ) ( L { θ 1 } + L { θ 2 } ) = s ψ 0 + s θ 0 ( s 2 + ω 2 ) ( L { θ 1 } + L { θ 2 } ) = s ( ψ 0 + θ 0 ) L { θ 1 } + L { θ 2 } = ( ψ 0 + θ 0 ) s ( s 2 + ω 2 ) Take inverse Laplace on both sides of the above equation. θ 1 ( t ) + θ 2 ( t ) = ( ψ 0 + θ 0 ) L − 1 { s ( s 2 + ω 2 ) } θ 1 ( t ) + θ 2 ( t ) = ( ψ 0 + θ 0 ) cos ω t (b) To determine The motion of the coupled pendulum in special cases using the obtained solution in subpart (a).

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 7, Problem 49RE

(a)

To determine

The solution of equations θ1′′+glθ1=−km(θ1−θ2) and θ2′′+glθ2=km(θ1−θ2) using Laplace transform when θ1(0)=θ0, θ1′(0)=0, θ2(0)=ψ0 and θ2′(0)=0.

(b)

To determine

The motion of the coupled pendulum in special cases using the obtained solution in subpart (a).

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Chapter 7, Problem 48RE

Chapter 7, Problem 50RE

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Chapter 7 Solutions

Differential Equations with Boundary-Value Problems (MindTap Course List)

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The solution of equations θ 1 ′ ′ + g l θ 1 = − k m ( θ 1 − θ 2 ) and θ 2 ′ ′ + g l θ 2 = k m ( θ 1 − θ 2 ) using Laplace transform when θ 1 ( 0 ) = θ 0 , θ 1 ′ ( 0 ) = 0 , θ 2 ( 0 ) = ψ 0 and θ 2 ′ ( 0 ) = 0 .

Solution Summary: The author explains the Laplace transform of the equations theta, _1prime, and .

Videos

The solution of equations θ1′′+glθ1=−km(θ1−θ2) and θ2′′+glθ2=km(θ1−θ2) using Laplace transform when θ1(0)=θ0, θ1′(0)=0, θ2(0)=ψ0 and θ2′(0)=0.

The motion of the coupled pendulum in special cases using the obtained solution in subpart (a).

Want to see the full answer?

Chapter 7 Solutions

The solution of equations θ 1 ′ ​ ′ + g l θ 1 = − k m ( θ 1 − θ 2 ) and θ 2 ′ ​ ′ + g l θ 2 = k m ( θ 1 − θ 2 ) using Laplace transform when θ 1 ( 0 ) = θ 0 , θ 1 ′ ( 0 ) = 0 , θ 2 ( 0 ) = ψ 0 and θ 2 ′ ( 0 ) = 0 .

Solution Summary: The author explains the Laplace transform of the equations theta, _1prime, and .

Videos

The solution of equations θ1′​′+glθ1=−km(θ1−θ2) and θ2′​′+glθ2=km(θ1−θ2) using Laplace transform when θ1(0)=θ0, θ1′(0)=0, θ2(0)=ψ0 and θ2′(0)=0.

The motion of the coupled pendulum in special cases using the obtained solution in subpart (a).

Want to see the full answer?

Chapter 7 Solutions

The solution of equations θ 1 ′ ′ + g l θ 1 = − k m ( θ 1 − θ 2 ) and θ 2 ′ ′ + g l θ 2 = k m ( θ 1 − θ 2 ) using Laplace transform when θ 1 ( 0 ) = θ 0 , θ 1 ′ ( 0 ) = 0 , θ 2 ( 0 ) = ψ 0 and θ 2 ′ ( 0 ) = 0 .

The solution of equations θ1′′+glθ1=−km(θ1−θ2) and θ2′′+glθ2=km(θ1−θ2) using Laplace transform when θ1(0)=θ0, θ1′(0)=0, θ2(0)=ψ0 and θ2′(0)=0.