How to Check a Physics Equation Using Dimensional Analysis

Two pendulums have the same string length and are released from the same small angle. One bob has three times the mass of the other. Which one completes a swing first?

A reasonable first guess is the lighter bob. It has less mass, so perhaps it should be easier to accelerate. But the heavier bob also experiences three times the gravitational force.

Before accepting either prediction, we need a way to check the equation describing the pendulum’s motion. One useful test is dimensional analysis: replacing every physical quantity with its fundamental dimensions to see whether the mathematics is physically possible.

This technique is particularly useful when you’ve derived an equation yourself, rearranged several relationships, or aren’t sure whether you’ve made an algebraic mistake.

01Why check an equation using dimensions?

Imagine deriving an expression for acceleration towards the centre of a circular path.

You obtain two possible results:

\[
a=\frac{v^2}{r}
\]

and

\[
a=\frac{v}{r}
\]

Both contain velocity, \(v\), and radius, \(r\). Both look reasonably plausible.

But only one can represent acceleration.

The first has units of metres per second squared. The second has units of inverse seconds. It cannot possibly represent an acceleration.

That’s the advantage of dimensional analysis. You can reject an impossible equation without substituting a single numerical value.

It also gives you a useful way to check a derivation before spending time calculating an answer.

What are dimensions?

A physical quantity has a dimension describing the type of quantity it represents.

For example:

• Distance has the dimension of length.
• Mass has the dimension of mass.
• Time has the dimension of time.
• Velocity has dimensions of length divided by time.

We represent these fundamental dimensions using:

For most mechanics equations, mass, length, and time are enough.

Electricity and magnetism may also require the dimension of electric current.

Dimensions and units are related, but they aren’t the same thing.

For example, velocity has SI units of \(\mathrm{m\,s^{-1}}\), but its dimensions are \([L][T]^{-1}\).

The units tell you how a quantity is measured. The dimensions tell you what kind of physical quantity it is.

Common physics quantities and their dimensions

You can build more complicated dimensions from the four fundamental quantities.

Notice how force can be expressed using fundamental units:

\[
1\ \mathrm{N}=1\ \mathrm{kg\,m\,s^{-2}}
\]

Therefore,

\[
[F]=[M][L][T]^{-2}
\]

This is the form you’ll use when checking a force equation.

02The experiment: finding a pendulum’s period

Return to the two pendulums.

To investigate whether mass affects their period, we can set up a simple pendulum experiment.

The length \(L\) is measured from the pivot to the centre of the bob.

We release the bob from a small angle without pushing it and measure the time taken for several complete oscillations. A light gate or video analysis can reduce the timing uncertainty.

The period, \(\tau\), is the time taken for one complete oscillation.

A sensible investigation would proceed in two stages.

First, investigate the effect of mass.

Keep the string length and release angle constant. Change the bob’s mass and measure the period again.

If mass affects the period, we should observe a change.

Second, investigate the effect of length.

Keep the bob’s mass and release angle constant. Change the string length and measure the period.

The initial experimental design matters here. If we change both mass and length simultaneously, we cannot identify which variable caused a change in the period.

We should change one variable at a time and repeat the measurements. Timing ten complete oscillations rather than just one also reduces the effect of reaction-time errors.

What does the experiment suggest?

The following is illustrative data, not a record of an actual experiment.

The ratio \(\tau^2/L\) is approximately constant.

This suggests:

\[
\tau^2\propto L
\]

or, equivalently,

\[
\tau\propto\sqrt{L}
\]

Repeating the investigation with different bob masses produces approximately the same period when the length and release angle are unchanged.

The observations support a model in which the period depends on length and gravitational acceleration, but not on the bob’s mass.

Now we have an experimental pattern. But how can we check whether a proposed equation is mathematically consistent with it?

03How to check a derived equation using dimensions

The central rule is:

Both sides of a physically valid equation must have the same dimensions.

There is an additional requirement when an equation contains addition or subtraction: every individual term being added or subtracted must also have the same dimensions.

For example:

\[
s=ut+\frac12at^2
\]

The left-hand side represents displacement, so its dimensions are \([L]\).

