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Building Design and Construction Handbook - Merritt - Ventech!

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5.144 SECTION FIVE<br />

�<br />

1 1 kg<br />

ƒ � � (5.240)<br />

T 2� W<br />

If, at time t � 0, the mass has an initial displacement y 0 <strong>and</strong> velocity v 0, substitution<br />

in Eq. (5.237) yields A � v 0/� <strong>and</strong> B � y 0. Hence, at any time t, the mass<br />

is completely located by<br />

v 0<br />

y � sin �t � y0 cos �t (5.241)<br />

�<br />

The stress in the spring can be computed from the displacement y.<br />

Vibrations of Lumped Masses. In multiple-degree systems, an independent differential<br />

equation of motion can be written for each degree of freedom. Thus, in<br />

an N-degree system with N masses, weighing W 1, W 2,...,W N lb, <strong>and</strong> N 2 springs<br />

with constants k rj (r � 1,2,...,N; j � 1,2,...,N), there are N equations of<br />

the form<br />

2 W d y<br />

N<br />

� rj j<br />

r r<br />

� k y � 0 r � 1,2,...,N (5.242)<br />

2 g dt j�1<br />

Simultaneous solution of these equations reveals that the motion of each mass can<br />

be resolved into N harmonic components. They are called the fundamental, second<br />

third, etc., harmonics. Each set of harmonics for all the masses is called a normal<br />

mode of vibration.<br />

There are as many normal modes in a system as degrees of freedom. Under<br />

certain circumstances, the system could vibrate freely in any one of these modes.<br />

During any such vibration, the ratio of displacement of any two of the masses<br />

remains constant. Hence, the solution of Eqs. (5.242) take the form<br />

N<br />

r � rn n n<br />

n�1<br />

y � a sin � (t � � ) (5.243)<br />

where a rn <strong>and</strong> � n are constants to be determined from the initial conditions of the<br />

system <strong>and</strong> � n is the natural circular frequency for each normal mode.<br />

To determine � n, set y 1 � A 1 sin �t; y 2 � A 2 sin �t ....Then, substitute these<br />

values of y r <strong>and</strong> their second derivatives in Eqs. (5.242). After dividing each equation<br />

by sin �t, the following N equations result:<br />

W1 2 � 11 � 1 12 2 1N N<br />

k � � A � k A � ���� k A � 0<br />

g<br />

W2 k21A1��k22 � � A2 � ���� k2NAN � 0<br />

g<br />

.................................................<br />

WN 2 � �<br />

kN1A1 � kN2A2 � ���� kNN � � AN � 0<br />

g<br />

(5.244)<br />

If there are to be nontrivial solutions for the amplitudes A 1, A 2,...,A N, the<br />

determinant of their coefficients must be zero. Thus,

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