Units and Dimensions - Deriving Relations Among Quantities

Description:

Deriving Time Period of a Simple Pendulum

Dimensional analysis helps us deduce relations between physical quantities.

Let’s assume that Time Period, ‘t’, of a simple pendulum depends on

• Mass of the bob, ‘M’,
• Length of the string, ’l’,
• gravitational acceleration, ‘g’ and,
• temperature of the room, ‘θ’.

If you wish, you can assume more dependencies.

Time period is proportional to arbitrary powers of each of the dependent quantities.

t ∝ M^a l^b g^c θ^d

Removing the proportionality sign −

t = K × M^a l^b g^c θ^d

This equality should be dimensionally consistent.

Dimensions of L.H.S

[T]

Dimensions of R.H.S

[M]^a[L]^b([L][T]^-2)^c[θ]^d = [M]^a[L]^{(b + c)}[T]^-2c[θ]^d

Dimensional equality implies

[T] = [M]^a[L]^{(b + c)}[T]^-2c[θ]^d

OR

[M]⁰L⁰[θ]⁰[T] = [M]^a[L]^{(b + c)}[T]^-2c[θ]^d

Hence, the equations we obtain by comparing exponents are

• a = 0
• b + c = 0
• -2c = 1
• d = 0

Solving these equations, we obtain

• a = 0
• b = 1/2
• C = -1/2
• d = 0

Putting the values of a, b, c and d in the original equation, we obtain

t = K × M⁰ l^1/2 g^-1/2 θ⁰

OR

t = K√l/g

Here, ‘K’ is the constant of proportionality and its value can be determined experimentally.

Value of ‘K’ comes out to be, K = 2π

Limitations of Dimensional Analysis

One CANNOT deduce dimensionless quantities using dimensional analysis.

• E.g., the value of ‘K’ can be only obtained by experiments.

It does not distinguish between physical quantities having same dimensions.

• E.g.: [Work] = [Torque] = [M][L]²[T]⁻²

One CANNOT deduce equations of the form: x₀ = ut + 1/2 at²