The method of dimensions can sometimes be used to deduce relation among the physical quantities. For this we should know the dependence of the physical quantity on other quantities (upto three physical quantities or linearly independent variables) and consider it as a product type of the dependence. Let us take an example.
Example 1
Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length (l), mass of the bob (m) and acceleration due to gravity (g). Derive the expression for its time period using method of dimensions.
Answer
The dependence of time period T on the quantities l, g and m as a product may be written as:
T = k$l^xg^ym^z$
where k is dimensionless constant and x, y and z are the exponents.
By considering dimensions on both sides, we have
[$L^0M^0T$] = [L]$^x$[LT]$^y$[M ]$^z$
= $L^{x+y}T^{-2y}M^z$
On equating the dimensions on both sides, we have
x + y = 0; –2y = 1; and z = 0
So that x = $\frac{1}{2}$, y = $-\frac{1}{2}$ and z = 0
Then, T = k$l^{1/2}g^{-1/2}$
or $T = k\sqrt{\frac{l}{g}}$
Note that value of constant k can not be obtained by the method of dimensions. Here it does not matter if some number multiplies the right side of this formula, because that does not affect its dimensions.
Actually, k = 2π so that $T = 2\pi\sqrt{\frac{l}{g}}$
Dimensional analysis is very useful in deducing relations among the interdependent physical quantities. However, dimensionless constants cannot be obtained by this method. The method of dimensions can only test the dimensional validity, but not the exact relationship between physical quantities in any equation. It does not distinguish between the physical quantities having same dimensions.
A number of exercises at the end of this chapter will help you develop skill in dimensional analysis.
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