AS 91523 · 2023 paper · Sam's violin

Question 1(d) Merit to Excellence

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

Microphone A is directly in front of the moving trailer, and microphone B directly behind it.

Explain how the trailer's motion, with Sam on it playing, affects: the frequency of the string; the speed of the sound in the air; the wavelength of the sound in front of and behind the violin; and the frequencies detected by microphones A and B.

The 2023 paper, page 3 ↗ Marking schedule, page 2 ↗

See it move: The Doppler effect: a moving source and a moving observer

Work it out

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

1 The string vibrates on a moving trailer. Does the motion change how fast the string vibrates?
No. It still vibrates at 196 Hz: its frequency is set by its length, tension and mass, and moving the trailer changes none of them.
2 Does the motion change the speed of the sound in the air?
No. The air sets the speed of sound: 342 m s⁻¹ ahead of the trailer and behind it.
3 Each wavefront spreads out from where the violin was when it was sent. What does that do ahead of the trailer, and behind it?
Ahead, each wavefront starts from a point a little nearer the one before, so they are closer together: a shorter wavelength, (342 − 5.30) / 196 = 1.718 m instead of 342 / 196 = 1.745 m. Behind, each starts further back, so they are spread out: a longer wavelength, (342 + 5.30) / 196 = 1.772 m.
4 Use v = fλ. What do A and B detect?
The speed is the same everywhere, so the shorter wavelength gives a higher frequency at A (199 Hz, from (c)) and the longer wavelength a lower frequency at B: 196 × 342 / (342 + 5.30) = 193 Hz.

Answer The string’s frequency is unchanged at 196 Hz, and so is the speed of sound in the air, 342 m s⁻¹. Because the violin moves towards A as it sends out each wavefront, the wavefronts are closer together in front (a shorter wavelength, 1.718 m) and further apart behind (a longer wavelength, 1.772 m). With the speed fixed, v = fλ gives a higher frequency at A (199 Hz) and a lower one at B (193 Hz).

What the marker wanted

Achieved Two of: the string’s frequency and the speed of sound are both unchanged (one point); the wavelength is shorter in front; it is longer behind; A detects a higher frequency; B detects a lower one.
Merit Explains one of: why the wavelength is shorter in front; why A detects a higher frequency; or why B detects a lower one.
Excellence All of it, explained. One minor error still earns the lower Excellence mark (E7).

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

Common mistake

Saying the sound is louder at A. NZQA’s 2023 report lists mixing up frequency and volume in the Doppler effect: the question is about frequency and wavelength.

How did it go?

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