MECHANICAL WAVES 15.1 15 IDENTIFY: v = f λ T = 1/f is the time for one complete vibration SET UP: The frequency of the note one octave higher is 1568 Hz v 344 m/s = 0.439 m T = = 1.28 ms EXECUTE: (a) λ = = f 784 Hz f v 344 m/s = = 0.219 m f 1568 Hz EVALUATE: When f is doubled, λ is halved IDENTIFY: The distance between adjacent dots is λ v = f λ The long-wavelength sound has the lowest (b) λ = 15.2 frequency, 20.0 Hz, and the short-wavelength sound has the highest frequency, 20.0 kHz SET UP: For sound in air, v = 344 m/s v 344 m/s EXECUTE: (a) Red dots: λ = = = 17.2 m f 20.0 Hz Blue dots: λ = 344 m/s = 0.0172 m = 1.72 cm 20.0 × 103 Hz (b) In each case the separation easily can be measured with a meterstick v 1480 m/s (c) Red dots: λ = = = 74.0 m 20.0 Hz f 1480 m/s = 0.0740 m = 7.40 cm In each case the separation easily can be measured 20.0 × 103 Hz with a meterstick, although for the red dots a long tape measure would be more convenient EVALUATE: Larger wavelengths correspond to smaller frequencies When the wave speed increases, for a given frequency, the wavelength increases IDENTIFY: v = f λ = λ /T SET UP: 1.0 h = 3600 s The crest to crest distance is λ Blue dots: λ = 15.3 800 × 103 m 800 km = 220 m/s v = = 800 km/h 3600 s 1.0 h EVALUATE: Since the wave speed is very high, the wave strikes with very little warning IDENTIFY: f λ = v SET UP: 1.0 mm = 0.0010 m v 1500 m/s EXECUTE: f = = = 1.5 × 106 Hz λ 0.0010 m EVALUATE: The frequency is much higher than the upper range of human hearing IDENTIFY: We want to relate the wavelength and frequency for various waves SET UP: For waves v = f λ EXECUTE: v = 15.4 15.5 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 15-1 15-2 Chapter 15 EXECUTE: (a) v = 344 m/s For f = 20,000 Hz, λ = λ= v 344 m/s = = 17 m The range of wavelengths is 1.7 cm to 17 m f 20 Hz (b) v = c = 3.00 × 108 m/s For λ = 700 nm, f = f = v 344 m/s = = 1.7 cm For f = 20 Hz, f 20,000 Hz c λ = 3.00 × 108 m/s 400 × 10−9 m c λ = 3.00 × 108 m/s 700 × 10−9 m = 4.3 × 1014 Hz For λ = 400 nm, = 7.5 × 1014 Hz The range of frequencies for visible light is 4.3 × 1014 Hz to 7.5 × 1014 Hz (c) v = 344 m/s λ = v 344 m/s = = 1.5 cm f 23 × 103 Hz v 1480 m/s = = 6.4 cm f 23 × 103 Hz EVALUATE: For a given v, a larger f corresponds to smaller λ For the same f, λ increases when v increases IDENTIFY: The fisherman observes the amplitude, wavelength, and period of the waves SET UP: The time from the highest displacement to lowest displacement is T /2 The distance from highest displacement to lowest displacement is 2A The distance between wave crests is λ , and the speed of the waves is v = f λ = λ /T (d) v = 1480 m/s λ = 15.6 EXECUTE: (a) T = 2(2.5 s) = 5.0 s λ = 6.0 m v = (b) A = (0.62 m)/2 = 0.31 m 15.7 6.0 m = 1.2 m/s 5.0 s (c) The amplitude becomes 0.15 m but the wavelength, period and wave speed are unchanged EVALUATE: The wavelength, period and wave speed are independent of the amplitude of the wave IDENTIFY: Use Eq (15.1) to calculate v T = 1/f and k is defined by Eq (15.5) The general form of the wave function is given by Eq (15.8), which is the equation for the transverse displacement SET UP: v = 8.00 m/s, A = 0.0700 m, λ = 0.320 m EXECUTE: (a) v = f λ so f = v/λ = (8.00 m/s)/(0.320 m) = 25.0 Hz T = 1/f = 1/25.0 Hz = 0.0400 s k = 2π /λ = 2π rad/0.320 m = 19.6 rad/m (b) For a wave traveling in the − x -direction, y ( x, t ) = A cos 2π ( x/λ + t/T ) (Eq (15.8).) At x = 0, y (0, t ) = A cos 2π (t/T ), so y = A at t = This equation describes the wave specified in the problem Substitute in numerical values: y ( x, t ) = (0.0700 m)cos(2π ( x/0.320 m + t/0.0400 s)) Or, y ( x, t ) = (0.0700 m)cos((19.6 m −1 ) x + (157 rad/s)t ) (c) From part (b), y = (0.0700 m)cos(2π ( x/0.320 m + t/0.0400 s)) Plug in x = 0.360 m and t = 0.150 s: y = (0.0700 m)cos(2π (0.360 m/0.320 m + 0.150 s/0.0400 s)) y = (0.0700 m)cos[2π (4.875 rad)] = +0.0495 m = +4.95 cm (d) In part (c) t = 0.150 s y = A means cos(2π ( x/λ + t/T )) = cosθ = for θ = 0, 2π , 4π ,… = n(2π ) or n = 0, 1, 2,… So y = A when 2π ( x/λ + t/T ) = n(2π ) or x/λ + t/T = n t = T ( n − x/λ ) = (0.0400 s)( n − 0.360 m/0.320 m) = (0.0400 s)( n − 1.125) For n = 4, t = 0.1150 s (before the instant in part (c)) © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Mechanical Waves 15-3 For n = 5, t = 0.1550 s (the first occurrence of y = A after the instant in part (c)) Thus the elapsed time is 0.1550 s − 0.1500 s = 0.0050 s EVALUATE: Part (d) says y = A at 0.115 s and next at 0.155 s; the difference between these two times is 0.040 s, which is the period At t = 0.150 s the particle at x = 0.360 m is at y = 4.95 cm and traveling upward It takes T/4 = 0.0100 s for it to travel from y = to y = A, so our answer of 0.0050 s is 15.8 reasonable IDENTIFY: Compare y ( x, t ) given in the problem to the general form of Eq (15.4) f = 1/T and v = f λ SET UP: The comparison gives A = 6.50 mm, λ = 28.0 cm and T = 0.0360 s EXECUTE: (a) 6.50 mm (b) 28.0 cm = 27.8 Hz 0.0360 s (d) v = (0.280 m)(27.8 Hz) = 7.78 m/s (e) Since there is a minus sign in front of the t/T term, the wave is traveling in the +x -direction EVALUATE: The speed of propagation does not depend on the amplitude of the wave IDENTIFY: Evaluate the partial derivatives and see if Eq (15.12) is satisfied ∂ ∂ SET UP: cos(kx + ω t ) = −k sin( kx + ω t ) cos( kx + ω t ) = −ω sin(kx + ω t ) ∂x ∂t ∂ ∂ sin( kx + ω t ) = k cos(kx + ω t ) sin(kx + ω t ) = ω cos(kx + ω t ) ∂t ∂x (c) f = 15.9 EXECUTE: (a) (b) ∂2 y ∂x ∂2 y ∂x = − Ak cos(kx + ω t ) = − Ak sin( kx + ω t ) ∂2 y ∂t ∂2 y ∂t = − Aω cos(kx + ω t ) Eq (15.12) is satisfied, if v = ω /k = − Aω sin(kx + ω t ) Eq (15.12) is satisfied, if v = ω /k ∂2 y ∂y ∂y ∂2 y = −kA sin(kx) = −ω A sin(ω t ) = −ω A cos(ω t ) Eq (15.12) is not = − k A cos(kx) ∂x ∂t ∂x ∂t satisfied ∂2 y ∂y = ω A cos(kx + ω t ) a y = = − Aω sin(kx + ω t ) (d) v y = ∂t ∂t EVALUATE: The functions cos( kx + ω t ) and sin(kx + ω t ) differ only in phase IDENTIFY: The general form of the wave function for a wave traveling in the −x -direction is