MYP Physics · topic MYP-09.03
Stationary waves, nodes, and antinodes
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
Identifying the missing fundamental on a string
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A string fixed at both ends is 1.20 m long. Its three lowest observed resonances are 50, 75 and 100 Hz. The 50 Hz pattern has two loops, each between adjacent nodes. The supplied graph relates the identified harmonic number to resonance frequency. Wave speed is unchanged between these observations. Supplied data (Harmonic number; Frequency / Hz): (2, 50), (3, 75), (4, 100). (a) Determine the unobserved fundamental, wave speed and node spacing at 75 Hz. [3 marks] (b) The tension is quadrupled while length and mass per length remain fixed. Predict the resonances and explain the roles of nodes and antinodes. [3 marks]
Identified string resonances
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Frequency / Hz
Horizontal axis (x): Harmonic number
Read the exact data as a table
| Harmonic number | Frequency / Hz |
|---|---|
| 2 | 50 |
| 3 | 75 |
| 4 | 100 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Determine the unobserved fundamental, wave speed and node spacing at 75 Hz.
Reveal the worked solution
- For both ends fixed, L = nλ/2. Two loops identify 50 Hz as the second harmonic, not the fundamental.
- The fundamental is 50/2 = 25 Hz. Wave speed is 2Lf₁ = 2(1.20)(25) = 60 m s⁻¹.
- At 75 Hz, λ = 60/75 = 0.80 m; adjacent nodes are λ/2 = 0.40 m apart.
- String wave speed follows v = √(T/μ), so multiplying tension by four doubles speed and all resonance frequencies.
- At a node, destructive superposition keeps displacement zero; at an antinode the displacement amplitude is maximum.
- The standing pattern arises from incident and reflected waves of equal frequency travelling in opposite directions. It is not a single wave carrying the pattern along the string.
Answer (a) Fundamental = 25 Hz; wave speed = 60 m s⁻¹; node spacing at 75 Hz = 0.40 m. (b) The three observed resonances become 100, 150 and 200 Hz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.