Laws of Pendulums
There are three laws in regard to the movement of simple pendulums that are well to remember. 1. The number of vibrations performed by pendulums in a given time are inversely as the square roots of the lengths. If the bob is displaced from the vertical and released, it will return, and ascend to an equal distance on the other side in virtue of its weight. The velocity of movement of the pendulum is in accordance with the laws of falling bodies for the moving pendulum is no more than a falling body under certain restrictions. If we assume the pendulum to be displaced laterally until its rod is in a horizontal position, it will be seen that the distance through which it descends is equal to the length of the pendulum. Hence it follows that the descent of a short pendulum will, in virtue of the laws above referred to, take place in much less time than that of a long one. Thus, consider the case of a short pendulum whose length is a quarter of that of the longer one; the shorter will travel twice as quickly as the longer, or, in other words, it will perform two oscillations while the longer performs one. The lengths are as 1:4 and the square roots of these numbers are 1 and 2; thus the number of oscillations are inversely as these square roots. The times occupied in the descent, or the periods of the oscillations, are proportional to the square roots of the length. If the longer pendulum fall in 2 seconds, the shorter falls twice as quickly and will therefore reach the vertical in 1 second; and 2 and 1 are the square roots of the length 4 and 1. 3. The lengths are inversely proportional to the squares of the number of oscillations in a given time. If we observe that:
The lengths are ___________________________________1 and 4. The corresponding number of oscillations __________2 and 1. The squares of these number _______________________4 and 1. we have some evidence of the truth of this law. These several laws will enable us to determine the length of pendulum for any case that presents itself. Saunier gives the following method for determining the length of a simple pendulum, the numbers of oscillations being given, or vice versa. Let the pendulum be required to perform 7,000 oscillations in an hour. The simple seconds pendulum (at Paris) measures 994 mm. (39.13in.), and it makes 60 x 60 or 3,600 oscillations per hour; we thus, from the law 3 above given, have the proportion: 7,000 x 7,000 : 3,600 x 3,600 : : 994 : x or 49,000,000 : 12,960,000 : : 994 : x Dividing the product of the means by the known extreme we obtain: X = 12,960,000 x 994/49,000,000 = 262.9 mm. (10.35ins.) If the length of a pendulum be given, say 121 mm. (4.764ins), the number of oscillations will be calculated in accordance with law 1: <??>121 : <??>994 : : 3,600 : x (The radix sign <??> indicates that the root is to be extracted from the number placed under it.) or 11 : 31.525 : : 3600 : x whence we obtain x=31.525 x 3,600/11=10,317, the required number of oscillations.
From the above it will be seen that it was very easy to find the length of the pendulum for a given number of vibrations, or vice versa, although such calculations are quite useless while we have the accompanying table of lengths of pendulums for any number of oscillations. However, it may often be necessary to solve such problems where you do not have access to such tables, and it is therefore valuable to know just how to proceed. M. Millet gives another method which is rather more simple. Take as a basis for calculation the pendulum that performs one oscillation in an hour, the length of which is 12,880,337.93 meters (507,109,080 inches) or in round numbers, 12,880,338 meters; by law 3 we obtain the following proportion: 12,880,338:x (the length): : V2 (the velocity): 12 Since the square of 1 is 1, it is only necessary to replace x by the length (if this is given), or V by the number of oscillations in an hour, (if they are pre-determined), and the value of the unknown quantity will be obtained. Example: How many oscillations will be made by a pendulum measuring 305mm. (12.008 inches)?
We have the proportion: 12,880,338: 0.305: : V2: 1.
Dividing the product of the extremes by the known mean, (To extract the square root of a whole number, place a point or dot over the units' place of the given number, and thence over every second figure to the left of that place, thus dividing the whole number into several periods. The number of points will show the number of figures in the required root. Find the greatest number whose square is contained in the first period at the left; this is the first figure in the root, and may be ascertained by the aid of the following table; Number 1, 4, 9, 16, 25, 36, 49, 64, 81. Square Root 1, 2, 3, 4, 5, 6, 7, 8, 9. Subtract the square of the number so determined from the first period and to the remainder bring down the second period. Divide the number thus formed, omitting the last figure, by twice the part of the root already obtained, and annex the quotient to the root and also to the divisor. Then multiply the divisor, as it now stands, by the part of the root last obtained, and subtract the product from the number formed, as above mentioned, by the first remainder and the second period. If there be more periods to be brought down the operation must be repeated, and if, when all the periods have been so brought down, there is a remainder, the given number has no exact square root. If the number be a decimal fraction, or a whole number and a decimal combined, proceed in a similar manner, but observe that a point must always occur over the units' figure and on alternate figures from it on either side to the right and left. A decimal point will be placed in the square root immediately before bringing down the first decimal period, and in cases where the given number has no exact root, it may be approximated to by bringing down successive pairs of ciphers. Example: Extract the square root of 273,529. 273529(523 25 102|235 |204 1043|3129 |3129 Applying the above rule, the square of 5 or 25, the largest contained in 27, is subtracted from the first period, and to the remainder, the second period, 35, is attached. The divisor for the dividend so formed is obtained by doubling the portion of the root already determined (5), and annexing 2 to the 10, since 10 will divide twice into 23, the dividend with the last figure omitted. The 2 is also added to the quotient as forming a figure in the root, and 102 multiplied by it as in ordinary division. The next period, 29, having been brought down to the remainder thus obtained, a similar operation is again gone through the entire quotient, so far as it has been determined, being each time doubled.) V2=12,880,338/0.305=42,230,616, and V will be the square root of this number, or 6,498 oscillations per hour. If the dimensions are given in English inches, the numbers 507,109,080 and 12,008 would be employed thus: 507,109,080: 12.008: : V2: 1; V2=507,109,080/12.008=42,230,936.
The slight difference in the results is due to the non-equality of the two approximate figures given above. Another example is given in which it will suffice to indicate the several stages of the calculation. Example: What should be the length of a pendulum to give 4,100 oscillations per hour? 12,880,338 : x : : 4,1002:1; 12,880,338 : x : : 16,810,000: 1; x=12,880,338/16,810,000= 0.766 meters (30.158 inches). Table Showing the Length of a Simple Pendulum That performs in one hour any given number of oscillations, from 1 to 20,000, and the variation in this length that will occasion a difference of 1 minute in 24 hours.
Calculated by E. Gourdin.
Number of Oscillations per Hour. Length in Millimeters. Variation in Length for One Minute in 24 Hours in Millimeters. Number of Oscillations per Hours. Length in Millimeters. Variation in Length for One Minute in 24 Hours in