Dimensional Analysis 

 

Dimensional analysis is a simple procedure whereby a specific equation that describes some physical phenomena can be identified by equating dimensions of the units involved. This is best explained by an example.

 

Suppose an equation is required that describes the period (t) of a pendulum. Assume the pendulum period will depend on its length (l), the acceleration due to gravity (g), the mass of the bob (m) and the angle though which it swings ()

 

Now the circular measure of an angle has no dimensions so the only quantities on which the pendulum period (t) will depend will be (l),  and (m)

 

                       Thus,                          

 

Where (k) will be a numerical quantity that has no dimensions.

 

The dimensions on the right hand side of the equation must equal the dimensions on the left hand side of the equation.

 

Now, (l) has +1 dimension in length, (g) has +1 dimensions in length and –2 dimensions in time and (m) has +1 dimensions in mass.

 

Thus,          for dimensions of length       a + b = 0

                   for dimensions of time             -2b = +1

            and for  dimensions of mass               c = 0

 

Solving for (a), (b), and (c),

 

                     Hence,                               

 

                            or,                               

 

 

The numerical value of (k) can be found to be (2) a constant by experimental measurement. It is interesting to note that the dimensional analysis confirms that, for a simple pendulum, the period of the pendulum is independent of the mass of the bob.

 

Another example would be the identification of the function of length (l), density () and Young’s modulus (q) of material of construction that defines the oscillation period (t) of tuning fork.

                                                                                                               Dimension

          Thus, linear dimension of the fork                    (l)                             L  

                    Young’s Modulus of the Material            (q)                         ML^-1T^-2

             And Density of the material                           ()                         ML^-3

 

                             Now assume                        

 

Consequently applying the law of dimensions

 

                                    For      M              0  =   +

                                                T               1  =  -2

                                                L                0 =  –– 3

 

Solving for ,  and                 

 

Thus substituting for ,  and