given by Eq (15.8) The time for one complete cycle to pass a point is the period T and the number that pass per second is the frequency f The speed of a crest is the wave speed v and the maximum speed of a particle in the medium is vmax = ω A SET UP: Comparison to Eq (15.8) gives A = 3.75 cm, k = 0.450 rad/cm and ω = 5.40 rad/s 2π rad 2π rad EXECUTE: (a) T = = = 1.16 s In one cycle a wave crest travels a distance 5.40 rad/s ω 2π rad 2π rad = = 0.140 m λ= k 0.450 rad/cm (b) k = 0.450 rad/cm f = 1/T = 0.862 Hz = 0.862 waves/second (c) 15.10 (c) v = f λ = (0.862 Hz)(0.140 m) = 0.121 m/s vmax = ω A = (5.40 rad/s)(3.75 cm) = 0.202 m/s 15.11 EVALUATE: The transverse velocity of the particles in the medium (water) is not the same as the velocity of the wave IDENTIFY and SET UP: Read A and T from the graph Apply Eq (15.4) to determine λ and then use Eq (15.1) to calculate v EXECUTE: (a) The maximum y is mm (read from graph) (b) For either x the time for one full cycle is 0.040 s; this is the period © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 15-4 Chapter 15 (c) Since y = for x = and t = and since the wave is traveling in the +x -direction then y ( x, t ) = A sin[2π (t/T − x/λ )] (The phase is different from the wave described by Eq (15.4); for that wave y = A for x = 0, t = 0.) From the graph, if the wave is traveling in the +x -direction and if x = and x = 0.090 m are within one wavelength the peak at t = 0.01 s for x = moves so that it occurs at t = 0.035 s (read from graph so is approximate) for x = 0.090 m The peak for x = is the first peak past t = so corresponds to the first maximum in sin[2π (t/T − x/λ )] and hence occurs at 2π (t / T − x/λ ) = π /2 If this same peak moves to t1 = 0.035 s at x1 = 0.090 m, then 2π (t/T − x/λ ) = π /2 Solve for λ: t1/T − x1/λ = 1/4 x1/λ = t1/T − 1/4 = 0.035 s/0.040 s − 0.25 = 0.625 λ = x1/0.625 = 0.090 m/0.625 = 0.14 m Then v = f λ = λ /T = 0.14 m/0.040 s = 3.5 m/s (d) If the wave is traveling in the − x-direction, then y ( x, t ) = A sin(2π (t/T + x/λ )) and the peak at t = 0.050 s for x = corresponds to the peak at t1 = 0.035 s for x1 = 0.090 m This peak at x = is the second peak past the origin so corresponds to 2π (t/T + x/λ ) = 5π /2 If this same peak moves to t1 = 0.035 s for x1 = 0.090 m, then 2π (t1/T + x1/λ ) = 5π /2 t1/T + x1/λ = 5/4 x1/λ = 5/4 − t1/T = 5/4 − 0.035 s/0.040 s = 0.375 λ = x1/0.375 = 0.090 m/0.375 = 0.24 m Then v = f λ = λ /T = 0.24 m/0.040 s = 6.0 m/s 15.12 EVALUATE: (e) No Wouldn’t know which point in the wave at x = moved to which point at x = 0.090 m ∂y IDENTIFY: v y = v = f λ = λ /T ∂t ∂ ⎛ 2π ⎞ ⎛ 2π v ⎞ ⎛ 2π ⎞ SET UP: ( x − vt ) ⎟ = + A ⎜ ( x − vt ) ⎟ A cos ⎜ ⎟ sin ⎜ ∂t ⎝ λ ⎠ ⎝ λ ⎠ ⎝ λ ⎠ 2π ⎛ λ ⎞ 2π λ ⎛x t⎞ EXECUTE: (a) A cos 2π ⎜ − ⎟ = + A cos ⎜ x − t ⎟ = + A cos ( x − vt ) where = λ f = v has been used ⎝λ T⎠ λ ⎝ λ T ⎠ T ∂y 2π v 2π = A sin ( x − vt ) λ λ ∂t (c) The speed is the greatest when the sine is 1, and that speed is 2π vA/λ This will be equal to v if A = λ /2π , less than v if A < λ /2π and greater than v if A > λ /2π EVALUATE: The propagation speed applies to all points on the string The transverse speed of a particle of the string depends on both x and t IDENTIFY: Follow the procedure specified in the problem SET UP: For λ and x in cm, v in cm/s and t in s, the argument of the cosine is in radians EXECUTE: (a) t = 0: 0.00 1.50 3.00 4.50 6.00 7.50 9.00 10.50 12.00 x(cm) 0.212 0.300 0.300 0.212 −0.212 −0.300 −0.212 y(cm) The graph is shown in Figure 15.13a (b) (i) t = 0.400 s: 0.00 1.50 3.00 4.50 6.00 7.50 9.00 10.50 12.00 x(cm) 0.203 0.300 0.221 0.0131 −0.221 −0.0131 −0.203 −0.300 −0.221 y(cm) The graph is shown in Figure 15.13b (ii) t = 0.800 s: 0.00 1.50 3.00 4.50 6.00 7.50 9.00 10.50 12.00 x(cm) 0.193 0.300 0.230 0.0262 0.0262 −0.193 −0.300 −0.230 −0.0262 y(cm) The graph is shown in Figure 15.13c (iii) The graphs show that the wave is traveling in the + x -direction (b) v y = 15.13 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Mechanical Waves 15-5 EVALUATE: We know that Eq (15.3) is for a wave traveling in the + x-direction, and y ( x, t ) is derived from this This is consistent with the direction of propagation we deduced from our graph Figure 15.13 15.14 IDENTIFY: v y and a y are given by Eqs (15.9) and (15.10) SET UP: The sign of v y determines the direction of motion of a particle on the string If v y = and a y ≠ the speed of the particle is increasing If v y ≠ 0, the particle is speeding up if v y and a y have the same sign and slowing down if they have opposite signs EVALUATE: (a) The graphs are given in Figure 15.14 (b) (i) v y = ω A sin(0) = and the particle is instantaneously at rest a y = −ω A cos(0) = −ω A and the particle is speeding up (ii) v y = ω A sin(π /4) = ω A/ 2, and the particle is moving up a y = −ω A cos(π /4) = −ω A/ 2, and the particle is slowing down ( v y and a y have opposite sign) (iii) v y = ω A sin(π /2) = ω A and the particle is moving up a y = −ω A cos(π /2) = and the particle is instantaneously not accelerating (iv) v y = ω A sin(3π /4) = ω A/ 2, and the particle is moving up a y = −ω A cos(3π /4) = ω A/ 2, and the particle is speeding up © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 15-6 Chapter 15 (v) v y = ω A sin(π ) = and the particle is instantaneously at rest a y = −ω A cos(π ) = ω A and the particle is speeding up (vi) v y = ω A sin(5π /4) = −ω A/ and the particle is moving down a y = −ω A cos(5π /4) = ω A/ and the particle is slowing down ( v y and a y have opposite sign) (vii) v y = ω A sin(3π /2) = − ω A and the particle is moving down a y = −ω A cos(3π /2) = and the particle is instantaneously not accelerating (viii) v y = ω A sin(7π /4) = −ω A/ 2, and the particle is moving down a y = −ω A cos(7π /4) = −ω A/ and the particle is speeding up ( v y and a y have the same sign) EVALUATE: At t = the wave is represented by Figure 15.10a in the textbook: point (i) in the problem corresponds to the origin, and points (ii)–(viii) correspond to the points in the figure labeled 1–7 Our results