The first term on the right has dimensions:

\[
[ut]=[L][T]^{-1}[T]=[L]
\]

The second term has dimensions:

\[
[at^2]=[L][T]^{-2}[T]^2=[L]
\]

Every term has dimensions of length. The equation passes the dimensional check.

A five-step method

Use this sequence whenever you want to test a derived equation.

Step 1: Identify the physical quantity on each side.

Write down the dimensions of the quantity you’re calculating.

For example, if you’re calculating force:

\[
[F]=[M][L][T]^{-2}
\]

Step 2: Replace every variable with its dimensions.

Use your reference table.

For example:

\[
[v]=[L][T]^{-1}
\]

\[
[a]=[L][T]^{-2}
\]

Step 3: Substitute the dimensions into the equation.

Treat the dimensions like algebraic variables.

Multiply dimensions when quantities are multiplied, and divide them when quantities are divided.

Step 4: Simplify the expression.

Collect the powers of mass, length, and time.

For example:

\[
\frac{[L]^2[T]^{-2}}{[L]}
=
[L][T]^{-2}
\]

Step 5: Compare both sides.

If the dimensions match, the equation is dimensionally consistent.

If they don’t match, the equation cannot be correct as written.

There is one important qualification: passing this test does not prove that an equation is physically correct. It only proves that its dimensions are compatible.

We’ll see why shortly.

05A more difficult example: checking Faraday’s law

Dimensional analysis isn’t restricted to mechanics.

Consider electromagnetic induction, where a changing magnetic flux through a coil produces an induced electromotive force (emf).

Faraday’s law for the average induced emf is:

\[
\varepsilon_{\mathrm{avg}}
=
-N\frac{\Delta\Phi}{\Delta t}
\]

where:

• \(\varepsilon_{\mathrm{avg}}\) is the average induced emf in volts,
• \(N\) is the number of turns in the coil,
• \(\Delta\Phi\) is the change in magnetic flux per turn in webers, and
• \(\Delta t\) is the time interval in seconds.

Let’s check the dimensions.

Step 1: Identify the dimensions of the left-hand side.

An emf is measured in volts.

Therefore:

\[
[\varepsilon]=[\mathrm{V}]
\]

Step 2: Identify the dimensions of magnetic flux.

One weber is equivalent to one volt-second:

\[
1\ \mathrm{Wb}=1\ \mathrm{V\,s}
\]

Therefore:

\[
[\Phi]=[\mathrm{V}][T]
\]

The number of turns, \(N\), is dimensionless.

Step 3: Substitute the dimensions.

The right-hand side has dimensions:

\[
\frac{[\mathrm{Wb}]}{[T]}
\]

Substituting:

\[
=\frac{[\mathrm{V}][T]}{[T]}
\]

Therefore:

\[
=[\mathrm{V}]
\]

Step 4: Compare both sides.

Both sides have dimensions of voltage.

Faraday’s law passes the dimensional check.

Step 5: Understand what the check cannot establish.

The negative sign represents Lenz’s law, which describes the direction of the induced emf relative to the change in magnetic flux.

Dimensional analysis cannot establish this negative sign.

It also cannot establish the factor \(N\). Even if you removed the number of turns, the equation would still have the correct dimensions because \(N\) is dimensionless.

The physical relationship and its numerical factors must be established through the underlying physics and experimental evidence, not dimensions alone.

06What dimensional analysis can and cannot tell you

This is probably the most important distinction to remember.

Dimensional analysis is a powerful way to reject an incorrect equation. It is not a complete method for proving one.

Why numerical constants can cause problems

Consider two possible expressions for kinetic energy:

\[
E_1=mv^2
\]

and

\[
E_2=\frac12mv^2
\]

Both have the dimensions:

\[
[M]\left([L][T]^{-1}\right)^2
\]

Simplifying:

\[
[M][L]^2[T]^{-2}
\]

Both expressions have the dimensions of energy.

Yet only the second gives the correct classical kinetic energy.

The factor of \(\frac12\) is dimensionless, so dimensional analysis cannot detect its omission.

This is why a successful dimensional check should never be treated as proof of a derivation.

Why dimensionless quantities still matter

There is another subtle limitation.

Angles are dimensionless in SI. This is because an angle measured in radians is the ratio of an arc length to a radius, both of which have dimensions of length.

