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New Century Senior PhysicsSecond EditionWorked solutionsChapter 16 - Soundfrom Richard WaldingQ1Sound waves are longitudinal waves producing compressions and rarefactions of the air particles in the direction the wave propagates. The collision of particles of the material medium that carries the energy. Q2Faster in high density. Air (low density) v = 340 m s-1; steel (high density) v = 5960 m s-1Q3Compression is a region in a longitudinal wave where the particles are closest together. Rarefaction is a region in a longitudinal wave where the particles are furthest apart.Q4Frequency of 4 Hz gives the biggest movement of organs.Q5As they would not be playing a single frequency, there would not be nodal and antinodal lines mixed.Q5(a) v = d/t; t = d/v = 3/340 = 0.0088 s (b) v = f; = v/f = 340/1000 = 0.340 mQ6For water: t = d/v = 50/1410 = 0.035 sFor Copper: t = d/v = 50/3560 = 0.014 sHeard (0.035 - 0.014) = 0.021 s earlier in the pipeQ7v = f = 800 x 0.42 = 336 m/s. Yes, can be heard (800 Hz is between 50-20000 Hz).Q8Assume light is instantaneous. Distance for sound = vt = 340 x 0.5 = 170 mQ9Sound takes 0.75 s to get to cliff. Distance = vt = 340 x 0.75 = 255 mQ10Minimum intensity = node. The distance between nodal lines is the distance from the 1st nodal to the 2nd nodal.Wavelength = v/f = 340/1700 = 0.2m(n-?) = xd/LFor n= 1: (1-?) 0.2 = x 15x = ? 0.2 5 =0.5 mFor n = 2: (2-?) 0.2 = x 15x = 1? 0.2 5 =1.5 mThe distance between nodes is 1.5m - 0.5 m = 1.0 mQ11Assume light (smoke) is instantaneous)v = d/t = 200/0.6 = 333 m s-1Q12Higher frequency give the bats more detailed information such as size, range, position, speed and direction of a prey's flight. This is because the smaller wavelengths can reflect off small objects with less diffraction than longer wavelengths.Q13(a) The 1 s reflection is off the fish and the 3 s reflection is off the seafloor much further away.(b) If the total time for the fish reflection is 1.0 s then that means 0.5 s from boat to fish and 0.5 s back to the receiver in the boat. Using s = vt = 1450 x 0.5 = 725 m. The fish are 725 m below the surface.Q14The minimum wavelength of ultrasound waves bats can emit is 3.3 mm. What is the highest frequency of sound that bats can emit?v= f f = v/. Assuming the speed of sound in air is 340 m s-1, then f = 340/(3.3 x 10-3) = 103000 Hz (103 kHz)Q15Explain the technique of submarine captains resting their vessels on the bottom to prevent them being detected by a surface vessel’s sonar.When at rest the boat would seem to be a part of the seafloor. When it moves the sonar would show it moving against the backdrop of the seafloor. Sonar can also be used to measure the rate of movement (away or towards) the observer using Doppler principles.Q16Q17Fundamental is 1st harmonic: = 4 L = 4 × 0.3 = 1.2 mf = v/ = 340/1.2 = 283.3 HzFor fifth harmonic f5 = 5 × f1 = 5 × 283.3 = 1416.6 HzQ18Q19(a) L = 1? = 2L/3 = 2 x 0.2/3 = 0.133 m; f = v/ = 340/0.133 = 2550 Hz(b)L = 3 ; = L/3 = 0.2/3 = 0.067 m. Freq = v/ = 340/0.067 = 5100 Hz(c)L = ? ; = 2L = 0.40 m. Freq = v/ = 340/0.4 = 850 HzQ20Q21Q22 = 10 log (I/Io) = 10 log (10-7/10-12) = 10 x 5 = 50 dBQ23 = 10 log (I/Io) = 10 log (5 10–8 /10-12) = 47 dBQ24 = 10 log (I/Io) 85 = 10 log (I/Io) 8.5 = log (I/Io) 108.5 = I/IoI = 108.5 x 10-12 = 3.16 x 10-4 W m-2Q25(a) increase, ((b) decrease, (c) decrease, (d) increaseQ26(a) vs = -60 km h-1 (approaching) = -16.67 m s-1f'=fv vov- vs = 1200v 0330- 16.67 = 1200 x 330/313.33 = 1264 Hz(b) vs = + 60 km h-1 (receding) = + 16.67 m s-1f'=fv vov+ vs = 1200v+ 0330+ 16.67 = 1200 x 330/346.5 = 1142 HzQ27(a) vo = 0; vs = 120 km h-1 = 33.33 m s-1; source approaching so vs is negative: f'=fv vov- vs = 1000v 0340- 33.3 = 1109 Hz(b) source receding; vs is + 33.3 m s-1 f'=fv vov+ vs = 1000340 0340+ 33.3 = 911 Hz(c) as there is no motion between observer and source, the sound will not change (= 1000 Hz).Q28Transverse: at right angles to the direction of propagation; Longitudinal: parallel to the direction of propagation.