agree with what is shown in the figure Figure 15.14 15.15 IDENTIFY and SET UP: Use Eq (15.13) to calculate the wave speed Then use Eq (15.1) to calculate the wavelength EXECUTE: (a) The tension F in the rope is the weight of the hanging mass: F = mg = (1.50 kg)(9.80 m/s ) = 14.7 N v = F/μ = 14.7 N/(0.0550 kg/m) = 16.3 m/s (b) v = f λ so λ = v/f = (16.3 m/s)/120 Hz = 0.136 m (c) EVALUATE: v = F/μ , where F = mg Doubling m increases v by a factor of 15.16 λ = v/f f remains 120 Hz and v increases by a factor of 2, so λ increases by a factor of IDENTIFY: The frequency and wavelength determine the wave speed and the wave speed depends on the tension F SET UP: v = μ = m/L v = f λ μ 0.120 kg ([40.0 Hz][0.750 m]) = 43.2 N 2.50 m EVALUATE: If the frequency is held fixed, increasing the tension will increase the wavelength IDENTIFY: The speed of the wave depends on the tension in the wire and its mass density The target variable is the mass of the wire of known length F SET UP: v = and μ = m/L EXECUTE: F = μ v = μ ( f λ ) = 15.17 μ EXECUTE: First find the speed of the wave: v = μ= F = 3.80 m = 77.24 m/s v = 0.0492 s F μ (54.0 kg)(9.8 m/s ) = 0.08870 kg/m The mass of the wire is v2 (77.24 m/s) m = μ L = (0.08870 kg/m)(3.80 m) = 0.337 kg EVALUATE: This mass is 337 g, which is a bit large for a wire 3.80 m long It must be fairly thick © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Mechanical Waves 15.18 15-7 IDENTIFY: For transverse waves on a string, v = F/μ The general form of the equation for waves traveling in the +x -direction is y ( x, t ) = A cos(kx − ω t ) For waves traveling in the −x-direction it is y ( x, t ) = A cos( kx + ω t ) v = ω /k SET UP: Comparison to the general equation gives A = 8.50 mm, k = 172 rad/m and ω = 4830 rad/s The string has mass 0.00128 kg and μ = m/L = 0.000850 kg/m EXECUTE: (a) v = ω k = 4830 rad/s 1.50 m d = 28.08 m/s t = = = 0.0534 s = 53.4 ms 172 rad/m v 28.08 m/s (b) W = F = μv = (0.000850 kg/m)(28.08 m/s) = 0.670 N 2π rad 2π rad = = 0.0365 m The number of wavelengths along the length of the string is k 172 rad/m 1.50 m = 41.1 0.0365 m (d) For a wave traveling in the opposite direction, y ( x, t ) = (8.50 mm)cos([172 rad/m]x + [4830 rad/s]t ) EVALUATE: We have assumed that the tension in the string is constant and equal to W This is reasonable since W 0.0125 N, so the weight of the string has a negligible effect on the tension (c) λ = 15.19 IDENTIFY: For transverse waves on a string, v = F/μ v = f λ SET UP: The wire has μ = m/L = (0.0165 kg)/(0.750 m) = 0.0220 kg/m EXECUTE: (a) v = f λ = (875 Hz)(3.33 × 10−2 m) = 29.1 m/s The tension is F = μ v = (0.0220 kg/m)(29.1 m/s) = 18.6 N (b) v = 29.1 m/s EVALUATE: If λ is kept fixed, the wave speed and the frequency increase when the tension is increased 15.20 IDENTIFY: Apply ΣFy = to determine the tension at different points of the rope v = F/μ SET UP: From Example 15.3, msamples = 20.0 kg, mrope = 2.00 kg and μ = 0.0250 kg/m EXECUTE: (a) The tension at the bottom of the rope is due to the weight of the load, and the speed is the same 88.5m/s as found in Example 15.3 (b) The tension at the middle of the rope is (21.0 kg)(9.80m/s ) = 205.8 N and the wave speed is 90.7 m/s (c) The tension at the top of the rope is (22.0 kg)(9.80 m/s ) = 215.6 N and the speed is 92.9 m/s (See 15.21 Challenge Problem (15.84) for the effects of varying tension on the time it takes to send signals.) EVALUATE: The tension increases toward the top of the rope, so the wave speed increases from the bottom of the rope to the top of the rope IDENTIFY: v = F/μ v = f λ The general form for y ( x, t ) is given in Eq (15.4), where T = 1/f Eq (15.10) says that the maximum transverse acceleration is amax = ω A = (2π f ) A SET UP: μ = 0.0500 kg/m EXECUTE: (a) v = F/μ = (5.00 N)/(0.0500) kg/m = 10.0 m/s (b) λ = v/f = (10.0 m/s)/(40.0 Hz) = 0.250 m (c) y ( x, t ) = A cos(kx − ω t ) k = 2π /λ = 8.00π rad/m; ω = 2π f = 80.0π rad/s y ( x, t ) = (3.00 cm)cos[π (8.00 rad/m)x − (80.0π rad/s)t ] (d) v y = + Aω sin(kx − ω t ) and a y = − Aω 2cos(kx − ω t ) a y , max = Aω = A(2π f ) = 1890 m/s (e) a y ,max is much larger than g, so it is a reasonable approximation to ignore gravity 15.22 EVALUATE: y ( x, t ) in part (c) gives y (0,0) = A, which does correspond to the oscillator having maximum upward displacement at t = IDENTIFY: Apply Eq (15.25) SET UP: ω = 2π f μ = m/L © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 15-8 Chapter 15 EXECUTE: (a) Pav = Pav = μ F ω A2 ⎛ 3.00 × 10−3 kg ⎞ −3 2 ⎜⎜ ⎟⎟ (25.0 N)(2π (120.0 Hz)) (1.6 × 10 m) = 0.223 W or 0.22 W to two figures 0.80 m ⎝ ⎠ (b) Pav is proportional to A2 , so halving the amplitude quarters the average power, to 0.056 W 15.23 EVALUATE: The average power is also proportional to the square of the frequency IDENTIFY: The average power carried by the wave depends on the mass density of the wire and the tension in it, as well as on the square of both the frequency and amplitude of the wave (the target variable) F SET UP: Pav = μ F ω A2 , v = μ ⎛ 2P EXECUTE: Solving Pav = μ F ω A2 for A gives A = ⎜ av ⎜ ω μF ⎝ 1/2 ⎞ ⎟⎟ ⎠ Pav = 0.365 W ω = 2π f = 2π (69.0 Hz) = 433.5 rad/s The tension is F = 94.0 N and v = μ= F v = 94.0 N (492 m/s)2 F μ so = 3.883 × 10−4 kg/m 1/2 15.24 ⎛ ⎞ 2(0.365 W) ⎟ = 4.51 × 10−3 m = 4.51 mm A=⎜ ⎜ (433.5 rad/s) (3.883 × 10−4 kg/m)(94.0 N) ⎟ ⎝ ⎠ EVALUATE: Vibrations of strings and wires normally have small amplitudes, which this wave does IDENTIFY: The average power (the target variable) is proportional to the square of the frequency of the wave and therefore it is inversely proportional to the square of the wavelength F SET UP: Pav = μ F ω A2 where ω = 2π f The wave speed is v = μ EXECUTE: ω = 2π f = 2π to v λ = 2π F λ μ so Pav = 4π ⎛ F ⎞ μ F ⎜ ⎟ A2 This shows that Pav is proportional λ ⎝μ⎠ 2 ⎛λ ⎞ ⎛ λ ⎞ Therefore Pav,1λ12 = Pav,2λ22 and Pav,2 = Pav,1 ⎜ ⎟ = (0.400 W) ⎜ ⎟ = 0.100 W λ ⎝ λ2 ⎠ ⎝ 2λ1 ⎠ EVALUATE: The wavelength is increased by a factor of 2, so the power is decreased by a factor of 22 = 15.25 IDENTIFY: For a