AS 91523 · 2023 paper · Sam's violin

Question 1(b) Merit

A violin's strings are fixed at both ends, 0.331 m apart, and Sam makes them vibrate with a bow. One string, the "G", plays a fundamental frequency of 196 Hz. Assume the speed of sound in air is 342 m s⁻¹.

L	0.331 m
f₁	196 Hz
v sound	342 m s⁻¹

The vibrating string is found to vibrate at 392 Hz and 588 Hz as well.

State the harmonic that causes the vibration at 588 Hz. Include a sketch showing where the nodes and antinodes are.

The 2023 paper, page 2 ↗ Marking schedule, page 1 ↗

See it move: Standing waves: string, open pipe, closed pipe

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1 How does 588 Hz compare with the fundamental?
588 / 196 = 3: three times the fundamental, so it is the third harmonic. (392 Hz is the second.)
2 How many loops does the third harmonic fit on a string fixed at both ends?
Three: the nth harmonic has n loops, so 0.331 m = 3λ/2.
3 Where do the nodes and antinodes go?
Four nodes: one at each fixed end and two between, every 0.331 / 3 = 0.110 m. Three antinodes, one in the middle of each loop.

Answer The third harmonic (588 = 3 × 196). Sketch three equal loops between the fixed ends: 4 nodes (both ends and two between, 0.110 m apart) and 3 antinodes at the middles of the loops.

What the marker wanted

Achieved The right harmonic, or a correct diagram with the nodes and antinodes labelled.
Merit Both: the third harmonic, and the correctly labelled diagram.

In our words, from NZQA’s 2023 marking schedule.

Common mistake

Not identifying the harmonic, or drawing a wave that does not end on a node at both fixed ends. NZQA’s 2023 report lists both among answers that did not reach Achieved. Divide by the fundamental: 588 / 196 = 3, so three loops.

How did it go?

Got it moves it further out. Review brings it back tomorrow.