Interactive practical

Stationary waves on a string

Tune a vibration generator to resonance and investigate how stationary-wave frequency depends on string length, tension and mass per unit length.

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A-Level Physics

Investigating standing waves on a string

Pearson Edexcel 9PH0 — matched to your course.

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Practical focus

Relate resonant length, frequency and wavelength while controlling tension.

Practical method

Aim

Investigate how the frequency of the first harmonic of a stretched string depends on its tension, and test f = \(\dfrac{1}{2L}\)\(\sqrt{\dfrac{T}{\mu }}\).

Variables

Independent variable: tension T (hanging mass)
Dependent variable: frequency of the first harmonic
Control variables: vibrating length L; the same string (mass per unit length μ).

Apparatus

String over a pulley with a mass hanger
Vibration generator and signal generator
Two bridges to set the length
Metre rule
Balance to find μ

Method

Find μ by weighing a measured length of the same string.
Set the vibrating length to 1.00 m between the vibrator and the pulley bridge.
Hang 0.10 kg and increase the frequency from zero until the string vibrates with one large loop (first harmonic). Record f.
Repeat for hanging masses up to 0.50 kg, approaching resonance from above and below and averaging.
Plot f against √T: a straight line through the origin with gradient \(\dfrac{1}{2L\sqrt{\mu }}\).

Sample results

L = 1.00 m, μ = 1.0 × 10⁻³ kg m⁻¹, T = mg.

Mass / kg	T / N	√T / N\(^{1/2}\)	f / Hz
0.10	0.981	0.990	15.7
0.20	1.962	1.401	22.1
0.30	2.943	1.716	27.1
0.40	3.924	1.981	31.3
0.50	4.905	2.215	35.0

Analysis

f is proportional to √T: gradient = \(\dfrac{35.0}{2.215}\) = 15.8 Hz N\(^{-1/2}\). Predicted gradient \(\dfrac{1}{2L\sqrt{\mu }}\) = \(\dfrac{1}{2 \times 1.00 \times 0.0316}\) = 15.8 Hz N\(^{-1/2}\), so the data agree with f = \(\dfrac{1}{2L}\)\(\sqrt{\dfrac{T}{\mu }}\).

Uncertainty

Judging the frequency of maximum amplitude is uncertain by about ±0.3 Hz (1–2%). Approaching from both sides and averaging reduces this. μ from weighing 2.00 m of string (±0.01 g in 2.0 g) is good to 0.5%.

Errors and improvements

Problem	Effect	Improvement
Judging maximum amplitude	Frequency uncertain	Approach from above and below and average
Vibrator end not a true node	Effective length differs	Measure between bridges / use a small vibrator amplitude
String stretches	μ and L change	Use a fairly inextensible string

Risk assessment

Hazard	Risk	Control measure
Masses falling	Injury to feet	Use a cushion below the hanger
String snapping under tension	Eye injury	Wear goggles; do not overload

Common mistakes

Plotting f against T rather than √T.
Using the mass instead of the weight for tension.
Counting the first harmonic as a full wavelength (it is half).

Worked analysis

T = 0.30 × 9.81 = 2.943 N. f = \(\dfrac{1}{2L}\)\(\sqrt{\dfrac{T}{\mu }}\) = \(\dfrac{1}{2.00}\)\(\sqrt{\dfrac{2.943}{1.0 \times 10^{-3}}}\) = 27.1 Hz, matching the measured value.

Common misconception

A stationary wave forms only at particular frequencies; between them the string vibrates with a small, irregular amplitude.

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