(For all questions unless specified use vsound = 340 m s–1.)Q29(a) f = 2.4 x 104/10 = 240 Hz(b) = 0.14 m. Wave equation v = f = 240 x 0.14 = 33.6 m s-1Q30v = s/t s = vt = 340 x 10 = 3400 mQ31Q32 Q33Assume it takes 0.6 s to floor and 0.6 s back to surface.v = s/t s = vt = 1400 x 0.6 = 840 mQ34.Q35Q36Q37Q38Distance travelled by sound from source (clapping) to hearing echo is 600 m. Ten lots of this is 6000 m and it takes 17.5 s. Speed = s/t = 6000/17.5 = 342.8 m s-1.Q39 Q40Q41Q42The speed of sound depends on the temperature of the air. As temperature rises the speed increases. The fundamental standing wave in an open pipe, for example, has a frequency f = v/2L, so as v increases so does frequency. Q43 Resonance is heard when frequency produced by the air column is the same as the tuning fork.The diagrams show the two simplest cases where standing waves will form. The difference between Diagram 1 and 2 is half a wavelength,?therefore as in one case the length is 21 cm and in the other it is 59 cm, then (59 - 21 = 38) cm is equal to ??. So??= 76 cm.since v = f??then f = v /? = 334 /0.76m = 439 Hz.PLEASE NOTE: It is not correct to say Diagram 1 represents 1/4??so??= 88 cm as there is no indication of end correction. The method of subtraction used above has to be done. Q44fx = 245 Hz, and fy = 247 HzZ with x gives 3 beats per second. Therefore fz is either 245 + 3 = 248 Hz or 245 - 3 = 242 HzZ with y gives 1 beat per second. Therefore fz is either 247 + 1 = 248 Hz or 247 - 1 = 246 Hz.The common answer is fz = 248 Hz.Q45 Tuning fork’s frequency = 442 HzWith the ‘A ‘ string a beat frequency of 5 Hz is heard.Therefore the ‘A’ string is either 442 + 5 = 447 or 442 - 5 = 437 Hz.With the rubber band the beat frequency of 3 Hz is heard. Since by adding the rubber band slows the vibrations and thus the frequency. The fork will be less. If the ‘A’ string is 447 then the beat frequency will be greater as frequency of the tuning fork will be further from the string when a rubber band is added.Since it is closer to the string’s frequency (the beat frequency is now 3 Hz) the frequency of the ‘A’ string is 437 Hz.Q46 Since the source of the sound is moving towards the cliff the frequency of the sound measured at the cliff will be10 km h-1?= 2.8 m s-1.f1 = f v/(v - vs) = (480 × 330)/(330 - 2.8) = 484.1 HzHowever the cliff reflects this sound back to a moving observer the frequency of the sound measured by this observer will be: f1?= f(v + v0)/vTherefore f1?= 484.1 (330 + 2.8)/330 = 488.2 HzQ47 f = 440 Hz, with the train moving towards the station.The person at the station would hear a frequency of f’ = f(v/(v - vs) and will be of high frequency.Since there is a beat frequency of 3 the observer on the station would measure the frequency of the trumpet on the moving train to be (440 + 3) Hz = 443 Hz.solving the equation :-443 = 440 × (340 /(340 - vs)) calculates to vs = 2.3 m s-1 or 8 km h-1.Q48This question is best done graphically as it would have been done in WW1 in France. Decide on a scale so that a line 2000 m long will fit on the