### ASA 127th Meeting M.I.T. 1994 June 6-10

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

**Peter J. Westervelt
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*Dept. of Phys., Brown Univ., Providence, RI 02912
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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]