Peter J. Westervelt
Dept. of Phys., Brown Univ., Providence, RI 02912
Following a suggestion by Professor M. Korman, the transient solution of
an undamped oscillator as a simple analogy to (rho)[sub s]=-W being the
solution of (open square)[sup 2]((rho)[sub s]+W)=0 has been used. Berntson et
al. [J. Acoust. Soc. Am. 86, 1968 (1989)] criticized the solution stating that
solutions to the homogeneous equation need to be appended in order to satisfy
boundary conditions. This is not the case here. Say the forcing junction F(t)
for the oscillator mx+kx=F(t) is derivable from F(t)=mG(t)+kG(t) then
x(t)=G(t). As an example, if the arbitrary transient starts at t[sub 0] and
ends T later then G(t)=[U(t-t[sub 0])-U(t-t[sub 0]-T)]H(t) and
F(t)=m[(delta)(t-t[sub 0])-(delta)(t-t[sub 0]-T)]H(t)+2m[(delta)(t-t[sub
0])-(delta)(t-t[sub 0]-T)]H(t)+[U(t-t[sub 0])-U(t-t[sub 0]-T)][mH(t)+kH(t)].
Let us obtain the solution for the case H(t)=b, a constant. In this case
F(t)=f[sub 1](t)+f[sub 2](t), where f[sub 1](t)=bk[U(t-t[sub 0])-U(t-t[sub
0]-T)] and f[sub 2](t)=bm[(delta)(t-t[sub 0]) -(delta)(t-t[sub 0]-T)]. f[sub
1](t) is a square wave and the response to it is x[sub 1](t)=bU(t-t[sub
0])(1-cos (omega)[sub 0]t) for t[sub 0]