Finding the modes of a stretched string

7 marks · A-level · original practice case

Write your working and explanations below or on paper. Your responses stay in this page; PhysicsHelp does not submit or save them. Copy anything you want to keep before closing it. Check calculations separately from method marks. Use unrounded working between parts.

The situation

A uniform string of length 0.80 m is fixed at both ends. Tension remains constant. The table shows idealised resonance frequencies with mode numbers not yet assigned. Treat these model values as exact.

Resonances
Resonance	Frequency / Hz
First	50
Second	100
Third	150

Part (a)

Determine the fundamental wavelength. [2 marks]

Give the numerical answer to 2 significant figures.

Final value / m

Working

Part (b)

Calculate the wave speed. [2 marks]

Give the numerical answer to 2 significant figures.

Final value / m s⁻¹

Working

Part (c)

Describe the nodes and antinodes at the second resonance. [2 marks]

Your explanation

Part (d)

Select the result if length is doubled while wave speed stays constant. [1 marks]

Choose one answer Fundamental frequency becomes 100 Hz. Fundamental frequency remains 50 Hz. Fundamental frequency becomes 25 Hz. Fundamental frequency becomes 12.5 Hz.

Marking rubric and worked answers

Indicative practice marks. Method marks assess valid working; accuracy marks require the associated valid method. Independent points stand on their own. Tick only points supported by your response. Do not award the same explanation twice. These ticks record your judgement, not an examiner’s result.

Part (a) marking points

Final answer: 1.6 m. Numerical checks assess the final value only; they do not check working or the number of digits typed.

For fixed ends, L = λ/2 for the fundamental. (1 method mark)
Obtain λ = 1.6 m. (1 accuracy mark)

No earlier numerical result is required. Award method credit for valid working and accuracy credit for the stated answer. Accept a defensible equivalent method; wrong physics cannot earn dependent accuracy credit.

Common errors: For the fundamental, the length is half a wavelength.

Part (b) marking points

Final answer: 80 m s⁻¹. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Use v = fλ with the fundamental. (1 method mark)
Obtain 80 m s⁻¹. (1 accuracy mark)

Use the wavelength from part (a) in speed = 50 Hz × wavelength.

Common errors: This uses string length as the fundamental wavelength.

Part (c) marking points

There are nodes at both ends and at the midpoint. (1 independent mark)
There are two antinodes, one in each half of the string. (1 independent mark)

Award each distinct physical point once. Equivalent scientifically correct wording is creditable; a contradictory statement cannot earn the same point.

Part (d) marking points

Final answer: Fundamental frequency becomes 25 Hz..

Fundamental frequency becomes 25 Hz. (1 independent mark)

One correct choice earns the mark; no partial credit.

Common errors: Frequency is inversely proportional to length, not proportional. Changing length changes the permitted fundamental wavelength. Doubling length halves rather than quarters frequency.

Fundamental frequency becomes 100 Hz. — Frequency is inversely proportional to length, not proportional.
Fundamental frequency remains 50 Hz. — Changing length changes the permitted fundamental wavelength.
Fundamental frequency becomes 25 Hz. — Doubling wavelength at constant speed halves frequency.
Fundamental frequency becomes 12.5 Hz. — Doubling length halves rather than quarters frequency.

Report a problem with this case

Related skills

waves interference practice

Scope and reference sources

AQA 3.3 and selected 3.6.1.2; OCR A 4.4 and selected 5.3. These are shared original exercises; wording, indicative marks and difficulty have not been endorsed by either board.

AQA specification · OCR A specification

Case reference: string-resonance · version 1.1.0 · seed 0. Choose another case