To determine The expression for the natural frequency of the system. Explanation Write the expression for the potential energy for the system. P E = k ( L 1 θ ) 2 − m g L 2 ( 1 − cos θ ) ...... (I) Here, the spring constant is k , and the angular deflection is θ , the acceleration due to gravity is g , the mass of the bob is m , distance of the bob from the pivot point is L 2 , and the distance of the spring from the pivot point is L 1 . For small angle approximation, cos θ ≈ 1 − θ 2 2 . Substitute 1 − θ 2 2 for cos θ in Equation (I). P E = k ( L 1 θ ) 2 − m g L 2 ( 1 − ( 1 − θ 2 2 ) ) = k ( L 1 2 ) θ 2 − m g L 2 θ 2 2 = ( k L 1 2 − m g L 2 2 ) θ 2 ...... (II) For the maximum potential energy, θ = θ max ; and for minimum potential energy, θ = 0 . Substitute θ max for θ and P E max for P E in Equation (II). P E max = ( k L 1 2 − m g L 2 2 ) θ max 2 ...... (III) Write the expression for the maximum angular deflection. θ max = A L 1 Here, the maximum deflection is A . Substitute A L 1 for θ max in Equation (III). P E max = ( k L 1 2 − m g L 2 2 ) ( A L 1 ) 2 Substitute 0 for θ and P E min for P E in Equation (II). P E min = 0 Write the expression for the kinetic energy of the system. K E = 1 2 I ( x ˙ L 2 ) 2 Here, the velocity of the system is x ˙ . Write the expression of energy balance by applying Rayleighs method. K E = P E max − P E min ...... (IV) Substitute 0 for P E min , ( k L 1 2 − m g L 2 2 ) ( A L 1 ) 2 for P E max , and 1 2 I ( x ˙ L 2 ) 2 for K E in Equation (IV). 1 2 I ( x ˙ L 2 ) 2 = ( k L 1 2 − m g L 2 2 ) ( A L 1 ) 2 − 0 1 2 I ( x ˙ L 2 ) 2 = ( k L 1 2 − m g L 2 2 ) ( A L 1 ) 2

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Author: Palm

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Free undamped vibrations

Whenever an object/system displaces in either side of its equilibrium position with the help of an external/internal means, then the object/system moves regularly or randomly on both sides of its equilibrium position. This motion is called vibration. Gen…

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Chapter 4, Problem 4.37P

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The expression for the natural frequency of the system.

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Chapter 4, Problem 4.36P

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EBK SYSTEM DYNAMICS

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