AS 91523 · 2025 paper · standing waves
Question 2(d)(ii) Merit to Excellence
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 |
|---|
Explain what will be heard as the length of string 1 continues to be gradually reduced.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
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1 Which way does string 1’s frequency go as its vibrating length is reduced?
Up. The tension, and so the wave speed, is unchanged; a shorter string has a shorter wavelength, so f = v / λ rises. -
2 A beat has just been heard. Is string 1 above or below 98.0 Hz?
Below. Shortening has been raising its frequency, and beats appear only once the two notes are close, so string 1 has been climbing towards 98.0 Hz from underneath. -
3 What happens to the beats as string 1 keeps rising towards 98.0 Hz?
fbeat = |f1 − 98.0| gets smaller, so the beats slow down: the gap between loud pulses gets longer. -
4 And when the two frequencies are equal?
fbeat = 0. There are no beats, just one steady note. -
5 What if the finger keeps moving and the string keeps getting shorter?
String 1 rises above 98.0 Hz, so the difference grows again and the beats speed up. When they come too fast to count, they stop sounding like pulses and you hear two separate notes.
What the marker wanted
- Achieved One of: the beats slow down (and then speed up again); or no beats are heard once the two frequencies are equal.
- Merit Explains how the beat frequency changes as string 1 is shortened.
- Excellence With (i), the complete answer: how beats form, and the beats slowing, stopping when the frequencies match, then quickening.
The schedule marks (d)(i) and (d)(ii) together as one item. Either explanation on its own reaches Merit; Excellence needs both, complete.
In our words, from NZQA’s 2025 marking schedule.
How did it go?
Got it moves it further out. Review brings it back tomorrow.