H211 Wave Motion - 2.1 Tutorial Solution (1718)

Topic 11: Waves Page 1 of 5 9749 H2 Physics Tutorial Solutions Waves Tutorial solution 1 (a) The principal difference is in the relative directions of oscillation and propagation/travel. These are parallel in longitudinal waves and perpendicular in transverse waves. (b) 1. Polarization 2. Need of a medium to propagate for longitudinal wave. (c) From the displacement-time graph, 3.5 waves occur in 17 ms. ∴ Period of wave, ܶ= ଵ଻×ଵ଴షయ ଷ.ହ = 0.0049 ݏ In the displacement-distance graph, 1.5 waves occupy 2.7 m. ∴ Wavelength, ߣ= ଶ.଻ ଵ.ହ = 1.8 ݉ Using , ݒ= ߣ݂= ఒ ் = ଵ.଼ ଴.଴଴ସଽ = 370 ݉ ݏିଵ [Remarks]: 1) Distinguishing transverse and longitudinal waves 2) Unique phenomenon of transverse waves 3) Taking average instead of direct reading of graph for period and wavelengths. 2 (a) Period, frequency and angular frequency (b) Period, ܶ= ்௢௧௔௟ ௧௜௠௘ ௧௔௞௘௡ ௙௢௥ ଶ ௖௢௠௣௟௘௧௘ ௪௔௩௘ ଶ = 1.0 ݏ c) Speed of wave, ݒ= ఒ ் = 0.05 1.0 = 0.05 ݉ ݏିଵ (d) Using ݒ= ߱ඥݔ଴ ଶ −ݔଶ, = 2ߨඥ0.12ଶ − 0ଶ = 0.75 ݉ ݏିଵ (e) Using ୼థ ଶగ = ୼௧ ் , Δ߶= 1 4 × 2ߨ= ߨ 2݀ܽݎ f) ݔ= −0.15 cos 2ݐߨ [Remarks]: 1) Finding period by taking average instead of direct reading of graph. 2) Calculating phase difference 3) Recognizing equation of graph 4) Distinguishing between wave speed and particle speed 3 To determine the phase difference between the two points, the formula ୼థ ଶగ = ୼௫ ఒ can be used. However, ߣ and Δݔ are not given in the qustion. To determine ߣwe can use the formula ݒ= ߣ݂ Hence ߣ= ௩ ௙ ߣ= 2.0 10 = 0.20 ݉ To find Δݔ Δݔ= (5.0 × 10ିଶ) sin 60° = √ଷ ସ଴ = 0.0433 ݉ Hence, Δ߶= ൬0.0433 0.20 ൰)(2ߨ) =1.36 ݀ܽݎ

Topic 11: Waves Page 2 of 5 9749 H2 Physics Tutorial Solutions 4 (a) (b) (i) D (transverse), A (longitudinal) (ii) B (transverse), C (longitudinal) (iii) C (transverse), D (longitudinal) (iv) A (transverse), B (longitudinal) [Remarks]: 1) Recognizing the direction of the particle when a wave passes through. 5 (a) Speed of microwaves, ݒ= 3.0 × 10଼ ݉ ݏିଵ ݐ= ݀ ݒ =4.35 × 10ଵଶ 3.0 × 10଼ = 1.45 × 10ସ ݏ b) Intensity of signal on Earth, ܫ= ௉ ସగ௥మ = 22.0 4ߨ(4.35 × 10ଵଶ)ଶ = 9.25 × 10ିଶ଺ ܹ ݉ିଶ Power received on Earth = IS = (9.25 × 10ିଶ଺)(260) = 2.41 × 10ିଶଷ ܹ c) The actual power received is greater because the signal from the satellite is directed towards Earth instead of being radiated uniformly in all directions, as assumed in (b). [Remarks]: 1) Microwave is an EM wave and hence it travels at speed of light. 2) Application of P = IS formula numerically. 3) Real life application versus assumptions. 6 (a) Using ܫ= ௉ ସగ௥మ and ratio method, ܫଶ ܫଵ = ൬ݎଵ ݎଶ ൰ ଶ ܫଶ = ൬ݎଵ ݎଶ ൰ ଶ ܫଵ = ൬1.0 5.0൰ ଶ (1 × 10ିହ) = 4.0 × 10ି଻ ܹ ݉ିଶ (b) Since ܣ∝ ଵ ௥ ܣଶ ܣଵ =ݎଵ ݎଶ ܣଶ = ൬1.0 5.0൰ (70) = 14 ݉ߤ [Remarks]: 1) The idea of ratio method in handling such problem involving proportions.

Topic 11: Waves Page 3 of 5 9749 H2 Physics Tutorial Solutions 7 By similar rectangles, The length of the second a