Explanation Given: Natural frequency is 2200 Hz . Equation of motion of box y = δ m sin ω f t . Acceleration of box δ m ω f 2 . Frequency of vibration f f . Concept used: Consider the system as shown below: Write the expression for amplitude of forced vibration. x m = δ m 1 − ( ω f 2 ω n 2 ) ...... (1) Here, x m is the amplitude of forced vibration, δ m is the maximum displacement, ω f is the forced frequency and ω n is the natural frequency of system. Write the Expression for total displacement. x = x m sin ω f t Substitute δ m 1 − ( ω f 2 ω n 2 ) for x m in the above expression. x = δ m sin ω f t 1 − ( ω f 2 ω n 2 ) ...... (2) Here, x is the total displacement. Write the expression for relative motion as. z = x − y Substitute δ m sin ω f t 1 − ( ω f 2 ω n 2 ) for x and δ m sin ω f t for y in above expression. z = ( δ m 1 − ( ω f 2 ω n 2 ) − δ m ) sin ω f t ...... (3) Here, z is the relative displacement. Write the general equation of relative displacement of motion. z = z m sin ω f t ...... (4) Here, z m is the maximum amplitude of relative displacement. Compare equation (4) and (3). z m = δ m ( ( ω f ω n ) 2 1 − ( ω f ω n ) 2 ) Multiply the above expression with ω n 2

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ISBN: 9781260265521

Author: BEER

Publisher: MCG

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Chapter 19.4, Problem 19.123P

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Percentage error with frequency of vibration of 600Hz.

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Chapter 19.4, Problem 19.122P

Chapter 19.4, Problem 19.124P

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Chapter 19 Solutions

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS

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