## 2pPA12. Answer to criticism of my treatment of nonscattering of sound by sound.

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]t[sub 0]+T. In f[sub 2] the (delta) generate displacement boundary conditions and x[sub 2](t)=bU(t-t[sub 0])cos (omega)[sub 0](t-t[sub 0]) for t[sub 0]t[sub 0]+T. The complete solution x=x[sub 1]+x[sub 2] is x(t)=0 for tt[sub 0]+T confirming the fact that x(t)=G(t). This one-dimensional analogy confirms the fact that there is no response outside the interaction region t[sub 0]