point source, I = P 4π r and I1 r22 = I r12 SET UP: μ W = 10−6 W EXECUTE: (a) r2 = r1 I1 10.0 W/m = (30.0 m) = 95 km I2 × 10−6 W/m 2 (b) ⎛r ⎞ I r32 = , with I = 1.0 μ W/m and r3 = 2r2 I = I ⎜ ⎟ = I /4 = 0.25 μ W/m I3 r2 ⎝ r3 ⎠ (c) P = I (4π r ) = (10.0 W/m )(4π )(30.0 m)2 = 1.1 × 105 W 15.26 EVALUATE: These are approximate calculations, that assume the sound is emitted uniformly in all directions and that ignore the effects of reflection, for example reflections from the ground IDENTIFY: Apply Eq (15.26) SET UP: I1 = 0.11 W/m r1 = 7.5 m Set I = 1.0 W/m and solve for r2 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Mechanical Waves EXECUTE: r2 = r1 15-9 I1 0.11 W/m = (7.5 m) = 2.5 m, so it is possible to move I2 1.0 W/m r1 − r2 = 7.5 m − 2.5 m = 5.0 m closer to the source 15.27 EVALUATE: I increases as the distance r of the observer from the source decreases IDENTIFY: and SET UP: Apply Eq (15.26) to relate I and r Power is related to intensity at a distance r by P = I (4π r ) Energy is power times time EXECUTE: (a) I1r12 = I 2r22 I = I1 (r1/r2 ) = (0.026 W/m )(4.3 m/3.1 m) = 0.050 W/m (b) P = 4π r I = 4π (4.3 m) (0.026 W/m ) = 6.04 W Energy = Pt = (6.04 W)(3600 s) = 2.2 × 104 J 15.28 EVALUATE: We could have used r = 3.1 m and I = 0.050 W/m in P = 4π r I and would have obtained the same P Intensity becomes less as r increases because the radiated power spreads over a sphere of larger area IDENTIFY: The tension and mass per unit length of the rope determine the wave speed Compare y ( x, t ) given in the problem to the general form given in Eq (15.8) v = ω /k The average power is given by Eq (15.25) SET UP: Comparison with Eq (15.8) gives A = 2.30 mm, k = 6.98 rad/m and ω = 742 rad/s EXECUTE: (a) A = 2.30 mm ω = 742 rad/s = 118 Hz 2π 2π π 2π (c) λ = = = 0.90 m 6.98 rad/m k (d) v = ω = 742 rad/s = 106 m/s k 6.98 rad/m (e) The wave is traveling in the −x-direction because the phase of y ( x, t ) has the form kx + ω t (b) f = (f) The linear mass density is μ = (3.38 × 10−3 kg)/(1.35 m) = 2.504 × 10−3 kg/m, so the tension is F = μ v = (2.504 × 10−3 kg/m)(106.3 m/s) = 28.3 N (g) Pav = 12 μ F ω A2 = 12 (2.50 × 10−3 kg/m)(28.3 N)(742 rad/s) (2.30 × 10−3 m) = 0.39 W EVALUATE: In part (d) we could also calculate the wave speed as v = f λ and we would obtain the same 15.29 result IDENTIFY: The intensity obeys an inverse square law P SET UP: I = , where P is the target variable 4π r EXECUTE: Solving for the power gives P = (4π r ) I = 4π (7.00 × 1012 m) (15.4 W/m ) = 9.48 × 1027 W 15.30 EVALUATE: The intensity of the radiation is decreased enormously due to the great distance from the star IDENTIFY: The distance the wave shape travels in time t is vt The wave pulse reflects at the end of the string, at point O SET UP: The reflected pulse is inverted when O is a fixed end and is not inverted when O is a free end EXECUTE: (a) The wave form for the given times, respectively, is shown in Figure 15.30a (b) The wave form for the given times, respectively, is shown in Figure 15.30b EVALUATE: For the fixed end the result of the reflection is an inverted pulse traveling to the left and for the free end the result is an upright pulse traveling to the left © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 15-10 Chapter 15 Figure 15.30 15.31 IDENTIFY: The distance the wave shape travels in time t is vt The wave pulse reflects at the end of the string, at point O SET UP: The reflected pulse is inverted when O is a fixed end and is not inverted when O is a free end EXECUTE: (a) The wave form for the given times, respectively, is shown in Figure 15.31a (b) The wave form for the given times, respectively, is shown in Figure 15.31b EVALUATE: For the fixed end the result of the reflection is an inverted pulse traveling to the left and for the free end the result is an upright pulse traveling to the left Figure 15.31 15.32 IDENTIFY: Apply the principle of superposition SET UP: The net displacement is the algebraic sum of the displacements due to each pulse EXECUTE: The shape of the string at each specified time is shown in Figure 15.32 EVALUATE: The pulses interfere when they overlap but resume their original shape after they have completely passed through each other Figure 15.32 15.33 IDENTIFY: Apply the principle of superposition SET UP: The net displacement is the algebraic sum of the displacements due to each pulse EXECUTE: The shape of the string at each specified time is shown in Figure 15.33 EVALUATE: The pulses interfere when they overlap but resume their original shape after they have completely passed through each other © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 15-18 15.51 Chapter 15 IDENTIFY and SET UP: Calculate v, ω , and k from Eqs (15.1), (15.5) and (15.6) Then apply Eq (15.7) to obtain y ( x, t ) A = 2.50 × 10−3 m, λ = 1.80 m, v = 36.0 m/s EXECUTE: (a) v = f λ so f = v/λ = (36.0 m/s)/1.80 m = 20.0 Hz ω = 2π f = 2π (20.0 Hz) = 126 rad/s k = 2π /λ = 2π rad/1.80 m = 3.49 rad/m (b) For a wave traveling to the right, y ( x, t ) = A cos(kx − ω t ) This equation gives that the x = end of the string has maximum upward displacement at t = Put in the numbers: y ( x, t ) = (2.50 × 10 −3 m)cos((3.49 rad/m)x − (126 rad/s)t (c) The left-hand end is located at x = Put this value into the equation of part (b): y (0, t ) = + (2.50 × 10−3 m)cos((126 rad/s)t ) (d) Put x = 1.35 m into the equation of part (b): y (1.35 m, t ) = (2.50 × 10 −3 m)cos((3.49 rad/m)(1.35 m) − (126 rad/s)t ) y (1.35 m, t ) = (2.50 × 10−3 m)cos(4.71 rad − (126 rad/s)t ) 4.71 rad = 3π /2 and cos(θ ) = cos(−θ ), so y (1.35 m, t ) = (2.50 × 10 −3 m)cos((126 rad/s)t − 3π /2 rad) (e) y = A cos(kx − ω t ) (part (b)) The transverse velocity is given by v y = ∂y ∂ = A cos( kx − ω t ) = + Aω sin( kx − ω t ) ∂t ∂t The maximum v y is Aω = (2.50 × 10−3 m)(126 rad/s) = 0.315 m/s (f) y ( x, t ) = (2.50 × 10−3 m)cos((3.49 rad/m)x − (126 rad/s)t ) t = 0.0625 s and x = 1.35 m gives y = (2.50 × 10−3 m)cos((3.49 rad/m)(1.35 m) − (126 rad/s)(0.0625 s)) = −2.50 × 10−3 m v y = + Aω