Physics Lesson 1.2.4 - Dimensional Analysis

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Welcome to our Physics lesson on Dimensional Analysis, this is the fourth lesson of our suite of physics lessons covering the topic of Length, Mass and Time, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Dimensional Analysis

Dimensional analysis is a method of analysis in which physical quantities are expressed in terms of their fundamental dimensions that is often used when there is not enough information to set up precise equations.

For example, let's suppose you don't remember the equation of velocity v in a uniform motion (the correct equation is v = Δx / t where Δx is the displacement and t is the moving time). You are somehow puzzled whether the correct formula is the abovementioned one, or it is v = Δx × t or v = t /Δx. In this case, you can use the dimensional analysis to remove the doubts if you know the unit of velocity. Thus, giving that the unit of velocity is metre per second [m / s] where metre is the unit of displacement Δx and second is the unit of time t, it is obvious the equation of velocity is v = Δx / t.

However, Dimensional Analysis has an important drawback. Through this method, we are not able to identify dimensionless constants in a formula. For example, if we know the unit of energy is joule [J], we are not able to find the correct formula of the kinetic energy (KE) as

[J] = [N × m] = [kg × m /s2 × m ]= [kg × m2 /s2 ] = [kg × (m /s)2 ]

(Remember that 1 N = 1 kg × m/s2). Giving that [kg] is the unit of mass m and [m/s] is the unit of velocity v, we may (wrongly) conclude that

KE=m × v2

However, the correct formula of kinetic energy (see the article: Kinetic Energy), is

KE=1/2 m × v2

Hence, we missed the constant dimensionless constant 1/2 when tried to find the equation based on the dimensional analysis. Therefore, to be correct in our response, we must write

KE ~ m × v2

(the symbol ̴ means "is proportional to"). Thus, the correct way to write the equation of KE found by using the dimensional analysis method is

KE=C × m × v2

where C is a dimensionless constant. In the specific case, C = 1/2.

Example:

A student believes the speed of sound v in a gas depends on three factors: pressure p density of the medium ρ and the volume V of the medium. He also has determined the approximate equation

v=C × px × ρy × Vz

where C is a dimensionless constant.

  1. Calculate the values of x, y and z.
  2. Does the speed of sound depend on all the above 3 factors?
  3. Write the correct formula for the speed of sound in a gas.

Useful formulae:

Pressure = Force/Area
Density = Mass/Volume
Force = Mass × Acceleration
Acceleration = Change in velocity/Time
Speed = Distance/Time

Solution

First, let's write all units in terms of metre, kilogram and second. We have

Unit of speed = Unit of distance/Unit of time

In symbols,

Unit of speed=[m/s]

Also,

Unit of pressure = Unit of force/Unit of area = Unit of mass × Unit of acceleration/Unit of area
= Unit of mass × Unit of velocity/(Unit of time))/(Unit of area) = Unit of mass × Unit of velocity/Unit of area × Unit of time

Giving that speed and velocity have the same unit [m/s], we obtain after substituting the symbols of the respective units

Unit of pressure = [kg × m/s)/(m2 × s ] = [kg/m × s2]

The same method is used for the other quantities as well. Thus,

Unit of density = Unit of mass/Unit of volume

In symbols, we have

Unit of density = [kg/m3]

where [m3] is obviously the unit of volume. Hence, combining all these units, we obtain

[m/s]=[kg/m × s2]x × [kg/m3]y × [m3/1]z
[m1/s1 ] = [kgx × kgy × m3z/mx × s2x × m3y]
[m1/s1 ] = [kgx+y × m3z)/mx+3y × s2x ]
[m1/s1 ] = [kgx+y × m3z-x-3y/s2x ]

Hence, we obtain three different equations:

Equation 1: related to kg

x + y = 0

Equation 2: related to m

3z - x - 3y = 1

Equation 3: related to s

2x = 1

From the third equation, we can find the value of x (x = 1/2). Also, since in the first equation we have x + y = 0=>y = - x = - 1/2.

Therefore, substituting the above two values in the second equation, we obtain

3z - 1/2 - 3 × - 1/2 = 1
3z - 1/2 + 3/2 = 1
3z + 2/2 = 1
3z + 1 = 1
3z = 0
z = 0

Since the volume's exponent is zero (z = 0), the speed of sound does not depend on the volume. This means the volume does not appear in the formula of the sound speed in a gas.

The correct formula is

v = C × p1/2 × ρ - 1/2 = C × p1/2/ρ-1/2 = C × (p/p)1/2

Giving that m1/2 = √x, we obtain

v = C × √(p/p)

where C is a dimensionless constant.

More Length, Mass and Time Lessons and Learning Resources

Units and Measurements Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial
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Notes
Revision
Questions
1.2Length, Mass and Time
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
1.2.1Length
1.2.2Mass
1.2.3Time
1.2.4Dimensional Analysis

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