Dimensional analysis

Objective. Use dimensions to constrain and check models.

1. Learn
2. Worked example
3. Practice
4. Feedback
5. Continue

Learn: the key idea

Dimensional analysis uses the physical dimensions of quantities (length, mass, time) to constrain the form of a relationship before solving any equations. Every term in a valid physical equation must share identical dimensions, so writing quantities as products of powers of base dimensions and demanding consistency fixes unknown exponents. Example: for a pendulum period T depending on length l and gravity g, dimensional homogeneity [T]=[l]^a[g]^b forces a=½, b=−½, giving T ∝ √(l/g) without solving any differential equation. This technique checks derived formulas for errors and predicts the structure of unknown relationships.

Worked example

Using dimensional analysis, find the exponents a and b if the period T of a pendulum satisfies T = k·l^a·g^b, where l is length and g is acceleration.

1 $Write dimensions: [T]=T, [l]=L, [g]=LT⁻²$
2 $Equate: T¹ = L^a (LT⁻²)^b = L^(a+b) T^(-2b)$
3 $Match time powers: 1 = -2b, so b = -1/2$
4 $Match length powers: 0 = a + b, so a = 1/2$

Answer: a = 1/2, b = −1/2

Concept mastery

Based on your past attempts at this lesson — the question types to practise again come first.

Complete a practice set to see your concept breakdown here.

Practice

Practice happens on its own screen, one question at a time. Answers stay hidden until you submit yours, and you can stop and pick up at the same question later.

Start practising

Common mistake. Students check only that an equation 'looks right' rather than verifying every additive term has matching dimensions; expressions like l + v (length plus speed) are dimensionally invalid and signal an error.

Your progress

Tick each step as you finish it. Your exact reading position is saved automatically, so Continue drops you back on the same line.

0 of 5 steps complete

Your place saves automatically as you read.

My dashboard