sin( kx − ω t ) = + (0.315 m/s)sin((3.49 rad/m) x − (126 rad/s)t ) t = 0.0625 s and x = 1.35 m gives v y = (0.315 m/s)sin((3.49 rad/m)(1.35 m) − (126 rad/s)(0.0625 s)) = 0.0 EVALUATE: The results of part (f) illustrate that v y = when y = ± A, as we saw from SHM in 15.52 Chapter 14 IDENTIFY: Compare y ( x, t ) given in the problem to the general form given in Eq (15.8) SET UP: The comparison gives A = 0.750 cm, k = 0.400π rad/cm and ω = 250π rad/s EXECUTE: (a) A = 0.750 cm, λ = = 5.00 cm, f = 125 Hz, T = 1f = 0.00800 s and 0.400 rad/cm v = λ f = 6.25 m/s (b) The sketches of the shape of the rope at each time are given in Figure 15.52 (c) To stay with a wavefront as t increases, x decreases and so the wave is moving in the −x-direction (d) From Eq (15.13), the tension is F = μ v = (0.50 kg/m)(6.25 m/s) = 19.5 N (e) Pav = 12 μ F ω A2 = 54.2 W EVALUATE: The argument of the cosine is (kx + ω t ) for a wave traveling in the −x-direction, and that is the case here © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Mechanical Waves 15-19 Figure 15.52 15.53 IDENTIFY: The speed in each segment is v = F/μ The time to travel through a segment is t = L/v SET UP: The travel times for each segment are t1 = L F , t2 = L μ1 4μ1 μ , and t3 = L F 4F μ1 μ μ + L = 72 L F F F (b) No The speed in a segment depends only on F and μ for that segment EVALUATE: The wave speed is greater and its travel time smaller when the mass per unit length of the segment decreases IDENTIFY: Apply Στ z = to find the tension in each wire Use v = F/μ to calculate the wave speed for each wire and then t = L/v is the time for each pulse to reach the ceiling, where L = 1.25 m 0.360 N m SET UP: The wires have μ = = = 0.02939 kg/m The free-body diagram for the L (9.80 m/s )(1.25 m) EXECUTE: (a) Adding the travel times gives ttotal = L 15.54 μ1 F + 2L beam is given in Figure 15.54 Take the axis to be at the end of the beam where wire A is attached © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 15-20 Chapter 15 EXECUTE: Στ z = gives TB L = w( L/3) and TB = w/3 = 583 N TA + TB = 1750 N, so TA = 1167 N TA vA = μ 1167 N 1.25 m = 0.00627 s = 6.27 ms = 199 m/s t A = 199 m/s 0.02939 kg/m = 583 N 1.25 m = 141 m/s t B = = 0.00888 s = 8.88 ms 0.02939 kg/m 141 m/s vB = Δt = t B − t A = 8.88 ms − 6.27 ms = 2.6 ms EVALUATE: The wave pulse travels faster in wire A, since that wire has the greater tension, so the pulse in wire A arrives first Figure 15.54 15.55 IDENTIFY and SET UP: The transverse speed of a point of the rope is v y = ∂y/∂t where y ( x, t ) is given by Eq (15.7) EXECUTE: (a) y ( x, t ) = A cos(kx − ω t ) v y = ∂y/∂t = + Aω sin(kx − ω t ) v y , max = Aω = 2π fA f = v λ and v = F ⎛ ⎞ FL , so f = ⎜ ⎟ ( M/L) ⎝λ⎠ M ⎛ 2π A ⎞ FL v y , max = ⎜ ⎟ ⎝ λ ⎠ M (b) To double v y , max increase F by a factor of 15.56 EVALUATE: Increasing the tension increases the wave speed v which in turn increases the oscillation frequency With the amplitude held fixed, increasing the number of oscillations per second increases the transverse velocity IDENTIFY: The maximum vertical acceleration must be at least g SET UP: amax = ω A EXECUTE: g = ω Amin and thus Amin = g/ω Using ω = 2π f = 2π v/λ and v = F/μ , this becomes gλ 2μ 4π F EVALUATE: When the amplitude of the motion increases, the maximum acceleration of a point on the rope increases IDENTIFY and SET UP: Use Eq (15.1) and ω = 2π f to replace v by ω in Eq (15.13) Compare this Amin = 15.57 equation to ω = k ′/m from Chapter 14 to deduce k ′ EXECUTE: (a) ω = 2π f , f = v/λ , and v = F/μ These equations combine to give ω = 2π f = 2π (v/λ ) = (2π /λ ) F/μ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Mechanical Waves 15-21 But also ω = k ′/m Equating these expressions for ω gives k ′ = m(2π /λ ) ( F/μ ) But m = μ Δx so k ′ = Δx(2π /λ ) F (b) EVALUATE: The “force constant” k ′ is independent of the amplitude A and mass per unit length μ , just as is the case for a simple harmonic oscillator The force constant is proportional to the tension in the string F and inversely proportional to the wavelength λ The tension supplies the restoring force and the 15.58 1/λ factor represents the dependence of the restoring force on the curvature of the string IDENTIFY: The frequencies at which a string vibrates depend on its tension, mass density and length T TL v SET UP: f1 = , where v = = T is the tension in the string, L is its length and m is its mass m 4L μ EXECUTE: (a) f1 = TL T v = = Solving for T gives L L m Lm T = (2 f1) Lm = 4(262 Hz)2 (0.350 m)(8.00 × 10−3 kg) = 769 N (b) m = T L(2 f1) = 769 N (0.350 m)(4)(466 Hz) = 2.53 g 8.00 × 10−3 kg v = 0.0229 kg/m T = 769 N and v = T/μ = 183 m/s f1 = gives 2L 0.350 m 183 m/s v L= = = 33.0 cm x = 35.0 cm − 33.0 cm = 2.00 cm f1 2(277 Hz) (c) For S1, μ = (d) For S , μ = 2.53 × 10−3 kg = 7.23 × 10−3 kg/m T = 769 N and v = T/μ = 326 m/s L = 0.330 m 0.350 m v 326 m/s = = 494 Hz L 2(0.330 m) EVALUATE: If the tension is the same in the strings, the mass densities must be different to produce sounds of different pitch IDENTIFY: The frequency of the fundamental (the target variable) depends on the tension in the wire The bar is in rotational equilibrium so the torques on it must balance v F SET UP: v = and f = Στ z = and f1 = 15.59 μ λ EXECUTE: λ = L = 0.660 m The tension F in the wire is found by applying the rotational equilibrium methods of Chapter 11 Let l be the length of the bar Then Στ z = with the axis at the hinge gives mg tan 30° (45.0 kg)(9.80 m/s ) tan 30° = = 127.3 N Fl cos30° = lmg sin 30° F = 2 v= 15.60 F μ = 127.3 N v 21.37 m/s = 21.37 m/s f = = = 32.4 Hz λ 0.660 m (0.0920 kg/0.330 m) EVALUATE: This is an audible frequency for humans IDENTIFY: The mass of the planet (the target variable) determines g at its surface, which in turn determines the weight of the lead object hanging from the string The weight is the tension in the string, which determines the speed of a wave pulse on that string mp F SET UP: At the surface of the planet g = G The pulse speed is v = μ Rp EXECUTE: On earth, v = v= Mg μ 