Question 2(c) Merit to Excellence

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.

When the 0.690 m string is played, its first harmonic is 147 Hz. A finger is then pressed onto the string, shortening it so that its first harmonic is 207 Hz. Assume the tension stays the same.

Calculate the new length of the string when the 207 Hz note is played.

The 2025 paper, page 6 ↗ Marking schedule, page 5 ↗

Work it out

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

1. 1 Which wavelength goes with the first harmonic of a string fixed at both ends?

   One loop fills the string, so the string is half a wavelength: λ = 2L = 2 × 0.690 = 1.38 m.

2. 2 How fast do waves travel along this string?

   v = fλ = 147 × 1.38 = 202.86 m s⁻¹

3. 3 The finger does not change the tension or the string. So what stays the same?

   The wave speed, 202.86 m s⁻¹. Only the vibrating length, and with it the wavelength, changes.

4. 4 Find the new wavelength, then the new length.

   λ = v / f = 202.86 / 207 = 0.980 m, and for a first harmonic L = λ / 2 = 0.490 m.

5. 5 Is there a quicker way to check it?

   With v fixed, f = v / 2L means f × L stays the same: L = 0.690 × 147 / 207 = 0.490 m. A shorter string gives a higher note, as it should.

What the marker wanted

• Achieved One of: the first harmonic’s wavelength, 1.38 m; a wave speed worked out from a wrong wavelength; or the right length.
• Merit The right wave speed on the string, 202.86 m s⁻¹, or the right method carried through from a wrong first-harmonic wavelength.
• Excellence The right length, 0.490 m, with the working.

In our words, from NZQA’s 2025 marking schedule.

Common mistake

How did it go?

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