Waves due to a floating sphere making heaving oscillations 4 
It may be noted that it is only the second equation in (15) which includes a term in 
f on the right. 
4. For large values of the frequency parameter, that is foo, we may expect 
the solution to approach that appropriate to the free surface condition ¢ = 0, namely, 
g= 108 WY coset. (17) 
The equations (15) are consistent with this if C, D approximate to zero, Ay, Az, ... 
being of order /~ and fA, approximating to unity. However, there are difficulties 
in evaluating some of the integrals involved for large values of /, due partly to 
having takenaconcentrated point source at the origin instead of a distributed source. 
We shall therefore limit the calculations to moderate values of f. For large values 
the problem is better treated separately, possibly by the method used by Ursell 
(1953) for the similar two-dimensional case. 
For small values of /, the free surface condition approximates to 0¢/0z = 0 and 
the solution is then 
{1 a> co qn 
b= orn + ont An Pon(t) cos at, (18) 
re (= 1)" (4n +1) Cn!) 
Pu) P, dats 
where 4, =I (1) Ponf ~ 22n(2n — 1) (2n + 1) (2n + 2) (n!)? ue) 
This is given by (15) and (16) with £ = 0, the coefficient Z) being then unity and the 
other L and M coefficients being zero. The series in (18) is convergent. We shall 
assume, in the general case, the convergence of the solution in (10) with the unknown 
coefficients derived from (15) and (16). 
5. The expression for J given in (12) may be put into a more suitable form for 
computation. If we write J for the integral in (12) we have 
@wsin (fu cos A) +cos(fucosf) ,/, .. 
ioe Pal, 14tu2 De (Busia PNG, (20) 
u sin (pu) + cos (pu) 
1+u? 
Further, ifwe put XX =|. K (qu) du, (21) 
X reduces to }7*{H)(q) — Yo(q)} for p = 0, ¢> 0, where H is the Struve function. Also 
we have ax 
op 
ai 1 
2 
(p+?) oe 
+X = =| Ky(qu) cos (pu) du = 
From this we deduce for the integral in (20) the form 
1 
47°{H,( 2 sin 0) — Yo( sin )} e708 + lr ef o8 al (tan? 0+ ¢2)-2e ftcostdt, (23) 
0 
Using (23) and (20), we find, after some reduction, 
L=1+/?—inf?e 4° {Hf sin @) + ¥)(f sin @)} cos 6 
+{H,(fsin@)+Y,(fsin @)} sin 0] — 62 cos Ae" (4 + f8Bcos@), (24) 
605 
