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638 ON THE RESULTANT OF A NUMBER OP UNIT VIBRATIONS [442
In the general case the limits for r and 9 are interdependent. The possible range for Q is from — \a. to -\-\OL (a < TT), but we require the range when r is prescribed. In virtue of the symmetry it suffices to consider a positive 6, and we begin by supposing a less than TT, so that the extreme values of r are 2 cos \ a and 2. We proceed to consider the relations by which the limiting values of r and 6 are connected.
For a given (positive) 9 less than %a the upper limit of r is 2 and the lower limit is 2 cos(^a — 0). When 0 >^a, there are no corresponding values of r. In Fig. 2, .where a is taken to be ^TT, the shaded area gives the possible values of r corresponding to any 6, or conversely the values of 6 corresponding to a prescribed ?\
In order to find the chances of a given 6, we integrate with respect to ?• in (29). We find
as the chance that 0, if positive, lies between 6 and (9 + dd. If we integrate (31) again with respect to 6 between 0 and |-a, we get £, the correct 'value, as there is an equal chance of d being negative.
Again, in order to find the chance of a prescribed r, when 6 is free to vary, we have to integrate (29) first with respect to 6. Referring to Fig. 2, we see that when r<2cos(^a); there are no corresponding values of 0, and that when r lies "between 2 cos Q a) and 2, the limits for Q are 0 and \ a. — cos"1 (^-r). In the first case there is no possibility of r lying between r and r + dr ; in the second case the probability is
which must be doubled when we admit, as we must, negative values of 0. If we integrate (32) as it stands again, with respect to r, we find the correct value, since
We may regard (31) and (32) as the solution of the problem in the case where a < TT.
When a>7r, d may lie outside the limits +£« applicable to 01 and $„, and the question becomes more complicated. It appears that we must distinguish two cases under this head, (i) where TT < a < 3vr/2, and (ii) where 37T/2 < a < 2-7T.
First for TT < a < 3?r/2, Fig. 4, where a is supposed to be o?r/4.
From 0 = 0 to '0 = £ (a - TT), r ranges from 0 to 2. From 0 = £ (a - -TT) to 0 = £a, r ranges from 2cos(£a-0) to 2. At 0 = %ct the lower and upper find