4.00 m 0.0280 kg = 1.0256 × 102 m/s μ = = 7.00 × 10−3 kg/m3 F = Mg, so 0.0390 s 4.00 m and the mass of the lead weight is © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 15-22 Chapter 15 ⎛ 7.00 × 10−3 kg/m ⎞ ⎛μ⎞ 2 M = ⎜ ⎟ v2 = ⎜ ⎟⎟ (1.0256 × 10 m/s) = 7.513 kg On the planet, ⎜ 9.8 m/s ⎝g⎠ ⎝ ⎠ ⎛ 7.00 × 10−3 kg/m ⎞ 4.00 m ⎛ μ ⎞ 2 = 66.67 m/s Therefore g = ⎜ ⎟ v = ⎜ v= ⎟⎟ (66.67 m/s) = 4.141 m/s ⎜ 0.0600 s 7.513 kg ⎝M ⎠ ⎝ ⎠ g =G 15.61 mp Rp2 and m p = gRp2 G = (4.141 m/s )(7.20 × 107 m) 6.6742 × 10−11 N ⋅ m /kg = 3.22 × 1026 kg EVALUATE: This mass is about 50 times that of Earth, but its radius is about 10 times that of Earth, so the result is reasonable IDENTIFY: The wavelengths of standing waves depend on the length of the string (the target variable), which in turn determine the frequencies of the waves v SET UP: f n = nf1 where f1 = 2L EXECUTE: f n = nf1 and f n +1 = ( n + 1) f1 We know the wavelengths of two adjacent modes, so 384 m/s v v for L gives L = = = 1.83 m 2L f 2(105 Hz) EVALUATE: The observed frequencies are both audible which is reasonable for a string that is about a half meter long IDENTIFY: Apply Στ z = to one post and calculate the tension in the wire v = F/μ for waves on the wire v = f λ The standing wave on the wire and the sound it produces have the same frequency For f1 = f n +1 − f n = 630 Hz − 525 Hz = 105 Hz Solving f1 = 15.62 2L n SET UP: For the 5th overtone, n = The wire has μ = m/L = (0.732 kg)/(5.00 m) = 0.146 kg/m The standing waves on the wire, λn = free-body diagram for one of the posts is given in Figure 15.62 Forces at the pivot aren’t shown We take the rotation axis to be at the pivot, so forces at the pivot produce no torque w 235 N ⎛L ⎞ EXECUTE: Στ z = gives w ⎜ cos57.0° ⎟ − T ( L sin 57.0°) = T = = = 76.3 N For tan 57 tan 57.0° ° ⎝ ⎠ waves on the wire, v = F μ = 76.3 N = 22.9 m/s For the 5th overtone standing wave on the wire, 0.146 kg/m v 22.9 m/s L 2(5.00 m) = = 1.67 m f = = = 13.7 Hz The sound waves have frequency 13.7 Hz and 6 λ 1.67 m 344 m/s wavelength λ = = 25.0 m 13.7 Hz EVALUATE: The frequency of the sound wave is just below the lower limit of audible frequencies The wavelength of the standing wave on the wire is much less than the wavelength of the sound waves, because the speed of the waves on the wire is much less than the speed of sound in air λ= Figure 15.62 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Mechanical Waves 15.63 15-23 IDENTIFY: The tension in the wires along with their lengths determine the fundamental frequency in each one (the target variables) These frequencies are different because the wires have different linear mass densities The bar is in equilibrium, so the forces and torques on it balance m F SET UP: Ta + Tc = w, Στ z = 0, v = , f1 = v/2L and μ = , where m = ρV = ρπ r L The densities of L μ copper and aluminum are given in a table in the text EXECUTE: Using the subscript “a” for aluminum and “c” for copper, we have Ta + Tc = w = 536 N Στ z = 0, with the axis at left-hand end of bar, gives Tc (1.40 m) = w(0.90 m), so Tc = 344.6 N Ta = 536 N − 344.6 N = 191.4 N f1 = v m ρπ r L μ= = = ρπ r 2L L L For the copper wire: F = 344.6 N and μ = (8.90 × 103 kg/m3 )π (0.280 × 10−3 m) = 2.19 × 10−3 kg/m, so v= F μ = 344.6 N 2.19 × 10 −3 kg/m = 396.7 m/s f1 = v 396.7 m/s = = 330 Hz L 2(0.600 m) For the aluminum wire: F = 191.4 N and μ = (2.70 × 103 kg/m3 )π (0.280 × 10−3 m)2 = 6.65 × 10−4 kg/m, so v = 15.64 F μ = 919.4 N 6.65 × 10 −4 kg/m = 536.5 m/s, which gives f1 = 536.5 m/s = 447 Hz 2(0.600 m) EVALUATE: The wires have different fundamental frequencies because they have different tensions and different linear mass densities IDENTIFY: The time it takes the wave to travel a given distance is determined by the wave speed v A point on the string travels a distance 4A in time T SET UP: v = f λ T = 1/f EXECUTE: (a) The wave travels a horizontal distance d in a time d d 8.00 m t= = = = 0.190 s v λ f (0.600 m)(70.0 Hz) (b) A point on the string will travel a vertical distance of 4A each cycle Although the transverse velocity v y ( x, t ) is not constant, a distance of h = 8.00 m corresponds to a whole number of cycles, n = h/(4 A) = (8.00 m)/[4(5.00 × 10−3 m)] = 400, so the amount of time is t = nT = n/f = (400)/(70.0 Hz) = 5.71 s 15.65 EVALUATE: (c) The time in part (a) is independent of amplitude but the time in part (b) depends on the amplitude of the wave For (b), the time is halved if the amplitude is doubled IDENTIFY: Follow the procedure specified in part (b) ∂u ∂u = 2v and = SET UP: If u = x − vt , then ∂t ∂x EXECUTE: (a) As time goes on, someone moving with the wave would need to move in such a way that the wave appears to have the same shape If this motion can be described by x = vt + b, with b a constant, then y ( x, t ) = f (b), and the waveform is the same to such an observer (b) ∂2 y ∂x speed v = d2 f du and ∂2 y ∂t = v2 d2 f du , so y ( x, t ) = f ( x − vt ) is a solution to the wave equation with wave 2 (c) This is of the form y ( x, t ) = f (u ), with u = x − vt and f (u ) = De − B (x −Ct/B ) The result of part (b) may be used to determine the speed v = C/B EVALUATE: The wave in part (c) moves in the + x -direction The speed of the wave is independent of the constant D © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 15-24 15.66 Chapter 15 IDENTIFY: The wavelengths of the standing waves on the wire are given by λn = changed the wavelength changes because the length of the wire changes; Δl = 2L When the ball is n Fl0 AY SET UP: For the third harmonic, n = For copper, Y = 11 × 1010 Pa The wire has cross-sectional area A = π r = π (0.512 × 10−3 m)2 = 8.24 × 10−7 m EXECUTE: (a) λ3 = 2(1.20 m) = 0.800 m (b) The increase in length when the 100.0 N ball is replaced by the 500.0 N ball is given by Δl = 15.67 (ΔF )l0 , AY where ΔF = 400.0 N is the increase in the force applied to the end of the wire (400.0 N)(1.20 m) Δl = = 5.30 × 10−3 m The change in wavelength is Δλ = 23 Δl = 3.5 mm (8.24 × 10−7 m )(11 × 1010 Pa) EVALUATE: The change in