NCEA Level 3 Physics (91523) 2017 — page 1 of 6

Assessment Schedule – 2017 Physics: Demonstrate understanding of wave systems (91523) Evidence Statement

Q Achievement Merit Excellence

ONE Correct labelled diagram. (a) (line doesn’t curve back or become parallel, continues to the end of the pipe, in not straight. Only one side needs to be correct)

(b) The clarinet is open / closed, so has a node and an antinode at the two ends. Open end must be an antinode node at closed end and The tube can sustain only odd harmonics because the tube “fits” only an odd and closed end must be a node. antinode at open end requires λ 3λ 5λ OR odd number of quarter number of quarter wavelengths (OR , , ). wavelengths 4 4 4 standing wave has odd number OR OR of ¼ wavelengths. OR even harmonics need nodes λ The fundamental has a pipe length of . The pipe length for even explanation good for only at both ends or antinodes at 4 second harmonic both ends because length is harmonics would require an even number of quarter wavelengths OR an even number of quarter L = etc. This requires a node / node OR antinode / antinode. Even harmonics require nodes at wavelengths both ends/antinodes at both ends.

(c) Shown using 137 Hz to show Correct formula and Formula can be written with ±. vw f ' = f 139 Hz substitution and solve. v ± v w s OR 341 139 = f Correct substitution 341− 5 OR 336 correctly shown with the wrong f = 139 = 137 Hz speed of sound 341 n.b. This is a “show” question. Clear working must be shown.

NCEA Level 3 Physics (91523) 2017 — page 2 of 6

(d) The clarinet on the train is moving towards the waves it has already Different frequencies cause Correct answer linking Correct answer linking higher produced. beats. higher frequency to shorter frequency to shorter wavelength This causes the wavelength to decrease and the frequency of the waves to OR wavelength with reason. with reason. increase. (The wave speed is constant.) Correct beat frequency. AND AND The two sounds arrived with slightly different frequencies, so sometimes OR Links loudness to phase or Explanation linking phase and they arrived in phase, interfering constructively causing a loud sound and type of interference to interference to loudness Links loudness to phase or type then they arrived out of phase, interfering destructively causing a quiet explain beats. changing to produce beats. of interference to explain beats. sound. OR AND OR The beat frequency is the difference between the two frequencies, so the Correct beat frequency. Correct beat frequency. beat frequency is: Waves bunch up / shorter wavelength. f − f = 139 −137 = 2 Hz 1 2 nb: not nodes and antinodes

NØ N1 N2 A3 A4 M5 M6 E7 E8

0 u 2u OR i OR c 3u OR i+u 4u OR 2u+i OR 2i 2i+u OR c+i 2i+2u OR 3i 1u + 1i + 1c 2i + 1c OR c+u NCEA Level 3 Physics (91523) 2017 — page 3 of 6

Q Achievement Merit Excellence

TWO Travelling waves transfer energy, standing waves don’t. Correct answer. Answers must be Not: (a) OR comparative, relevant and not Travelling waves move but wrong standing waves stay still The amplitude is the same at different places on a travelling wave. The Standing wave has a reflected amplitude is different at different places on a standing wave. wave / reflects back on itself. OR Standing waves have nodes and antinodes but travelling waves do not. OR Travelling waves have one source but a standing wave requires 2 sources. OR Standing waves require interference but travelling waves do not.

(b) v 3λ 2L Correct relationship between L working linking wavelength f = for 3rd harmonic, L = → λ = to L, substituted into the λ 2 3 and l for 1st or 3rd harmonic. v wave equation, 3rd harmonic. f = 2L 3 3v f = 2L λ OR 1st harmonic: L = 2 v f = 2L 3v 3rd harmonic = 3 fi = 2L

(c) 1 1 Correct frequency, wavelength or Correct answer and unit f = = Hz λ = 2L = 48.0 m T 1.8 distance travelled in one period. (at least 3sf). v = f λ 1 v = × 48.0 = 26.7 m s−1 1.8

