SEISMIC METHODS 791 



Since ^_ £ma, + £mi« 



£».*» = 



J5mo» = 1/2 m IT max 



and Fma» = Aw 



To find the maximum average kinetic energy, differentiate E with respect to ix? and 

 set the derivative equal to zero. This gives 



«/="« = n== (151) 



Therefore, it appears that the free resonant frequency without damping gives the maxi- 

 mum kinetic energy. Equation 150 may now be written in the following form: 



£ = J4 



— - — ) +4»' 



W 



When w = kn or when w = — the above equation has the same value. This is inter- 



k 

 preted as indicating that the kinetic energy is the same for an octave above its maximum 

 as it is for an octave below. However, the value of A is less for the higher than for the 

 lower frequencies. 



The necessary work (JV) to maintain the forced vibration is that done by the 

 force C cos wt. The expression for this work may be written as 



However, 



= i C cos zvt dy = am I 



PF = i C cos zvt dy = am I cos zvt dy 



dy = d \A cos (zvt — <p)] = — Azv sin (zvt — 4>) dt 

 azv 



V (n^ - zjiff + WnW 

 azv 



sin (zvt — <p) dt 



{cos (p sin zvt — sin <p cos zvt} dt 



— TT-r.- { (n" — w^) sin zvt — 2nhzjv cos zvt} dt 



(n- - zv'y + AhhiW 



i 



therefore : JV = \ {2hnzv cos" zvt — (n" — ztf) sin zvt cos zvt} dt 



(n^ — zif)^ + 4h-n~zv" ^ 



In order to avoid fluctuations occurring during the course of a single cycle, it is neces- 

 sary to integrate over a whole number of periods. 



Thus 



/*. 



I sin zvt cos zvt dt ^0 



i 



COS" zvt dt =■ % t 



