Explanation Given information: The equation of ( y ) is δ m sin ω f t . The natural frequency ( ω n ) is 2200 Hz . The frequency of vibration ( ω f ) is 600 Hz . The maximum acceleration ( a m ) is δ m ω f 2 . Calculation: The expression or the displacement transmitted ( x m ) as follows: x m = δ m 1 − ( ω f ω n ) 2 Calculate the forced response ( x ) using the relation: x = x m sin ω f t Substitute δ m 1 − ( ω f ω n ) 2 for x m . x = ( δ m 1 − ( ω f ω n ) 2 ) sin ω f t Calculate the relative motion ( z ) using the relation: z = x − y Substitute ( δ m 1 − ( ω f ω n ) 2 ) sin ω f t for x and δ m sin ω f t for y . z = ( δ m 1 − ( ω f ω n ) 2 ) sin ω f t − δ m sin ω f t z = [ δ m 1 − ( ω f ω n ) 2 − δ m ] sin ω f t z m = δ m 1 − ( ω f ω n ) 2 − δ m z m = δ m [ 1 1 − ( ω f ω n ) 2 − 1 ] Simplify the Equation

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ISBN: 9781259633133

Author: BEER

Publisher: MCG

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Chapter 19.4, Problem 19.123P

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Find the percent error (E) when the frequency of the vibration is 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...,STAT.+DYNA.(LL)-W/ACCESS

Ch. 19.1 - A particle moves in simple harmonic motion....Ch. 19.1 - Prob. 19.2P Ch. 19.1 - Prob. 19.3P Ch. 19.1 - Prob. 19.4P Ch. 19.1 - Prob. 19.5P Ch. 19.1 - Prob. 19.6P Ch. 19.1 - Prob. 19.7P Ch. 19.1 - Prob. 19.8P Ch. 19.1 - Prob. 19.9P Ch. 19.1 - Prob. 19.10P

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Find the percent error ( E ) when the frequency of the vibration is 600 Hz .

Solution Summary: The author calculates the percent error ( E ) when the frequency of the vibration is 600Hz.

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Find the percent error (E) when the frequency of the vibration is 600Hz.

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