Table of contents | |
Dimensional Analysis | |
Dimensional Formula | |
Applications of Dimensional Analysis | |
Limitations of Dimensional Analysis |
Dimensions of a physical quantity are the power to which the fundamental quantities must be raised to represent the given physical quantity.
Dimensional AnalysisFor example, density = (mass/volume) = mass/(length)3
or density = (mass) (length)-3
Thus, the dimensions of density are 1 in mass and -3 in length. The dimensions of all other fundamental quantities are zero.
For convenience, the fundamental quantities are represented by one-letter symbols.
Dimensional Formula
It is an expression that shows how and which of the fundamental units are required to represent the unit of physical quantity.
For example:
By definition, Velocity is a derived quantity. We can write its dimensional formula using:
Velocity = Displacement/Time
Now, displacement is measured in Length, [L1], and time is measured in [T1] which are both fundamental quantities. Hence, we can represent the dimensional formula of velocity as:
[v] = [s/t] = [L/T] = [LT-1]
Similarly, we can evaluate the dimensional formula of all physically derived quantities. The table given below shows some of the most occurring physical quantities and their dimensions.
This is based on the fact that the product of the numerical value (n) and its corresponding unit (u) is a constant, i.e.,
n[u] = constant
or n1[u1] = n2 [u2]
Suppose the dimensions of a physical quantity are a in mass, b in length, and c in time. If the fundamental units in one system are M1, L1, and T1 and in the other system are M2, L2, and T2 respectively. Then we can write.
Here n1 and n2 are the numerical values in two systems of units respectively.
Using Eq. (i), we can convert the numerical value of a physical quantity from one system of units into the other system.
Example 1. The value of the gravitation constant is G = 6.67 × 10-11 Nm2/kg2 in SI units. Convert it into CGS system of units.
Solution. The dimensional formula of G is [M-1 L3 T-2].
Using equation number (i), i.e.,
Here, n1 = 6.67 x 10-11
M1 = 1 kg, M2 = 1 g = 10-3 kg, L1 = 1 m, L2 = 1 cm = 10-2 m, T1 = T2 = 1s
Substituting in the above equation, we get
or n2 = 6.67 x 10-8
Thus, the value of G in the CGS system of units is 6.67 x 10-8 dyne cm2/g2.
Example 2. Show that the expression of the time period T of a simple pendulum of length l given by
is dimensionally correct.
Solution.
As in the above equation, the dimensions of both sides are the same. The given formula is dimensionally correct.
This principle states that the dimensions of all the terms in a physical expression should be the same. For example, in the physical expression s = ut + 1/2 at2, the dimensions of s, ut, and 1/2 at2 all are the same.
Note: The physical quantities separated by the symbols +, -, =, >, <, etc., have the same dimensions.
Example 3. The velocity v of a particle depends upon the time t according to the equation
.
Write the dimensions of a, b, c, and d.
Solution.
From the principle of homogeneity
If we know the factors on which a given physical quantity may depend, we can find a formula relating the quantity to those factors. Let us take an example.
Example.4. The frequency (f) of a stretched string depends upon the tension F (dimensions of force), length l of the string, and the mass per unit length m of string. Derive the formula for frequency.
Solution. Suppose, that the frequency f depends on the tension raised to the power a, length raised to the power b, and mass per unit length raised to the power c. Then.
f ∝ [F]a[l]b[μ]c
or f = k[F]a[l]b[μ]c
Here, k is a dimensionless constant. Thus,
[f] = [F]0[l]b[μ]c
or [M0L0T-1] = [MLT-2]a[L]b[ML-1]c
or [M0L0T-1] = [Ma+c La+b-c T-2a]
For dimensional balance, the dimensions on both sides should be the same.
Thus, a + c = 0 ...(ii)
a + b - c = 0 ...(iii)
and - 2a = - 1 ...(iv)
Solving these three equations, we get
a = 1/2, c = (-1/2) and b = -1
Substituting these values in Eq. (i), we get
f = k(F)1/2(l)-1(μ)-1/2
or
Experimentally, the value of k is found to be 1/2
The method of dimensions has the following limitations:
(a) By this method the value of dimensionless constant can not be calculated.
(b) By this method the equation containing trigonometrical, exponential, and logarithmic terms cannot be analyzed.
(c) If a physical quantity depends on more than three factors, then a relation among them cannot be established because we can have only three equations by equalizing the powers of M, L, and T.
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