AS 91523 · 2025 paper · standing waves
Question 2(b) Merit
Practises Standing waves & resonance
A cello's strings are held at each end by the bridge and the nut, 0.690 m apart, and can vibrate in different harmonics.
| L | 0.690 m |
|---|
After the cello has been in storage, the tension in its strings has decreased and it is out of tune: the strings play a frequency other than the desired one. The paper gives v = √(tension ÷ mass per unit length) for a wave on a string. No calculations are required.
Use the equation to explain how this could make the frequency higher or lower than the desired frequency.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 What does the equation say happens to the wave speed when the tension drops?
v = √(tension ÷ mass per unit length). The string itself has not changed, so its mass per unit length is the same, and a smaller tension gives a smaller speed. -
2 Has the wavelength of each harmonic changed?
No. It is fixed by the length between the bridge and the nut (λ = 2L for the first harmonic), and that length has not changed. -
3 So which way does the frequency go?
f = v / λ. A smaller speed over the same wavelength gives a lower frequency, for every harmonic the string plays.
What the marker wanted
- Achieved One of: the wave speed decreases, or the frequency decreases.
- Merit All three: the speed decreases, the wavelength stays the same, so the frequency decreases.
In our words, from NZQA’s 2025 marking schedule.
✕Common mistake
Saying the frequency “changes”. NZQA’s 2025 report lists answers that said a quantity “changed”, without saying whether it went up or down, among those that did not reach Achieved. Give the direction for the speed and the frequency, and say the wavelength stays the same.How did it go?
Got it moves it further out. Review brings it back tomorrow.