NCEA Level 3 Physics (91523) 2017 — page 4 of 6

(d) Mike and Kate are at antinodes 2nd harmonic correctly 2nd harmonic correctly drawn N A N Kate OR drawn including labelling of including labelling of an an antinode/ (Mike and Kate antinode / (Mike and Kate Mike and Kate in antiphase Mike drawn at antinodes and drawn at antinodes and OR statements that they are at statements that they are at By jumping up and down, they send (transverse) waves along the bridge. Description of standing wave antinodes). antinodes). These waves hit a closed end and reflect inverted. The two sets of waves formation. AND AND interfere and set up a standing wave. OR Mike and Kate are at Mike and Kate are at They must stand at the ¼ and ¾ positions because this is where the 2nd harmonic correctly drawn antinodes. antinodes. antinodes are for the second harmonic. including labelling of nodes and OR AND antinodes. They must jump up and down 180 º out of phase (because adjacent Mike and Kate in antiphase. Mike and Kate in antiphase. antinodes are out of phase). OR OR AND Mike and Kate must jump at the correct frequency (1.11 Hz or T = 0.9s). Mike and Kate must jump at the Mike and Kate must jump at Mike and Kate must jump at correct frequency(1.11Hz). the correct frequency the correct frequency (1.11 Hz OR (1.11Hz or T = 0.9s). or T = 0.9 s) Initially the amplitude/energy of OR the standing wave increases each Initially the amplitude / cycle. energy of the standing wave increases each cycle.

NØ N1 N2 A3 A4 M5 M6 E7 E8

0 u 2u OR i OR c 3u OR i+u 4u OR 2u+i OR 2i 2i+u OR c+i 2i+2u OR 3i 1u + 1i + 1c 2i + 1c OR c+u NCEA Level 3 Physics (91523) 2017 — page 5 of 6

Q Achievement Merit Excellence

THREE nλ = d sinθ Correct formula and substitution Don’t need to substitute in (a) −6 and answer. n = 1. n = 1, λ = 2 ×10 sin15.4° −7 λ = 5.31×10 m

(b) The light passes through the narrow slits and diffracts. The path difference is n Concepts of path difference Crest + crest / trough + trough The diffracted waves overlap and interfere. wavelengths so the interference of 1 wavelength, phase and ok for achieved only. is constructive / antinode. constructive interference The path difference for light reaching the first antinode is one explained and clearly linked. wavelength. OR So the light waves arrive in phase and add constructively (causing a The waves arrive in phase to bright spot). produce constructive interference / antinode.

(c) Diagram shows pattern with antinodes further apart, and central antinode Antinodes get further apart Antinodes get further apart in same place. (stated or in diagram). (stated or in diagram). Red light has a lower frequency than green so it has a longer wavelength. AND AND nλ Red light has a longer Explanation using either sinθ = so a longer wavelength means the angle between the d wavelength. equation. (NOT λ is antinodes is bigger, so the antinodes are further apart. proportional to θ). OR correctly uses path difference creating constructive interference explanation. NCEA Level 3 Physics (91523) 2017 — page 6 of 6

(d) The maximum angle that the spectrum can be formed is 90°. Uses 90 degrees. Correct answer. Correct working and answer. The light which has its antinode at the biggest angle is red because it has OR OR AND the longest wavelength. Red has a longer wavelength. Both correct explanations. Explanations of use of red So we calculate the slit separation that gives an angle of 90° for the red (replacement for 3c only) wavelength. light. OR AND v 3.00 ×108 λ = = = 6.98 ×10−7m Correct wavelength for red light. 90 degree angle (or maximum red f 4.30 ×1014 path difference = 3 nλ wavelength). sinθ = and n = 3 d nλ d = = 3× 6.98 ×10−7 sinθ d 2.09 10−6 m = ×

NØ N1 N2 A3 A4 M5 M6 E7 E8

0 u 2u OR i OR c 3u OR i+u 4u OR 2u+i OR 2i 2i+u OR c+i 2i+2u OR 3i 1u + 1i + 1c 2i + 1c OR c+u OR c+3u

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