Units and Measurements · Lesson 6 of 7
Finding Relationships by Dimensional Analysis
“Infer a possible physical relationship by solving for the powers of relevant variables.”
• Construct a product assumption from relevant variables. • Equate powers of independent dimensions. • Derive the pendulum scaling relationship. • Explain the physical and mathematical limits of the method.
From checking a formula to proposing one
Dimensional analysis can do more than test an existing expression. If physical reasoning identifies the quantities on which an outcome depends, a product of powers can reveal a possible relationship. The list of variables is an assumption: dimensions cannot tell us that an omitted physical effect is unimportant. We also assume that the dependence can be represented by a product multiplied by a dimensionless factor.
The procedure is to name the dependent quantity, select plausible independent variables, introduce unknown exponents, replace each variable by its dimensional formula, and equate the powers of each independent base dimension. Solving those algebraic equations gives the exponents when enough independent information is available. The remaining dimensionless factor must come from experiment or a fuller theory.
A simple pendulum
A simple pendulum idealises a small bob on a light, inextensible string, swinging under gravity. For small swings and negligible drag, suppose its period depends on string length, bob mass and gravitational acceleration. Call these l, m and g. We will use τ for the period so that the time dimension T is easy to distinguish from the physical quantity.
Problem
Use dimensions to find how the period depends on length, gravity and mass.
- 1.The period has dimensions . Substitution gives .
- 2.Collect powers: .
- 3.Equating powers gives , and .
- 4.Therefore , and . Hence .
- 5.The zero mass exponent means that mass does not affect this idealised period under the stated assumptions.
Reading the relationship
A longer pendulum takes longer to complete an oscillation, but period is not directly proportional to length. Multiplying length by four multiplies the period by two. Stronger gravity shortens the period: multiplying g by four halves it. The result is independent of bob mass within the model, so replacing the bob with a heavier one does not by itself change the predicted small-amplitude period.
Problem
A pendulum’s length becomes nine times as large at the same location. How does its period change?
- 1.Write the ratio because the same k and g cancel.
- 2.Substitute : the ratio is .
- 3.The period becomes three times as long. No numerical value of g is needed for this comparison.
Why assumptions still matter
The swing amplitude is an angle and therefore dimensionless. At larger amplitudes, the period depends on that angle through a dimensionless correction that dimensions cannot determine. Drag, string flexibility or a bob of significant size can introduce additional parameters. The derived scaling is valuable only within a model whose neglected effects are small enough.
If several proposed variables have dependent dimensional formulae, equating exponents may leave more unknowns than independent equations. There is then no unique product law from dimensions alone. Nor can this procedure decide whether a relationship contains a sum or a complicated dimensionless function. Treat the result as a physically motivated scaling relation and check it against experiment or theory.
Quiz
At the same location, quadrupling pendulum length changes its ideal small-amplitude period by what factor?
Key Takeaways
• Choose relevant variables before introducing unknown powers. • Equating exponents can reveal a scaling relationship. • For the ideal small-amplitude pendulum, period scales as the square root of length divided by gravity. • Dimensionless constants and dimensionless-variable effects require additional physics.