Stationary Wave Equation and Practical Investigation Revision Notes | A Level AQA Physics

1. Why a Plucked String Sings at One Specific Pitch

Pluck a guitar string and it vibrates at a very precise frequency — not a random one. That frequency depends on exactly three things: how long the string is, how tightly it is stretched, and how heavy it is per metre. The equation that captures all three is the first harmonic equation, and it is the heart of this lesson.

Before the equation makes sense, you need a clear picture of what is actually happening on the string. A stationary wave forms when two identical progressive waves travel in opposite directions along the string and superpose. The result is a pattern of fixed nodes (points of zero displacement) and antinodes (points of maximum displacement) that appears to stand still.

2. The First Harmonic and Its Geometry

The first harmonic — also called the fundamental mode — is the simplest stationary wave pattern a string can support. Picture a string fixed at both ends. At the first harmonic, there is exactly one antinode in the middle and a node at each fixed end. The distance between the two nodes is the full length $l$ of the string, and that distance equals exactly half a wavelength: $l = \frac{\lambda}{2}$, so $\lambda = 2l$.

Because $\lambda = 2l$ at the first harmonic, and wave speed $v = f\lambda$, you can write $v = f \times 2l$, giving $f = \frac{v}{2l}$. The wave speed on a string is determined by the tension $T$ and the mass per unit length $\mu$ through $v = \sqrt{\frac{T}{\mu}}$. Substituting this expression for $v$ directly into $f = \frac{v}{2l}$ produces the first harmonic equation.

Notice the structure of the equation: $f$ is proportional to $\frac{1}{l}$, proportional to $\sqrt{T}$, and inversely proportional to $\sqrt{\mu}$. Each of those proportionalities is directly testable in the required practical, which is why understanding the equation's structure matters as much as being able to use it numerically.

3. Understanding Each Variable's Effect

The equation tells a clear physical story for each variable. Increasing tension $T$ increases the restoring force on displaced string elements, so the wave travels faster and the frequency rises. Increasing mass per unit length $\mu$ makes the string more sluggish — it resists acceleration — so the wave travels more slowly and the frequency falls. Increasing length $l$ means the half-wavelength that fits the string is longer, so the frequency falls.

A useful way to remember the square-root relationships: doubling the tension does not double the frequency — it multiplies it by $\sqrt{2} \approx 1.41$. You need to quadruple the tension to double the frequency. This is a favourite examiner trap.

4. Worked Calculation Using the First Harmonic Equation

Theory is only useful when you can apply it under exam conditions. The step-by-step below walks through a typical AQA-style calculation, annotated with method marks so you can see exactly where marks are awarded.

Worked example: Calculating the fundamental frequency of a vibrating string

A string of length 0.80 m is fixed at both ends and placed under a tension of 36 N. The mass per unit length of the string is $4.0 \times 10^{-3}$ kg m$^{-1}$. Calculate the frequency of the first harmonic. [3 marks]

1. $f = \frac{1}{2l}\sqrt{\frac{T}{\mu}}$ — [M1] Correct formula stated. Identify every symbol: $l = 0.80$ m, $T = 36$ N, $\mu = 4.0 \times 10^{-3}$ kg m$^{-1}$.
2. $\frac{T}{\mu} = \frac{36}{4.0 \times 10^{-3}} = \frac{36}{0.004} = 9000 \text{ m}^2\text{s}^{-2}$ — Dividing 36 by 0.004: $36 \div 0.004 = 9000$. Keep the full value — do not round here.
3. $\sqrt{9000} = \sqrt{9000}$. Note: $94^2 = 94 \times 94 = 8836$ and $95^2 = 95 \times 95 = 9025$, so $\sqrt{9000} \approx 94.9$ m s$^{-1}$ — This is the wave speed $v$ on the string. Checking: $94.9 \times 94.9 \approx 9006 \approx 9000$ — consistent to 3 s.f.
4. $f = \frac{1}{2 \times 0.80} \times 94.9 = \frac{94.9}{1.60}$ — [M1] Correct substitution with units consistent throughout.
5. $f = \frac{94.9}{1.60} = 59.3$ Hz — [A1] Answer to 3 significant figures with correct unit. Sanity check: a frequency in the tens of Hz is physically reasonable for a string of this length and tension.

