<p> NCEA Level 3 Physics (91523) 2016 — page 1 of 6</p><p>Assessment Schedule – 2016 Physics: Demonstrate understanding of wave systems (91523) Evidence Statement</p><p>NØ N1 N2 A3 A4 M5 M6 E7 E8</p><p>0 1a 2a 3a 4a OR 1a + 1m 2a + 1m 1a + 2m 1a + 1m + 1e 1a + 2m + 1e</p><p>Q Evidence Achievement Merit Excellence</p><p>ONE Correct diagram and labels. (a)</p><p>(b) The first pipe produces a lower frequency, so it has a longer wavelength and a Lower frequency pipe / 350 Hz / Statement relating wavelength longer pipe length. first pipe is longer to pipe length. The wave speed is constant. AND one of AND v l = correct description of relationship Statement relating wavelength f so a lower frequency means a longer wavelength. between frequency and wavelength to frequency to explain that the first pipe is longer. The wavelength of a standing wave in a pipe is proportional to the pipe length. OR Longer pipes have longer wavelengths</p><p>(c) The phenomenon is called beating (beats). Beating occurs / beats occur. AND Explanation linking phase and The two pipes produce similar but different frequencies. ONE of: interference to loudness At one time, two waves arrive in phase and add constructively (loud), a short changing over time to produce Links loudness to phase or type of time later two waves arrive out of phase and add destructively (quiet). beats interference NCEA Level 3 Physics (91523) 2016 — page 2 of 6</p><p>(d) Correct value for 2nd pipe Correct value for 2nd pipe Correct answer and working. f1 = 762 Hz beat frequency = 4 Hz frequency. frequency AND realisation that 3l \ f2 = 766 Hz 2nd pipe contains . 3rd harmonic = 766 Hz 4 OR v 343 so l = = = 0.4478 m Correct calculation of L with f 766 incorrect frequency. 3l 3rd harmonic....pipe length = 4 3l 3´ 0.4478 L = = = 0.336 m 4 4</p><p>ALTERNATE SOLUTION</p><p> f3 = 762 Hz</p><p> f3' = f3 + 4 = 766 Hz ( f1 - f2 = fbeat )</p><p>3 f0 = 766 Hz 766 f = 0 3 v l = f0 l v 343´ 3 L = = = = 0.336 m 4 4 f 4 ´ 766 0 NCEA Level 3 Physics (91523) 2016 — page 3 of 6</p><p>Q Evidence Achievement Merit Excellence</p><p>TWO At X, the plane is moving towards Mike. The plane is catching up to waves Correct answer linking higher (a) Since the plane is moving, the centre of each new wavefront is slightly already emitted. frequency to shorter wavelength displaced to the right. As a result, the wavefronts bunch up on the right OR with reason. side, this reduces the wavelength. Shorter wavelength Shorter wavelength means a higher frequency OR æ v ö higher frequency. ç f = ÷ v is constant. è l ø</p><p>(b) At Y, the frequency heard is “normal” (185 Hz). Answer correct.</p><p>(c) As the plane accelerates to the right: Frequency gradually decreases. Frequency gradually decreases This will cause the wavelength of the waves behind the plane to gradually OR AND full explanation. increase. Wavelength gradually increases. AND the frequency will gradually decrease. (General description of the Doppler effect for a receding source can be used as a replacement for 2a.) NCEA Level 3 Physics (91523) 2016 — page 4 of 6</p><p>(d) v + v Correct frequency (171.5 Hz). Correct substitution into Correct answer with correct l¢ = w s f Doppler equation. working. 343+ v OR OR 2.00 = s –1 185 Correct Doppler calculation –27 m s with explanation that using incorrect frequency. the negative means the plane is v = 370 - 343 = 27 m s-1 s OR moving away. OR Correct Doppler calculation v 343 f = = = 171.5 Hz except for –sign in equation. l 2.00 (produces –27 m s–1) v f ¢ = f w vw ± vs 343 171.5 = 185 343+ vs 185 ´ 343 343+ v = s 171.5 185 ´ 343 v = - 343 s 171.5 -1 vs = 27.0 m s (accept 27) ALTERNATIVE SOLUTION: f = 185 v 343 f ¢ = = l' 2 æ ö vw f ¢ = f ç ÷ è vw + vs ø f ¢ v = w f vw + vs 343 343 2 = 185 343+ vs 343 0.9270 = 343+ vs</p><p>(343+ vs ) ´ 0.9270 = 343</p><p>318.0 + 0.9270vs = 343</p><p>0.9270vs = 25.0 v = 27 m s-1 s NCEA Level 3 Physics (91523) 2016 — page 5 of 6</p><p>Q Evidence Achievement Merit Excellence</p><p>THREE dx Correct formula and substitution. nl = (a) l 1.28 ´10-4 ´ 0.0100 l = 2.10 -7 l = 6.10 ´10 m</p><p>(b) 500 lines mm-1 = 5´105 l m-1 Correct calculation of d. Correct formula, substitution, answer and units. 1 -6 OR d = 5 = 2 ´10 m 5´10 Correct use of equation with nl 6.10 ´10-7 incorrect d. sinq = = = 0.305 d 2 ´10-6 q = 17.5°</p><p>(c) nl Antinodes get closer. Antinodes get closer sinq = d OR AND one of: So if wavelength decreases and d is constant, sin will decrease,(so  will Describes relationship between λ • Explanation using either decrease, so antinodes / bright spots get closer. and θ or x. equation. OR • Correctly uses path difference dx creating constructive nl = , so if  decreases, x will decrease, meaning the antinodes will get interference explanation. l closer. OR If the wavelength decreases, then the path difference to the antinodes will decrease. This will cause the angle between the antinodes to decrease, and the bright spots to get closer together. NCEA Level 3 Physics (91523) 2016 — page 6 of 6</p><p>(d) White light contains all the wavelengths of visible light. All the wavelengths White seen is central antinode. Correct explanation for: Comprehensive explanation of diffract as they pass through the diffraction grating and spread out through OR different colours have 1st White in centre / middle white central antinode AND 180º. Each slit acts as a point source of all the wavelengths of light in the antinode at different angle. caused by antinode / maxima coloured spectra AND dark visible spectrum. regions due to destructive OR different colours have different constructive interference / of interference of all visible Waves from each slit interfere with waves of the same wavelength from wavelengths. all colours / frequencies. wavelengths other slits. Interference is predominantly destructive due to the many closely OR dark regions are nodes AND one of: spaced sources. (attributed to destructive • Colours with larger In the dark region all visible wavelengths are experiencing destructive interference). wavelengths have 1st interfere. Only wavelengths invisible to the eye constructively interfere. No antinode at larger angles / light is seen. larger x. • Dark regions due to Individual colours are seen where constructive interference is occurring for destructive interference of all that colour but destructive interference is occurring for all other colours. The (visible) wavelengths. red antinode is at a larger angle than the violet, because the red light has a nl dx • Dark regions due to longer wavelength. As shown by sinq = or nl = . constructive interference of d l wavelengths not visible. The centre of the pattern is white because there is a central antinode for all colours. Here each wavelength of light constructively interferes with other sources of the same wavelength. The resulting white light is a composite maximum of all the colours.</p><p>Cut Scores</p><p>Not Achieved Achievement Achievement with Merit Achievement with Excellence</p><p>0 – 6 7 – 13 14 – 19 20 – 24</p>