Therefore:

\[
[\theta]=1
\]

But being dimensionless does not mean a quantity is physically irrelevant.

For example, the angle between a force and a displacement affects the work done by that force:

\[
W=Fd\cos\theta
\]

The cosine term has no dimensions, but it can change the answer completely.

When the force is perpendicular to the displacement, \(\cos90^\circ=0\), so the work done is zero.

A dimensional check alone cannot tell you that this angular factor is necessary.

07Choosing the right equation before checking dimensions

There’s a common mistake in physics calculations: choosing an equation simply because its variables look familiar.

Dimensional analysis doesn’t solve this problem.

You must first identify the physical situation and check whether the equation’s assumptions apply.

For example:

A formula can pass a dimensional check but still be inappropriate for the question.

For example, \(v=u+at\) is dimensionally correct. But if acceleration changes throughout the motion, you cannot automatically use this constant-acceleration equation.

Likewise, \(a_c=v^2/r\) describes the radial component of acceleration in circular motion. If the object’s speed is also changing, there is a tangential component of acceleration as well.

Before checking dimensions, ask yourself:

• What quantity am I calculating?
• What physical process is occurring?
• Which variables are known?
• What assumptions does the equation require?
• Does the equation describe this particular situation?

For a more detailed approach to identifying equations from known variables, assumptions, and the physical situation, see How to Use the HSC Physics Data and Formulae Sheets.

08Common mistakes when checking dimensions

Mistake 1: Checking only the left-hand and right-hand sides

Consider:

\[
s=ut+at
\]

Both sides of the equation have dimensions of length:

\[
[s]=[L]
\]

and:

\[
[ut]=[L]
\]

\[
[at]=[L][T]^{-1}
\]

Actually, the second term has dimensions of velocity, not length.

Therefore, the equation is inconsistent.

The important point is to check every individual term, particularly when an equation contains addition or subtraction.

Mistake 2: Assuming every constant is dimensionless

Some constants have their own physical dimensions.

For example, the universal gravitational constant \(G\) has dimensions:

\[
[G]=[L]^3[M]^{-1}[T]^{-2}
\]

Planck’s constant \(h\) has dimensions:

\[
[h]=[M][L]^2[T]^{-1}
\]

You must include these dimensions when checking equations containing physical constants.

Don’t confuse a numerical coefficient, such as \(2\pi\), with a physical constant such as \(G\).

Mistake 3: Confusing a dimensionally correct equation with a proven equation

Suppose you derive:

\[
F=3ma
\]

This equation has the correct dimensions for force.

But dimensional analysis cannot tell you whether the factor of 3 is justified.

You need to return to the original derivation and check the physics.

Mistake 4: Forgetting that dimensions don’t track direction

Velocity, displacement, and momentum are vector quantities.

Dimensional analysis can check their units, but it cannot determine whether their directions or signs are correct.

For example, a momentum equation can be dimensionally consistent while incorrectly assigning a positive sign to a westward component.

For this separate skill, see Conservation of Momentum in Two Dimensions for HSC Physics.

09Where this fits in the NSW HSC Physics syllabus

Dimensional analysis is a general scientific checking technique rather than a standalone formula to memorise.

There is also an important syllabus transition.

As at November 2026:

• The 2017 Physics Stage 6 syllabus still applies to students in Year 12.
• The 2025 Physics 11-12 syllabus begins teaching for Year 11 in Term 1, 2027.
• The new syllabus begins teaching for Year 12 in Term 4, 2027.
• The first HSC examination under the 2025 syllabus is in 2028.

You can check the official NESA syllabus implementation information.

The examples in this resource demonstrate a transferable physics skill. They should not be interpreted as a claim that every illustrated derivation is a prescribed syllabus requirement.

10Questions and solutions

Question 1

A student derives the following equation for displacement under constant acceleration:

\[
s=ut+\frac12at^2
\]

where:

• \(s\) is displacement,
• \(u\) is initial velocity,
• \(a\) is acceleration, and
• \(t\) is elapsed time.

Use dimensional analysis to determine whether the equation is dimensionally consistent.

Solution 1

The equation is dimensionally consistent because every term has dimensions of length.

