NCEA Level 3 Physics 91523 Wave Systems: past papers, practice and marking
- NCEA Level 3
- Physics
- AS91523
- 4 credits
- External exam
About 91523
AS91523, Demonstrate understanding of wave systems. Standing waves, beats, the Doppler effect and interference: the 4-credit external, reduced to the six question types it keeps reusing.
What 91523 covers
- Standing wavesStrings and pipes: end conditions, harmonics, λ and f from length, tension and wave speed
- Doppler effectMoving source, stationary observer: f′, v_s or v_w, wavefronts, frequency–time graphs
- BeatsBeat frequency, tuning by beats, why beats form (not path difference)
- Two-source interferencePath difference, phase difference and superposition; nλ = dx/L
- Diffraction gratingsGrating equation nλ = d sin θ, number of maxima, white-light spectra, grating versus double slit
- ResonanceDriving an air column, string or object at its natural frequency
91523 past papers
Open a 91523 past paper beside your own Ako page, then get it marked, part by part.
Work through a past paper All NCEA Level 3 Physics past papers
Practice questions
Question 1: Standing waves
A pan-flute pipe is open at the top and closed at the bottom by a cork. The air column is 0.200 m long. The player checks the pipe against a 440 Hz tuning fork. The speed of sound in air is 343 m s⁻¹.
- Show that the pipe’s fundamental frequency is about 429 Hz.(up to Achieved)
- Calculate the beat frequency heard when the pipe and the tuning fork are sounded together.(up to Achieved)
Check your answer
Part a: what gets the marks
- Closed pipe: a quarter wavelength fits the air column: λ = 4L = 4 × 0.200 = 0.800 m
- Wave equation: f = v / λ = 343 / 0.800 = 428.75 Hz
A full answer: f = 428.75 Hz ≈ 429 Hz. A pipe closed at one end fits a quarter wavelength (node at the closed end, antinode at the open end), so λ = 4L = 0.800 m and f = v/λ = 343/0.800 = 428.75 Hz.
Part b: what gets the marks
- The beat frequency is the difference between the two frequencies: f_beat = 440 − 428.75 = 11.25 Hz
A full answer: f_beat = 11 Hz (11.25 Hz). The loudness rises and falls about 11 times each second.
What gets each grade
Question 2: Standing waves
A string of length 0.648 m is fixed at both ends and vibrates in its fundamental at 82.4 Hz (a bass guitar’s low E).
- Calculate the speed of the wave on the string.(up to Achieved)
- The player wants to raise the note by tightening the string. Explain what happens to the wave speed, the wavelength and the frequency.(up to Merit)
Check your answer
Part a: what gets the marks
- Fundamental wavelength: λ = 2L = 1.296 m
- Wave equation: v = f λ = 82.4 × 1.296 = 107 m s⁻¹
A full answer: v = 107 m s⁻¹
Part b: what gets the marks
- The wave speed on a string increases with tension (v = √(T/μ)), so tightening the string increases v.
- The string length is unchanged and both ends are still fixed, so the fundamental still fits half a wavelength. λ stays at 1.296 m.
- Effect on frequency: f = v/λ, with λ fixed and v larger → f increases
A full answer: Wave speed: increases. The speed of a wave on a string is set by the properties of the medium, the tension in it and its mass per unit length. Tightening the string raises the tension while leaving the mass per unit length unchanged, so v increases.
Wavelength: unchanged, at 1.296 m. The string is the same length and is still fixed at both ends, so the fundamental must still fit exactly half a wavelength between the nut and the bridge. The geometry has not changed, so λ = 2L = 1.296 m as before.
Frequency: increases. From f = v/λ, with λ held fixed and v larger, the frequency rises, which is what raising the pitch means.
What gets each grade
Question 3: Standing waves
The same 0.650 m guitar string, with a wave speed of 325 m s⁻¹, is touched lightly at a point one third of the way along and plucked, so that it vibrates in its third harmonic.
- State the number of nodes and antinodes on the string in the third harmonic.(up to Achieved)
- Calculate the wavelength and frequency of the third harmonic.(up to Achieved)
- Explain why a string fixed at both ends can only produce frequencies that are whole-number multiples of the fundamental.(up to Excellence)
Check your answer
Part a: what gets the marks
- The n-th harmonic fits n half-wavelengths onto the string. For n = 3 there are three antinodes, with nodes between them as well as at the two fixed ends.
- Count: antinodes = 3 nodes = 4 (two at the ends, two inside)
A full answer: Three antinodes and four nodes, counting the fixed ends.
Part b: what gets the marks
- Three half-wavelengths fit the string: L = 3λ/2 → λ = 2L/3 = 1.300/3 = 0.4333 m
- Frequency from the wave equation: f = v/λ = 325 / 0.4333 = 750 Hz
A full answer: λ = 0.433 m and f = 750 Hz, which is exactly three times the fundamental.
Part c: what gets the marks
- Both ends are clamped, so both must be nodes. Nodes on a standing wave are spaced exactly half a wavelength apart.
- Follow the arithmetic: L must equal a whole number of half-wavelengths: L = n λ/2 → λ = 2L/n → f = v/λ = n v / 2L = n f₁
- Any other frequency would require a node somewhere other than at the ends, or a fraction of a half-wavelength to fit, which the fixed ends forbid. Waves at those frequencies interfere destructively with their own reflections and die away.
A full answer: Both ends are clamped, so both must be nodes, and adjacent nodes are half a wavelength apart. The string must hold a whole number of half-wavelengths:
L = nλ/2, so λ = 2L/n
f = v/λ = nv/2L = nf₁ (n = 1, 2, 3, …)
Any other frequency would need a fraction of a half-wavelength to fit, which the fixed ends forbid. Its reflections return out of phase, interfere destructively and the wave dies away.