Explanation Given: The mass of the block m is 120 kg .Displacement of beam at A is δ = δ m sin ω f t .Static deflection is 15 mm . Force frequency is ω f = 18 rad/s . Maximum deflection δ m is 10 mm . Concept used: Expression for the natural frequency of the system ω n = k m ...... (1) Here, ω n is the natural frequency of system, k is the spring constant and m is the mass of block. Newton’s second law of motion. ∑ F = m a ...... (2) Here, mass of the block is m , Acceleration of the block is a and Net force is ∑ F . Expression for the amplitude of forced vibration y m = P m k ( 1 − ω f 2 ω n 2 ) ...... (3) Calculation: Stiffness of the beam is calculated as follows: k = m g δ s k = 120 × 9.81 15 × 10 − 3 k = 78480 N/m Substitute 78480 N/m for k and 120 kg for m in equation (1). ω n = 78480 120 ω n = 25.57 rad/s From the above figure, deflection of the beam at C is calculated as follows: y c y a = 3 6 y c = 1 2 y a y c = 1 2 δ m sin ω f t Substitute k 1 2 δ m sin ω f t − k y for ∑ F in equation (2)

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ISBN: 9781259977237

Author: BEER

Publisher: MCG

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Forced undamped vibrations

In mechanical engineering, vibration is expressed as the term that occurs due to an object's backward and forward movement about a point of equilibrium. Generally, there are three types of vibrations which are,

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Chapter 19.4, Problem 19.106P

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Maximum acceleration of the block C.

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Maximum acceleration of the block C.

Solution Summary: The author explains how Newton's second law of motion is used to calculate the amplitude of forced vibration.

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