DYNAMICS OF PACKAGE CUSHIONING 399 



must find y and j at / = tr . Since Ws is motionless at / = /r , the relative 

 displacement and velocity at that time are identical with xo and .f2 respec- 

 tively. The former is simply the stretch of the spring necessary to just 

 pull the mass nis off the floor, i.e., 



{y]t^tr = i^^i-'r = -~- (2.4.7) 



To find the velocity at / = /r , substitute (2.4.7) in (2.2.4) and also substi- 

 tute Ir for / in the latter. This gives an equation for determining/,- . Then, 

 returning to (2.2.4), differentiate it once to obtain .fo and substitute for / the 

 value tr just found. The result is 



t=,, = [y],=,^ = - a/ 2gk - 

 The solution of (2.4.3) with initial conditions (2.4.7) and (2.4 



i,l,.„ = ly,,., = - ^/ 2,, _ ^-"^1<2^^^ . (2.4.8) 



y = -I /l/V' -^^ sin (u.( - n, (2.4.9) 



where f = co/r — tan 



k2 



-1 o}[y]i^t^ 



[yU. 



We are now in a position to find the acceleration of the packaged item 

 after rebound. Substitute y of (2.4.9) for .vo — Xs in (2.4.1) to obtain 



*2 = g + - i/lgh - ^^ sin (co/ - f ). (2.4.10) 



To obtain a simple formula for the ratio of the maximum accelerations 

 after and before rebound, let us assume that both are much greater than 

 gravitational acceleration. Then if 



Gr — maximum number of g's after rebound, (maximum of (2.4.10)) 

 Gm = A/ Yr^ ~ maximum number of g's before rebound, 

 we find, from (2.4.10), neglecting the term g outside the radical, 



G„. W2 + Ws y 2hk 



/- 



& W. ./. W3 (2,^11) 



il 



Hence, the maximum acceleration after rebound is always less than the 

 maximum acceleration before rebound. Therefore, conditions after re- 

 bound need only be examined when the frequency after rebound (see equa- 

 tion (2.4.6)) is near the natural frequency of vibration of a critical element 

 of the packaged item (see Section i.5). 