Start with the left-hand side:

\[
[s]=[L]
\]

Now consider the first term:

\[
[ut]=[L][T]^{-1}[T]
\]

Simplifying:

\[
[ut]=[L]
\]

For the second term:

\[
[at^2]=[L][T]^{-2}[T]^2
\]

Therefore:

\[
[at^2]=[L]
\]

The numerical factor \(\frac12\) is dimensionless.

All three terms have dimensions of length, so the equation passes the dimensional check.

Remember that this result does not independently establish that acceleration is constant. That condition comes from the physics of the kinematics equation.

Question 2

A constant force \(F\) acts on an object, causing it to move through a displacement \(d\).

A student derives the equation:

\[
W=Fd
\]

where \(W\) is work done.

A dimensional check confirms that both sides have dimensions of energy.

However, the force acts at an angle \(\theta\) to the displacement.

(a) Explain why the dimensional check cannot establish whether the student’s equation is complete.

(b) Write the correct expression for work done by the force.

(c) Use the cases \(\theta=0^\circ\) and \(\theta=90^\circ\) to explain why the angular factor matters.

Solution 2

The student’s equation is dimensionally consistent but physically incomplete when the force is not parallel to the displacement.

Part (a):

First, check the dimensions of the student’s equation.

Work has dimensions:

\[
[W]=[M][L]^2[T]^{-2}
\]

Force has dimensions:

\[
[F]=[M][L][T]^{-2}
\]

Displacement has dimensions:

\[
[d]=[L]
\]

Therefore:

\[
[Fd]=[M][L]^2[T]^{-2}
\]

The dimensions match.

However, this doesn’t account for the angle between the force and displacement.

Part (b):

The correct expression is:

\[
W=Fd\cos\theta
\]

Only the component of the force parallel to the displacement contributes to the work done.

Part (c):

When the force and displacement are parallel:

\[
\theta=0^\circ
\]

Therefore:

\[
W=Fd\cos0^\circ=Fd
\]

The student’s equation works in this particular situation.

Now consider perpendicular force and displacement:

\[
\theta=90^\circ
\]

Therefore:

\[
W=Fd\cos90^\circ=0
\]

The force does no work through that displacement.

The student’s original expression would incorrectly predict \(W=Fd\), even though the force has no component along the direction of motion.

This is a good example of why dimensional consistency is necessary but not sufficient. Both equations have the correct dimensions, but only one describes the general physical situation.

Question 3

A coil containing 150 turns is connected to a voltage sensor.

A bar magnet moves along the axis of the coil, changing the magnetic flux through each turn.

The magnetic flux per turn decreases from \(0.50\ \mathrm{mWb}\) to \(0.10\ \mathrm{mWb}\) over \(0.020\ \mathrm{s}\).

(a) Use Faraday’s law to calculate the magnitude of the average induced emf.

(b) Demonstrate that Faraday’s law is dimensionally consistent.

(c) Explain why dimensional analysis cannot determine the polarity of the induced emf.

Solution 3

The average induced emf has a magnitude of \(3.0\ \mathrm{V}\), and Faraday’s law is dimensionally consistent.

Part (a):

Start with Faraday’s law:

\[
\varepsilon_{\mathrm{avg}}
=
-N\frac{\Delta\Phi}{\Delta t}
\]

The change in magnetic flux is:

\[
\Delta\Phi=\Phi_f-\Phi_i
\]

Substituting:

\[
\Delta\Phi=0.10-0.50
\]

\[
\Delta\Phi=-0.40\ \mathrm{mWb}
\]

Convert to webers:

\[
\Delta\Phi=-4.0\times10^{-4}\ \mathrm{Wb}
\]

Now substitute:

\[
\varepsilon_{\mathrm{avg}}
=
-\frac{150(-4.0\times10^{-4})}{0.020}
\]

Therefore:

\[
\varepsilon_{\mathrm{avg}}=+3.0\ \mathrm{V}
\]

The magnitude of the average induced emf is:

\[
|\varepsilon_{\mathrm{avg}}|=3.0\ \mathrm{V}
\]

Part (b):

The left-hand side has dimensions of voltage:

\[
[\varepsilon]=[\mathrm{V}]
\]

The right-hand side has dimensions:

\[
\left[\frac{N\Delta\Phi}{\Delta t}\right]
=
\frac{[\mathrm{Wb}]}{[T]}
\]

Since:

\[
1\ \mathrm{Wb}=1\ \mathrm{V\,s}
\]

we obtain:

\[
\frac{[\mathrm{V}][T]}{[T]}
=
[\mathrm{V}]
\]

Both sides have the same dimensions.

