Assessment Schedule 2011 Physics: Demonstrate understanding of wave systems (90520)

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NCEA Level 3 Physics (90520) 20 page of 5 Assessment Schedule 20 Physics: Demonstrate understanding of wave systems (90520) Evidence Statement Q Evidence Achievement Merit Excellence ONE (a)(i) The incident wave is reflected at the bottom, and so the reflected wave travels up through the incident wave. Correct answer. Must give the idea of reflected wave travelling up and interfering or combining with the incident wave (ii) At a node the medium is fixed. The top of the string is fixed. The bottom of the string is free to move, and so will not be a node, The top end is a node because the two waves combine destructively and the bottom end is an antinode because the two waves combine constructively. Answer describes how the restriction on the movement of the string at the top causes a node. AND the bottom end is free to move/ not fixed, so will not be a node / be an antinode Achievement + since the top end is fixed so destructive interference (or crest on a trough) happens making it a node. Achievement + since the bottom end is free so constructive interference (or crest on a crest) happens making it an antinode. 3! = 2 λ = 6 cm 4 v = f λ f = 2.5 0.6 = 5.625 = 6 Hz 2 Correct wavelength. 6 cm. Incorrect wavelength is used to calculate the frequency but λ must be in metres. 2 Correct answer. 6 Hz.

NCEA Level 3 Physics (90520) 20 page 2 of 5 (c) f (2nd) = 2 f (st) = 2 (5.625 3) 0.526 Hz λ(2nd) = v f = 2.5 0.526 = 0.2375 = 24 cm 2 Correct fundamental = 5.2Hz Correct answer for the wavelength of the second harmonic wave can supply replacement evidence. 2 Correct answer,0 Hz or consequential to. ½ Correct frequency and correct explanation. The string is 2 cm long, which is half of the wavelength of the second harmonic (λ 2 = 24cm). The top end of the string has to be a node and 2 cm from the top will be another node and as the bottom of the string has to be an antinode, it is not possible to fit a wave of this length into the string. ONE idea is described correctly. BOTH ideas are linked correctly. Some other correct and valid explanation Do not accept closed pipe has only odd harmonics. TWO (a)(i) A spectrum is produced because the amount of bending as the light goes through the grating depends on the wavelength of the light (nλ = dsinθ). Correct statement of why a spectrum is produced. Differentwavelengths/frequencies /colour bend/diffract at different angles (ii) Because red light has the longest wavelength, it is bent away from the straight-through direction the most, so each spectrum will have its red side furthest from the centre. Correct idea that red end of the spectrum is furthest out from the central position. or The longest wavelength is furthest out. Correct explanation that red has longest wavelength. AND Longer (not long) wavelength diffracts more. n λ = d sinθ and so, for a particular wavelength, d is directly related to θ. This means that, as d gets smaller, each order spectrum is bent more and so there will be fewer spectra produced. Smaller d linked to spectra being more spread out. Smaller d is linked to fewer n without the evidence of the equation. Achievement, plus fewer spectra produced. Relates that n is directly proportional to d because nλ = d sinθ / λ is constant.

NCEA Level 3 Physics (90520) 20 page 3 of 5 (c) nλ = d sinθ, n =, λ = f c = 7.05 0 7 m 2 Correct λ 7.05 0 7 m 2 Correct sinθ =.070 sinθ = sinθ =.070 3.00! 0 4.25! 0 8 4 6.6" 0! 7 This angle does not exist (so light of this frequency does not form a spectrum). Sinθ is substituted as and then n is calculated to have a value of 0.935, and since this is not a whole number so no spectra will be seen. Correct answer. Do not accept mathematical error. (d)(i) The waves having wavelengths greater than 5.2 0 7 cannot be diffracted through an angle less than 90, so these wavelengths cannot contribute to the flash / spectrum. Some idea of why only part of a full spectrum is produced. Clear idea of why only part of a full spectrum is produced. Do not accept,light is reflected back. ½ Correct explanation and correct answer. (ii) nλ = d sinθ, n =, sin 90 = d = λ = 5.2 0 7 m = 5.2 0 4 mm N = = 923 = 900 lines mm! 4 5.2" 0 2 Correct answer without the evidence. Value of N is given in metres. 2 Correct answer 900 lines per mm with the evidence, evidence that : d = λ = 5.2 0 7 m, because sin 90 = evidence can come from any part of the question.

NCEA Level 3 Physics (90520) 20 page 4 of 5 THREE (a) When a source of waves is moving towards a detector, each time the source produces a wave crest, it is slightly nearer to the detector than it was when it produced the previous crest. This means the effective wavelength of the waves reaching the detector is less than the wavelength of the waves produced by the source. Some idea of why the wavelength is shorter, eg: Because waves bunch up. Waves get compressed. Source is catching up on the waves. Correct diagram. Do not accept, since frequency is higher so λ is shorter. Clear idea of why the wavelength is shorter. As given in the evidence. The decreasing distance between the source and detector is linked to successive wave is emitted closer to the detector so apparent λ is shorter. Catching up of the source is linked to, each successive wave is emitted closer to the detector so apparent λ is shorter. Waves bunch up because velocity of the waves relative to the source has decreased, so waves have less space to occupy.

NCEA Level 3 Physics (90520) 20 page 5 of 5 f' = v! = 545 =.0537 "06 #3.467 "0 v W f' = f v W # v S v S = v W f'# f f' (.0532 #.0522) "06 = 545 =.47 m s #.0522 "0 6! = v f = 545 =.4684 "0#3 6.0522 "0 $! = (.4684 #.4670) "0 #3 =.4 "0 #6 m T = f =.0522 "0 6 v = $! T =.4 "0#6 ".0522 "0 6 =.47 m s # 2 Correct f.0537 0 6 Hz. Wrong f,but rest of the substitution is correct. λ is calculated =.4684 0 3 m velocity is calculated as v = 543.5774 m s but does not mention that it is the relative velocity. 2 Correct f. + Correct substitution into formula. λ is calculated =.4684 0 3 m AND T =/f = /.0522 0 6 But does not calculate. v = change in λ /T =,423) Relative velocity is calculated as v = 543.5774 m s mentions that it is the relative velocity. 2 Correct answer..423 m s Allow answers that have correctly used the double Doppler equation, typically used by radiologists, or from first principles. Relative v = f λ =.0522 0 6.4670 0-3 = 543.5774 m s Speed of sound in blood = 545 m s Difference between relative v and v = 545 543.577 =.4226 m s Judgement Statement Achievement Achievement with Merit Achievement with Excellence 2 A + 2 A2 2 M + 2 M2 + A 2 E + 2 M + M2 + A Note: where the criterion is not specified, the required grade(s) can be from either.