# 9.1.4: The Pendulum Class Problem

The only known problem that dimensional analysis can solve (to some degree) is the pendulum class problem. In this section several examples of the pendulum type problem are presented. The first example is the classic Pendulum problem.

Example 9.4 Fig. 9.2 Figure for example

length of pendulum [$$\ell$$]. Assume that no other parameter including the mass affects the problem. That is, the relationship can be expressed as
$\label{pendulum:form} \omega = f \left(\ell, g \right) \tag{10}$
Notice in this problem, the real knowledge is provided, however in the real world, this knowledge is not necessarily given or known. Here it is provided because the real solution is already known from standard

Solution 9.4

The solution technique is based on the assumption that the indexical form The Indexical form

$\label{pendulum:basic} \omega = C_1\times \ell^{a} g^{b} \tag{11}$

The solution functional complexity is limited to the basic combination which has to be in some form of multiplication of $$\ell$$ and $$g$$ in some power. In other words, the multiplication of $$\ell\,g$$ have to be in the same units of the frequency units. Furthermore, assuming, for example, that a trigonometric function relates $$\ell$$ and $$g$$ and frequency. For example, if a $$\sin$$ function is used, then the functionality looks like $$\omega = \sin (\ell\,g)$$. From the units point of view, the result of operation not match i.e. ($$sec \neq \sin\,(sec)$$). For that reason the form in equation (11) is selected. To satisfy equation (11) the units of every term are examined and summarized the following table.

Table 9.2 Units of the Pendulum Parameters

 Parameter Units Parameter Units Parameter Units $$\omega$$ $$t^{-1}$$ $$\ell$$ $$L^{1}$$ $$g$$ $$L^{1}t^{-2}$$

Thus substituting of the Table 9.2 in equation (11) results in

\label{pendulum:govIndalign} t^{-1} = C_1\left( L^1 \right)^a \, \left(L^1 \,t^{-2}\right)^b \Longrightarrow L ^{a +b} t^{-2\,b} \tag{12}
after further rearrangement by multiply the left hand side by $$L^0$$ results in
$\label{pendulum:govRe} L^{0}t^{-1} = C\,L ^{a +b} t^{-2\,b} \tag{13}$
In order to satisfy equation (13), the following must exist
$\begin{array}{ccc} \label{pendulum:reEq} 0 = a + b & and & -1 = \dfrac{-2}{b} \end{array} \tag{14}$
The solution of the equations (14) is $$a= -1/2$$ and $$b= -1/2$$. Thus, the solution is in the form of
$\label{pendulum:solution} \omega = C_1\, \ell^{1/2} \, g^{-1/2} = C_1 \, \sqrt{\dfrac{g}{\ell}} \tag{15}$
It can be observed that the value of $$C_1$$ is unknown. The pendulum frequency is known to be
$\label{pendulum:frequency} \omega = \dfrac{1}{2\pi} \sqrt{\dfrac{g}{\ell}} \tag{16}$

What was found in this example is the form of the solution's equation and frequency. Yet, the functionality e.g. the value of the constant was not found. The constant can be obtained from experiment for plotting $$\omega$$ as the abscissa and $$\sqrt{\ell/g}$$ as ordinate. According to some books and researchers, this part is the importance of the dimensional analysis. It can be noticed that the initial guess merely and actually determine the results. If, however, the mass is added to considerations, a different result will be obtained. If the guess is relevant and correct then the functional relationship can be obtained by experiments.

## Contributors

• Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.

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