Part (c):

Dimensional analysis cannot determine the polarity because the negative sign in Faraday’s law represents Lenz’s law.

The polarity depends on the direction of the change in magnetic flux and the chosen positive orientation for the coil.

You need the physical direction of the induced effect, not just the dimensions of the equation, to establish the sign.

Question 4

An investigation suggests that the period \(\tau\) of a simple pendulum depends on:

• Its length \(L\),
• Gravitational acceleration \(g\), and
• The bob’s mass \(m\).

Assume the relationship has the form:

\[
\tau=CL^ag^bm^c
\]

where \(C\) is a dimensionless constant and \(a\), \(b\), and \(c\) are unknown powers.

Use dimensional analysis to determine the values of \(a\), \(b\), and \(c\).

State the resulting relationship and explain one important limitation of this derivation.

Solution 4

The exponents are:

\[
a=\frac12,\qquad b=-\frac12,\qquad c=0
\]

Therefore:

\[
\tau=C\sqrt{\frac{L}{g}}
\]

Step 1: Write the dimensions of each quantity.

The period has dimensions:

\[
[\tau]=[T]
\]

Length has dimensions:

\[
[L]=[L]
\]

Gravitational acceleration has dimensions:

\[
[g]=[L][T]^{-2}
\]

Mass has dimensions:

\[
[m]=[M]
\]

Step 2: Substitute into the proposed relationship.

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

Collect the dimensions:

\[
[T]=[M]^c[L]^{a+b}[T]^{-2b}
\]

Step 3: Equate the powers of each fundamental dimension.

The mass dimensions give:

\[
c=0
\]

The length dimensions give:

\[
a+b=0
\]

The time dimensions give:

\[
-2b=1
\]

Rearranging the time equation:

\[
b=-\frac12
\]

Substitute this into the length equation:

\[
a-\frac12=0
\]

Therefore:

\[
a=\frac12
\]

Step 4: Write the resulting relationship.

Substitute the three exponents into the original expression:

\[
\tau=CL^{1/2}g^{-1/2}m^0
\]

Since \(m^0=1\):

\[
\tau=C\sqrt{\frac{L}{g}}
\]

The bob’s mass does not appear in the resulting relationship.

For small oscillations, the simple pendulum model gives \(C=2\pi\).

However, dimensional analysis cannot determine this numerical constant.

There is another limitation: the derivation assumes that the listed variables are sufficient. A dimensionless quantity such as the initial release angle could also affect the period. We have not established that amplitude is irrelevant simply because it wasn’t included in the original expression.

Question 5

A student investigates the effect of bob mass and release angle on the period of a simple pendulum.

The string length is kept at \(0.80\ \mathrm{m}\), and gravitational acceleration is \(9.8\ \mathrm{m\,s^{-2}}\).

The following illustrative measurements are obtained. Each period has an estimated measurement uncertainty of \(\pm0.01\ \mathrm{s}\).

The student makes the following argument:

“The dimensional derivation shows that the period is independent of mass. The release angle is also dimensionless, so it cannot affect the period. Therefore, the equation \(\tau=2\pi\sqrt{L/g}\) must apply to all four measurements.”

Assess this argument.

Your response should:

(a) Evaluate whether the measurements support independence from bob mass.

(b) Calculate the predicted small-angle period and compare it with the measurements at \(60^\circ\).

(c) Explain why a dimensionless release angle can still affect the period.

(d) Write the most general dimensionally compatible form for the period under the stated idealised assumptions.

Solution 5

The student’s argument is incorrect. The measurements show no detectable mass dependence, but they show a clear increase in period at the larger release angle.

The mistake is assuming that dimensionless quantities cannot affect a physical relationship.

Part (a): Mass dependence

Compare the measurements at \(5^\circ\):

\[
T_{0.10}=1.80\ \mathrm{s}
\]

\[
T_{0.30}=1.79\ \mathrm{s}
\]

The difference is:

\[
\Delta T=0.01\ \mathrm{s}
\]

This is comparable to the measurement uncertainty.