graph paper. Suggest 10 cm = 1000 m (a scale of 1:10000).Assume the three microphones A, B, C are on a straight line 1000 m apart as shown below. The gun must be a bit further away from B than A as the sound is first heard at A, then B, then C. Draw the gun in position as shown with A, B and C 10 cm apart. You cannot draw in the position of the gun as that is what you are trying to find. The sound radiates out from the gun in a spherical wavefront (or circular if drawn on paper in 2-D) as shown.The sound takes 1.2 seconds to travel to B after being heard at A.? Assume that the speed of sound in air is 330 m s-1, this corresponds to a distance of 330 × 1.2 = 396 m. If we draw a circle centered at B with a radius of 396 m (3.96 cm) then all points on that line represent the possible locations of a? sound that is 396 m away from B at the same time.The question also says that the sound at C is heard 3.7 s after it is heard at A. In this time a sound will have travelled 330 × 3.7 = 1221 m. So draw a circle centered on C with a radius of 12.21 cm (= 1221 m at a scale of 1:10000). Mark the intersection as point X.The sound of the gun is heard at point X at the same time as point A so draw a circle centered at X with a radius of XA. Then draw a circle centered at A with the same radius. Where these lines intersect is the location of the gun G. The distance from the gun to A (GA) is measured as 9 cm ( = 900 m). The angle BAG is 82°.?The angle BCG is about 25°.The question implied that there may be more than one solution. This is correct. An imaginary source of sound (in a mirror-image position on the opposite side of the line of microphones as in the diagram below) would also give rise to similar conditions at the microphones. This imaginary source would be ruled out by common sense.?????Algebraic methodOn the diagram above, the circular wavefront is shown as it reaches microphone A. Three lines are drawn from the Gun to microphones A, B and C. As well, a perpendicular "y" is dropped from the Gun to the line of the microphones. It is "x" metres from A. We have three right angled triangles each sharing a common side (y). Three simultaneous equations can be proposed with the two unknowns x and y.Equation 1 for?DGCX:? y2?= (1221 + s)2?- (2000 - x)2Equation 2 for?DGBX:? y2?= (396 + s)2?- (1000 - x)2Equation 3?for?DGAX:? y2?= s2?- x2Equating Eq 1 and 3:?12212?+ 2442s + s2?- 20002?+ 4000x? +x2? = s2?- x2x = 42 mEquating Eq 2 and 3:3962?+ 792s + s2?- 10002?+ 2000x - x2?= s2?- x2s = 958 mThe angle GAX = cos-1?(x/s) = cos-1?(42/958) = 87.5°***72f = 12LTMT = 1.00 × 9.8 = 9.8 NM = m/L = 3.2 × 10-3 kg/2.0 = 1.6 × 10-3 kg/mf = 12 1.29.81.6 10-3 = 0.417 6.125 ×103 = 0.417 78.26 = 32.63 Hz Standing waves will occur at all multiples of the fundamental:n = 2, f = 2 32.6 = 65.2 Hzn = 3, f = 3 32.6 = 97.89 Hz n = 4, f = 4 32.6 = 130.5 HzA standing wave is formed as an interference pattern between 2 waves of appropriate frequency. The interference for a standing wave results in specific areas of destructive and constructive interference to cause nodes and antinodes occurring at points along the string length that are unchanging. However, the 1kg mass tensioning the string does not allow a standing wave to be formed. None of the frequencies match the frequency of the vibrator therefore there will be no standing waves formed. There is still interference but not at constant points along the string. This can be seen in that the n value is not a whole number. ................
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