tension changes the wave speed and that in turn changes the frequency of the standing wave, but the problem asks only about the wavelength IDENTIFY and SET UP: Use Eq (15.13) to replace μ , and then Eq (15.6) to replace v EXECUTE: (a) Eq (15.25): Pav = 12 μ F ω A2 v = F/μ says μ = F /v so Pav = 12 ( F /v) F ω A2 = 12 Fω A2 /v ω = 2π f so ω /v = 2π f/v = 2π /λ = k and Pav = 12 Fkω A2 , as was to be shown (b) IDENTIFY: For the ω dependence, use Eq (15.25) since it involves just ω , not k: Pav = 12 μ F ω A2 SET UP: Pav , μ , A all constant so F ω is constant, and F1ω12 = F2 ω22 EXECUTE: ω2 = ω1 ( F1/F2 )1/ = ω1 ( F1/4 F1 )1/4 = ω1 (4)−1/4 = ω1/ ω must be changed by a factor of 1/ (decreased) IDENTIFY: For the k dependence, use the equation derived in part (a), Pav = 12 Fkω A2 SET UP: If Pav and A are constant then Fkω must be constant, and F1k1ω1 = F2 k2ω2 ⎛ F ⎞⎛ ω ⎞ ⎛ F ⎞ ⎛ ω1 ⎞ 2 EXECUTE: k2 = k1 ⎜ ⎟⎜ ⎟ = k1 ⎜ ⎟ ⎜⎜ = k1 = k1/ ⎟⎟ = k1 F F 16 ω / ω ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ 15.68 k must be changed by a factor of 1/ (decreased) EVALUATE: Power is the transverse force times the transverse velocity To keep Pav constant the transverse velocity must be decreased when F is increased, and this is done by decreasing ω IDENTIFY: The phase angle determines the value of y for x = 0, t = but does not affect the shape of the y ( x, t ) versus x or t graph ∂ cos(kx − ω t + φ ) = −ω sin( kx − ω t + φ ) ∂t EXECUTE: (a) The graphs for each φ are sketched in Figure 15.68 ∂y (b) = −ω Asin(kx − ω t + φ ) ∂t SET UP: (c) No φ = π /4 or φ = 3π /4 would both give A/ If the particle is known to be moving downward, the result of part (b) shows that cos φ < 0, and so φ = 3π /4 (d) To identify φ uniquely, the quadrant in which φ lies must be known In physical terms, the signs of both the position and velocity, and the magnitude of either, are necessary to determine φ (within additive multiples of 2π ) EVALUATE: The phase φ = corresponds to y = A at x = 0, t = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Mechanical Waves 15-25 Figure 15.68 15.69 IDENTIFY and SET UP: The average power is given by Eq (15.25) Rewrite this expression in terms of v and λ in place of F and ω EXECUTE: (a) Pav = 12 μ F ω A2 v = F/μ so F = v μ ω = 2π f = 2π (v/λ ) Using these two expressions to replace F and ω gives Pav = μπ 2v3 A2 /λ ; μ = (6.00 × 10−3 kg)/(8.00 m) 1/2 ⎛ 2λ P ⎞ A = ⎜ 3av ⎟ ⎜ 4π v μ ⎟ ⎝ ⎠ = 7.07 cm (b) EVALUATE: Pav ~ v3 so doubling v increases Pav by a factor of Pav = 8(50.0 W) = 400.0 W 15.70 IDENTIFY: The wave moves in the +x direction with speed v, so to obtain y ( x, t ) replace x with x − vt in the expression for y ( x,0) SET UP: P( x, t ) is given by Eq (15.21) EXECUTE: (a) The wave pulse is sketched in Figure 15.70 (b) for ( x − vt ) < − L ⎧0 ⎪ h( L + x − vt )/L for − L < ( x − vt ) < ⎪ y ( x, t ) = ⎨ ⎪ h( L − x + vt )/L for < ( x − vt ) < L ⎪⎩0 for ( x − vt ) > L (c) From Eq (15.21): ⎧ − F (0)(0) = ⎪ ∂y ( x, t ) ∂y ( x, t ) ⎪ − F ( h/L)(− hv/L) = Fv(h/L) P ( x, t ) = − F =⎨ ∂x ∂t ⎪ − F (− h/L)(hv/L) = Fv(h/L) ⎪ − F (0)(0) = ⎩ for ( x − vt ) < − L for − L < ( x − vt ) < for < ( x − vt ) < L for ( x − vt ) > L Thus the instantaneous power is zero except for − L < ( x − vt ) < L, where it has the constant value Fv(h/L) EVALUATE: For this pulse the transverse velocity v y is constant in magnitude and has opposite sign on either side of the peak of the pulse © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 15-26 Chapter 15 Figure 15.70 15.71 IDENTIFY: Draw the graphs specified in part (a) SET UP: When y ( x, t ) is a maximum, the slope ∂y/∂x is zero The slope has maximum magnitude when y ( x, t ) = EXECUTE: (a) The graph is sketched in Figure 15.71a (b) The power is a maximum where the displacement is zero, and the power is a minimum of zero when the magnitude of the displacement is a maximum (c) The energy flow is always in the same direction ∂y = − kA sin( kx + ω t ) and Eq (15.22) becomes P ( x, t ) = − Fk ω A2sin (kx + ω t ) The power (d) In this case, ∂x is now negative (energy flows in the −x -direction ), but the qualitative relations of part (b) are unchanged The graph is sketched in Figure 15.71b EVALUATE: cosθ and sin θ are 180° out of phase, so for fixed t, maximum y corresponds to zero P and y = corresponds to maximum P Figure 15.71 15.72 IDENTIFY: The time between positions and is equal to T/2 v = f λ The velocity of points on the string is given by Eq (15.9) ⎛ 60 s ⎞ SET UP: Four flashes occur from position to position 5, so the elapsed time is ⎜ ⎟ = 0.048 s The ⎝ 5000 ⎠ figure in the problem shows that λ = L = 0.500 m At point P the amplitude of the standing wave is 1.5 cm EXECUTE: (a) T/2 = 0.048 s and T = 0.096 s f = 1/T = 10.4 Hz λ = 0.500 m (b) The fundamental standing wave has nodes at each end and no nodes in between This standing wave has one additional node This is the 1st overtone and 2nd harmonic (c) v = f λ = (10.4 Hz)(0.500 m) = 5.20 m/s (d) In position 1, point P is at its maximum displacement and its speed is zero In position 3, point P is passing through its equilibrium position and its speed is vmax = ω A = 2π fA = 2π (10.4 Hz)(0.015 m) = 0.980 m/s FL FL (1.00 N)(0.500 m) and m = = = 18.5 g m v (5.20 m/s)2 EVALUATE: The standing wave is produced by traveling waves moving in opposite directions Each point on the string moves in SHM, and the amplitude of this motion varies with position along the string (e) v = F μ = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Mechanical Waves 15.73 15-27 IDENTIFY and SET UP: There is a node at the post and there must be a node at the clothespin There could be additional nodes in between The distance between adjacent nodes is λ /2, so the distance between any two nodes is n(λ /2) for n = 1, 2, 3, … This must equal 45.0 cm, since there are nodes at the post and clothespin Use this in Eq (15.1) to get an expression for the possible frequencies f EXECUTE: 45.0 cm = n(λ /2), λ = v/f , so f = n[v/(90.0 cm)] = (0.800 Hz) n, n = 1, 2, 3, … 15.74 EVALUATE: