170 REPORT OF THE CALIFORNIA EARTHQUAKE COMMISSION. 
If the movements of the pendulum are simple harmonic, and due to tilts alone without 
linear displacements, we merely interchange — Vga, for Vd?&/dt’ in equation (71); we get 
wy = — S cos (pt — x) ¥ p' (82) 
As @, does not enter the equation as a derivative, no integration is necessary. p’ changes 
its value suddenly from + p’ to — p’, or vice versa, when pt is zero or any multiple of 7; 
therefore w, consists of parts of a simple harmonic curve separated by sudden discontinu- 
ities at these times. But as we can not admit discontinuities in the value of ,, we must 
conclude that when p’ has an appreciable value, a simple harmonic movement of the 
pointer can not be produced by tilts of the ground. 
We are therefore led to reverse the process and determine what movement of the 
pointer would be produced by a simple harmonic tilt of the ground. We must replace 
Vyo, in equation (67) by EH cos(pt—¥,), and integrate the equation after omitting 
‘Vd’E/dt?. (The same solution would apply to the case of simple harmonic linear displace- 
ments if we omitted Vgw,, and replaced Vd?E/dt? by E cos (pt—,); that is, if we made 
&= —(E/Vp’) cos(pt—¥,).) The solution of the equation would then be very simple if 
we could neglect p’, but when we consider this term it becomes rather complicated; but 
it can be found. From the nature of the disturbing force, and on account of the damp- 
ing and friction, it is evident that after a short time the movement of the pendulum 
must become periodic and have the same period as the force. We can therefore write the 
solution in the general form 
a = a, (cos pt — ~,) + a, cos (2 pt — 2) ++ - + ete. = da, cos (mpt — Wp») (83) 
where m represents all positive integers. It is also evident that the arms of the broken 
curve in figure 46 (which represents the movements of the pointer; the continuous 
curve represents the disturbance) from a, to a, 
and from a, to a,, must be perfectly similar, 
as the forces when the pendulum is moving 
in one direction are exactly the negative of 
those when it is moving in the opposite direc- 
tion. Therefore the time the pendulum takes 
to swing from a, to a, will be exactly half 
its period, and if we take the time as zero 
when the pendulum is at a), its maximum 
displacement, we can develop p’ as a series 
of sines of the form of equation (70). Sub- 
stituting these values in equation (67), after omitting Vd?&/dé, and requiring the equation 
to be identically satisfied, we have, with the equation da/dt=0 when t=0, a sufficient 
number of conditions to determine the values of the amplitudes a,, a,, etc., and the phases 
W,, VY, ete., of equation (83), and thus completely to determine this solution. The work 
is rather long and it will be sufficient to give the result. We find for the solution 

a=VEFTSp 
On = & cos pt+ S sin pt— > 4 n*r 2 cmp cos mpt + (m*p* — n*) sin mpt (84) 
p p rm (m*p* — n?)? + (2 cmp)? 
where m has all odd positive integral values greater than 1. 4,, = 0, when m is even. 
4 n?r mp? — n? 
S= Se eee er eee eee 
2 rm (mp? —n*)? + (2 nmp)? 
with the same values of m. 