Now compare the measurements at \(60^\circ\):

\[
T_{0.10}=1.91\ \mathrm{s}
\]

\[
T_{0.30}=1.90\ \mathrm{s}
\]

Again, the difference is only:

\[
\Delta T=0.01\ \mathrm{s}
\]

The measurements therefore provide no evidence of a significant mass dependence.

This supports the ideal pendulum model, in which the bob’s mass cancels out.

It does not prove that every real pendulum is completely independent of mass. The model assumes a light string, a point-like bob, and negligible air resistance.

Part (b): Calculate the small-angle prediction.

The small-angle period equation is:

\[
\tau=2\pi\sqrt{\frac{L}{g}}
\]

Substitute:

\[
L=0.80\ \mathrm{m}
\]

\[
g=9.8\ \mathrm{m\,s^{-2}}
\]

Therefore:

\[
\tau=2\pi\sqrt{\frac{0.80}{9.8}}
\]

\[
\tau=2\pi\sqrt{0.08163}
\]

\[
\tau=2\pi(0.2857)
\]

\[
\tau\approx1.80\ \mathrm{s}
\]

The predicted small-angle period is approximately \(1.80\ \mathrm{s}\).

At \(5^\circ\), the measured periods agree with this prediction within experimental uncertainty.

At \(60^\circ\), however, the measured period is approximately \(1.90\ \mathrm{s}\).

The difference is approximately:

\[
1.90-1.80=0.10\ \mathrm{s}
\]

This is much larger than the stated measurement uncertainty.

The small-angle equation therefore does not accurately describe the larger-angle measurements.

Part (c): Why can the release angle matter?

The student has confused being dimensionless with being irrelevant.

An angle measured in radians is dimensionless, but it can still appear in an equation.

For a pendulum, the period can depend on the initial release angle even though that angle contributes no fundamental dimensions.

This is precisely why dimensional analysis cannot establish that a variable has no effect simply because it is dimensionless.

At small angles, the angle dependence is approximately negligible, giving:

\[
\tau\approx2\pi\sqrt{\frac{L}{g}}
\]

At larger angles, the period increases.

The small-angle approximation is therefore a restriction on the model, not an exact description of every possible pendulum motion.

Part (d): Write the more general relationship.

Under the idealised assumptions, the period can be expressed as:

\[
\tau=\sqrt{\frac{L}{g}}F(\theta_0)
\]

where \(F(\theta_0)\) is a dimensionless function of the initial release angle \(\theta_0\).

For small angles:

\[
F(\theta_0)\approx2\pi
\]

For larger angles, the function changes, increasing the predicted period.

The bob’s mass does not appear under the stated idealised assumptions.

This is a more complete conclusion than simply writing the small-angle equation and assuming it applies universally.

The important lesson is that dimensional analysis determines which combinations of dimensions are possible. It cannot tell you that your original list of variables was complete, nor can it rule out dimensionless functions.

10A reliable checklist for HSC calculations

When you’ve derived or rearranged a physics equation, use this checklist before substituting numbers.

• [ ] Have I identified the correct physical quantity?
• [ ] Have I selected an equation that describes the situation?
• [ ] Have I checked the assumptions and limitations of the equation?
• [ ] Have I written the dimensions of every physical variable?
• [ ] Have I included the dimensions of any physical constants?
• [ ] Do both sides of the equation have matching dimensions?
• [ ] Do all terms in any addition or subtraction have matching dimensions?
• [ ] Have I remembered that numerical coefficients and dimensionless factors cannot be checked through dimensions?
• [ ] Have I checked signs and vector directions separately?

If an equation fails the dimensional check, go back through your derivation. Look for a missing power, an incorrect rearrangement, or a quantity that you’ve substituted incorrectly.

If it passes, move on to checking the physical assumptions and the numerical factors.

That’s the distinction worth keeping: dimensional consistency tells you whether an equation is possible, while physics tells you whether it is correct.

Once you can make that distinction confidently, the next step is using experimental relationships and graphs to test whether a mathematically valid model actually describes the behaviour of a physical system.