Explanation Given: The point X 1 = ( x 1 , 0 ) is a critical point which is a center and the system is undamped when f ′ ( x 1 ) = 0 and f ″ ( x 1 ) > 0 . Calculation: It is given that the mass of the bead is m and the bead slides along the wire whose shape function is z = f ( x ) . From the figure 1 it is observed that the tangential force F due the weight of the bead is as follows: F = m g sin θ The horizontal component of the tangential force is as follows: F x = − m g sin θ cos θ … … ( 1 ) The function for the curve of the wire is z = f ( x ) so the slope at any point is obtained as follows: tan θ = f ′ ( x ) The trigonometric ratio tan θ for any right angle triangle with base angle as θ is defined as the ratio of the perpendicular and base. As per the above statement it is concluded that the length of the perpendicular is f ′ ( x ) and the length of the base is 1 . As per the Pythagoras theorem the length of hypotenuse is calculated as follows: H = 1 + ( f ′ ( x ) ) 2 The expression for sin θ and cos θ is as follows: sin θ = f ′ ( x ) 1 + ( f ′ ( x ) ) 2 cos θ = 1 1 + ( f ′ ( x ) ) 2 Substitute the expression of sin θ and cos θ in equation ( 1 ) . F x = − m g f ′ ( x ) 1 + ( f ′ ( x ) ) 2 1 1 + ( f ′ ( x ) ) 2 = − m g f ′ ( x ) 1 + ( f ′ ( x ) ) 2 Consider the damping force D which acts in the opposite direction to the motion of bead and the damping force is a constant multiple of the velocity of the bead. The horizontal component of the damping force is as follows: D x = − β x ′ The equation of motion for the bead is formed as follows: m x ″ = − m g f ′ ( x ) 1 + ( f ′ ( x ) ) 2 − β x ′ x ″ = − g f ′ ( x ) 1 + ( f ′ ( x ) ) 2 − β m x ′ Label the above equation as follows: x ″ = − g f ′ ( x ) 1 + ( f ′ ( x ) ) 2 − β m x ′ … … ( 2 ) Consider that y = x ′ . Substitute y for x ′ in equation ( 2 ) . ( x ′ ) ′ = − g f ′ ( x ) 1 + ( f ′ ( x ) ) 2 − β m x ′ y ′ = − g f ′ ( x ) 1 + ( f ′ ( x ) ) 2 − β m y ( y = x ′ ) The plane autonomous system is formed as follows: x ′ = y y ′ = − g f ′ ( x ) 1 + ( f ′ ( x ) ) 2 − β m y Consider that P ( x , y ) = x ′ and Q ( x , y ) = y ′ . The equations formed are as follows: P ( x , y ) = y Q ( x , y ) = − g f ′ ( x ) 1 + ( f ′ ( x ) ) 2 − β m y Differentiate P ( x , y ) partially with respect to x . ∂ P ( x , y ) ∂ x = ∂ y ∂ x = 0 Differentiate P ( x , y ) partially with respect to y . ∂ P ( x , y ) ∂ y = ∂ y ∂ y = 1 Differentiate Q ( x , y ) partially with respect to x . ∂ Q ( x , y ) ∂ x = ∂ ( − g f ′ ( x ) 1 + ( f ′ ( x ) ) 2 − β m y ) ∂ x = ∂ ( − g f ′ ( x ) 1 + ( f ′ ( x ) ) 2 ) ∂ x = − g ( f ″ ( x ) + f ″ ( x ) ( f ′ ( x ) ) 2 − 2 f ″ ( x ) ( f ′ ( x ) ) 2 ( 1 + ( f ′ ( x ) ) 2 ) 2 ) = − g f ″ ( x ) [ 1 − ( f ′ ( x ) ) 2 ] ( 1 + ( f ′ ( x ) ) 2 ) 2 Differentiate Q ( x , y ) partially with respect to y

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 11.4, Problem 4E

To determine

The time period of the bead of mass (m) when x(0) is near x1 and x′(0)=0 such that the bead slides along the wire and the shape function of the wire is z=f(x).