Answer: $f \approx 59$ Hz (to 2 s.f.) or $59.3$ Hz (to 3 s.f.)

5. Required Practical: Stationary Waves on a String

The required practical asks you to investigate how the frequency of stationary waves on a string varies with length, tension, and mass per unit length. The standard setup uses a string stretched over a pulley with hanging masses providing the tension, driven by a vibration generator connected to a signal generator.

The choice of graph axes is critical for AQA mark schemes. Plotting $f$ against $\frac{1}{l}$ (not $f$ against $l$) linearises the relationship, giving a straight line through the origin. A curve through the origin would only earn partial credit because it does not allow you to extract the gradient and verify the equation quantitatively.

Uncertainty handling matters too. The main sources of uncertainty are reading the resonance frequency (the string's amplitude peaks over a small range of frequencies, not at a single sharp value) and measuring the vibrating length $l$ (the node positions are not perfectly sharp). Repeating measurements and taking a mean frequency at resonance reduces random error. Using a longer string reduces the percentage uncertainty in $l$.

6. Rearranging the Equation and Graphical Analysis

Exam questions often ask you to rearrange the first harmonic equation to find $T$ or $\mu$, or to interpret a graph whose gradient contains the equation. Both skills require you to be comfortable manipulating the square root algebraically.

Worked example: Rearranging to find tension $T$ from a graph gradient

A student plots $f$ against $\frac{1}{l}$ for a string with $\mu = 2.5 \times 10^{-3}$ kg m$^{-1}$ under constant tension. The gradient of the straight-line graph is 48 Hz m. Calculate the tension $T$ in the string.

1. $f = \frac{1}{2l}\sqrt{\frac{T}{\mu}} = \frac{1}{2}\sqrt{\frac{T}{\mu}} \times \frac{1}{l}$ — Written in the form $y = mx$, the gradient $m = \frac{1}{2}\sqrt{\frac{T}{\mu}}$.
2. $\frac{1}{2}\sqrt{\frac{T}{\mu}} = 48$ Hz m — The gradient has units of Hz m because $f$ is in Hz and $\frac{1}{l}$ is in m$^{-1}$, so gradient = Hz $\div$ m$^{-1}$ = Hz m.
3. $\sqrt{\frac{T}{\mu}} = 2 \times 48 = 96$ m s$^{-1}$ — Multiply both sides by 2. The units confirm this is a wave speed.
4. $\frac{T}{\mu} = 96^2 = 96 \times 96 = 9216$ m$^2$ s$^{-2}$ — Expanding: $96 \times 96 = (90 + 6)(90 + 6) = 8100 + 540 + 540 + 36 = 9216$. Always expand longhand to avoid errors.
5. $T = 9216 \times \mu = 9216 \times 2.5 \times 10^{-3}$ — Multiply: $9216 \times 2.5 = 23040$, then $\times 10^{-3}$ gives $23.04$ N.
6. $T = 23$ N (to 2 s.f.) — [A1] Units are newtons. Sanity check: a tension of ~23 N for a light string is physically reasonable for this type of experiment.

Answer: $T = 23$ N (to 2 s.f.)

7. Exam Technique and Common Mistakes

AQA mark schemes for this topic reward precise use of the equation, correct unit handling, and clear graphical analysis. The mistakes below account for the majority of lost marks in past papers.

For a 'describe the graph you would expect' question, the AQA mark scheme typically awards one mark for the correct axes (including the linearised variable), one mark for a straight line through the origin, and one mark for a positive gradient. All three are needed for full credit.