Explanation Given information: The length of the cord BC is l . The length of the rod AB is L . The mass of the rod is m . The acceleration due to gravity is g . Calculation: Show the position 1 of the system with displacement as in Figure (1). For position 1 (Maximum deflection): Let θ m be the small angle between the cord CB . When the rod AB moves from the equilibrium position the center of gravity ( G ) moves upward then the vertical displacement of the G is y m . The vertical displacement of point B is y B . Express the vertical displacement ( y B ) of the point B : y B = l − l cos θ m = l ( 1 − cos θ m ) ( ∵ ( 1 − cos θ m = 2 sin 2 θ m 2 ) ) = l ( 2 sin 2 θ m 2 ) For small oscillation 2 sin 2 θ m 2 ≈ θ m 2 2 . y B = l θ m 2 2 Express the vertical displacement ( y m ) of the center of gravity: y m = y B 2 Substitute 1 / 2 l θ m 2 for y B . y m = 1 / 2 l θ m 2 2 = 1 4 l θ m 2 Write the expression for the kinetic energy ( T 1 ) : T 1 = 0 Write the expression for the potential energy ( V 1 ) : V 1 = m g y m Substitute 1 / 4 l θ m 2 for y m . V 1 = m g ( 1 / 4 l θ m 2 ) = 1 4 m g l θ m 2 For position 2: Show the position 2 of the system with velocity as in Figure (2).

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Author: BEER

Publisher: MCGRAW-HILL HIGHER EDUCATION

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Chapter 19.3, Problem 19.93P

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The frequency (fn) of oscillation when the end B of the rod is given a small horizontal displacement and released.

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Chapter 19.3, Problem 19.92P

Chapter 19.3, Problem 19.94P

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Chapter 19 Solutions

VECTOR MECHANIC

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The frequency ( f n ) of oscillation when the end B of the rod is given a small horizontal displacement and released.

Solution Summary: The author calculates the frequency of oscillation when the end B of the rod is given a small horizontal displacement and released.

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The frequency (fn) of oscillation when the end B of the rod is given a small horizontal displacement and released.

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Chapter 19 Solutions