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Chapter 11.4, Problem 3E

Chapter 11.4, Problem 5E

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Chapter 11 Solutions

Advanced Engineering Mathematics

Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E Ch. 11.1 - Prob. 6E Ch. 11.1 - Prob. 7E Ch. 11.1 - Prob. 8E Ch. 11.1 - Prob. 9E Ch. 11.1 - Prob. 10E

Ch. 11.1 - Prob. 11E Ch. 11.1 - Prob. 12E Ch. 11.1 - Prob. 13E Ch. 11.1 - Prob. 14E Ch. 11.1 - Prob. 15E Ch. 11.1 - Prob. 16E Ch. 11.1 - Prob. 23E Ch. 11.1 - Prob. 24E Ch. 11.1 - Prob. 25E Ch. 11.1 - Prob. 26E Ch. 11.1 - Prob. 27E Ch. 11.1 - Prob. 28E Ch. 11.1 - Prob. 29E Ch. 11.1 - Prob. 30E Ch. 11.2 - Prob. 17E Ch. 11.2 - Prob. 18E Ch. 11.2 - Prob. 19E Ch. 11.2 - Prob. 20E Ch. 11.2 - Prob. 21E Ch. 11.2 - Prob. 22E Ch. 11.2 - Prob. 23E Ch. 11.2 - Prob. 24E Ch. 11.2 - Prob. 25E Ch. 11.2 - Prob. 26E Ch. 11.3 - Prob. 1E Ch. 11.3 - Prob. 2E Ch. 11.3 - Prob. 3E Ch. 11.3 - Prob. 4E Ch. 11.3 - Prob. 5E Ch. 11.3 - Prob. 6E Ch. 11.3 - Prob. 7E Ch. 11.3 - Prob. 8E Ch. 11.3 - Prob. 9E Ch. 11.3 - Prob. 10E Ch. 11.3 - Prob. 11E Ch. 11.3 - Prob. 12E Ch. 11.3 - Prob. 13E Ch. 11.3 - Prob. 14E Ch. 11.3 - Prob. 15E Ch. 11.3 - Prob. 16E Ch. 11.3 - Prob. 17E Ch. 11.3 - Prob. 18E Ch. 11.3 - Prob. 19E Ch. 11.3 - Prob. 20E Ch. 11.3 - Prob. 21E Ch. 11.3 - Prob. 22E Ch. 11.3 - Prob. 23E Ch. 11.3 - Prob. 24E Ch. 11.3 - Prob. 25E Ch. 11.3 - Prob. 26E Ch. 11.3 - Prob. 27E Ch. 11.3 - Prob. 28E Ch. 11.3 - Prob. 29E Ch. 11.3 - Prob. 30E Ch. 11.3 - Prob. 31E Ch. 11.3 - Prob. 32E Ch. 11.3 - Prob. 33E Ch. 11.3 - Prob. 34E Ch. 11.3 - Prob. 35E Ch. 11.3 - Prob. 36E Ch. 11.3 - Prob. 37E Ch. 11.3 - Prob. 38E Ch. 11.3 - Prob. 39E Ch. 11.4 - Prob. 1E Ch. 11.4 - Prob. 2E Ch. 11.4 - Prob. 3E Ch. 11.4 - Prob. 4E Ch. 11.4 - Prob. 5E Ch. 11.4 - Prob. 6E Ch. 11.4 - Prob. 7E Ch. 11.4 - Prob. 8E Ch. 11.4 - Prob. 9E Ch. 11.4 - Prob. 11E Ch. 11.4 - Prob. 12E Ch. 11.4 - Prob. 13E Ch. 11.4 - Prob. 14E Ch. 11.4 - Prob. 15E Ch. 11.4 - Prob. 16E Ch. 11.4 - Prob. 17E Ch. 11.4 - Prob. 18E Ch. 11.4 - Prob. 19E Ch. 11.4 - Prob. 20E Ch. 11.4 - Prob. 21E Ch. 11.4 - Prob. 22E Ch. 11 - Prob. 1CR Ch. 11 - Prob. 2CR Ch. 11 - Prob. 3CR Ch. 11 - Prob. 4CR Ch. 11 - Prob. 5CR Ch. 11 - Prob. 6CR Ch. 11 - Prob. 7CR Ch. 11 - Prob. 8CR Ch. 11 - Prob. 11CR Ch. 11 - Prob. 12CR Ch. 11 - Prob. 13CR Ch. 11 - Prob. 15CR Ch. 11 - Prob. 16CR Ch. 11 - Prob. 17CR

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The time period of the bead of mass ( m ) when x ( 0 ) is near x 1 and x ′ ( 0 ) = 0 such that the bead slides along the wire and the shape function of the wire is z = f ( x ) .

Solution Summary: The author calculates the tangential force F due to the weight of the bead, which is as follows: x_1=(0).

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The time period of the bead of mass (m) when x(0) is near x1 and x′(0)=0 such that the bead slides along the wire and the shape function of the wire is z=f(x).

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Chapter 11 Solutions