Higher frequencies have smaller wavelengths, so more node-to-node segments fit between the post and clothespin IDENTIFY: The displacement of the string at any point is y ( x, t ) = ( ASW sin kx)sin ω t For the fundamental mode λ = L, so at the midpoint of the string sin kx = sin(2π /λ )( L/2) = 1, and y = ASW sin ω t The transverse velocity is v y = ∂y/∂t and the transverse acceleration is a y = ∂v y /∂t SET UP: Taking derivatives gives v y = ay = ∂y = ω ASW cos ω t , with maximum value v y , max = ω ASW , and ∂t ∂v y = −ω ASW sin ω t , with maximum value a y , max = ω ASW ∂t EXECUTE: ω = a y , max /v y , max = (8.40 × 103 m/s )/(3.80 m/s) = 2.21 × 103 rad/s, and then ASW = v y , max /ω = (3.80 m/s)/(2.21 × 103 rad/s) = 1.72 × 10−3 m (b) v = λ f = (2 L)(ω /2π ) = Lω /π = (0.386 m)(2.21 × 103 rad/s) /π = 272 m/s 15.75 EVALUATE: The maximum transverse velocity and acceleration will have different (smaller) values at other points on the string IDENTIFY: Carry out the derivation as done in the text for Eq (15.28) The transverse velocity is v y = ∂y/∂t and the transverse acceleration is a y = ∂v y /∂t (a) SET UP: For reflection from a free end of a string the reflected wave is not inverted, so y ( x, t ) = y1( x, t ) + y2 ( x, t ), where y1( x, t ) = A cos(kx + ω t ) (traveling to the left) y2 ( x, t ) = A cos(kx − ω t ) (traveling to the right) Thus y ( x, t ) = A[cos(kx + ω t ) + cos( kx − ω t )] EXECUTE: Apply the trig identity cos( a ± b) = cos a cos b ∓ sin a sin b with a = kx and b = ω t: cos(kx + ω t ) = cos kx cos ω t − sin kx sin ω t and cos(kx − ω t ) = cos kx cos ω t + sin kx sin ω t Then y ( x,t ) = (2 A cos kx)cos ωt (the other two terms cancel) (b) For x = 0, cos kx = and y ( x, t ) = A cos ωt The amplitude of the simple harmonic motion at x = is 2A, which is the maximum for this standing wave, so x = is an antinode (c) ymax = A from part (b) ∂y ∂ ∂ cos ω t = [(2 A cos kx)cos ω t ] = A cos kx = −2 Aω cos kx sin ω t ∂t ∂t ∂t At x = 0, v y = −2 Aω sin ω t and (v y ) max = Aω vy = ay = ∂2 y ∂t = ∂v y ∂t = −2 Aω cos kx ∂ sin ω t = −2 Aω cos kx cos ω t ∂t At x = 0, a y = −2 Aω cos ω t and (a y ) max = Aω EVALUATE: The expressions for (v y ) max and (a y ) max are the same as at the antinodes for the standing 15.76 wave of a string fixed at both ends IDENTIFY: The standing wave is given by Eq (15.28) SET UP: At an antinode, sin kx = v y ,max = ω A a y ,max = ω A EXECUTE: (a) λ = v/f = (192.0 m/s)/(240.0 Hz) = 0.800 m, and the wave amplitude is ASW = 0.400 cm The amplitude of the motion at the given points is © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 15-28 Chapter 15 (i) (0.400 cm)sin(π ) = (a node) (ii) (0.400 cm) sin(π /2) = 0.400 cm (an antinode) (iii) (0.400 cm) sin(π /4) = 0.283 cm (b) The time is half of the period, or 1/(2 f ) = 2.08 × 10−3 s (c) In each case, the maximum velocity is the amplitude multiplied by ω = 2π f and the maximum acceleration is the amplitude multiplied by ω = 4π f 2: (i) 0, 0; (ii) 6.03 m/s, 9.10 × 103 m/s ; (iii) 4.27 m/s, 6.43 × 103 m/s 15.77 EVALUATE: The amplitude, maximum transverse velocity, and maximum transverse acceleration vary along the length of the string But the period of the simple harmonic motion of particles of the string is the same at all points on the string ⎛ v ⎞ IDENTIFY: The standing wave frequencies are given by f n = n ⎜ ⎟ v = F/μ Use the density of steel ⎝ 2L ⎠ to calculate μ for the wire SET UP: For steel, ρ = 7.8 × 103 kg/m3 For the first overtone standing wave, n = EXECUTE: v = μ= Lf = (0.550 m)(311 Hz) = 171 m/s The volume of the wire is V = (π r ) L m = ρV so m ρV = = ρπ r = (7.8 × 103 kg/m3 )π (0.57 × 10−3 m) = 7.96 × 10−3 kg/m The tension is L L F = μ v = (7.96 × 10−3 kg/m)(171 m/s) = 233 N EVALUATE: The tension is not large enough to cause much change in length of the wire 15.78 IDENTIFY: The mass and breaking stress determine the length and radius of the string f1 = v , with v = 2L F μ SET UP: The tensile stress is F/π r EXECUTE: (a) The breaking stress is F = 7.0 × 108 N/m and the maximum tension is F = 900 N, so πr 900 N solving for r gives the minimum radius r = = 6.4 × 10−4 m The mass and density are π (7.0 × 108 N/m2 ) fixed, ρ = M2 so the minimum radius gives the maximum length πr L L= M π r 2ρ = 4.0 × 10−3 kg π (6.4 × 10−4 m) (7800 kg/m3 ) = 0.40 m F = F Assuming the maximum length of (b) The fundamental frequency is f1 = F = L μ L M/L ML the string is free to vibrate, the highest fundamental frequency occurs when F = 900 N and f1 = 15.79 900 N = 375 Hz (4.0 × 10−3 kg)(0.40 m) EVALUATE: If the radius was any smaller the breaking stress would be exceeded If the radius were greater, so the stress was less than the maximum value, then the length would be less to achieve the same total mass IDENTIFY: At a node, y ( x, t ) = for all t y1 + y2 is a standing wave if the locations of the nodes don’t depend on t SET UP: The string is fixed at each end so for all harmonics the ends are nodes The second harmonic is the first overtone and has one additional node EXECUTE: (a) The fundamental has nodes only at the ends, x = and x = L (b) For the second harmonic, the wavelength is the length of the string, and the nodes are at x = 0, x = L/2 and x = L (c) The graphs are sketched in Figure 15.79 (d) The graphs in part (c) show that the locations of the nodes and antinodes between the ends vary in time EVALUATE: The sum of two standing waves of different frequencies is not a standing wave © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Mechanical Waves 15-29 Figure 15.79 15.80 v The buoyancy force B that the water exerts on the object reduces the tension in the 2L wire B = ρfluidVsubmerged g IDENTIFY: f1 = SET UP: For aluminum, ρa = 2700 kg/m3 For water, ρ w = 1000 kg/m3 Since the sculpture is completely submerged, Vsubmerged = Vobject = V © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 15-30 Chapter 15 EXECUTE: (a) L is constant, so f air f w and the fundamental frequency when the sculpture is = vair vw ⎛v ⎞ F v Fw submerged is f w = f air ⎜ w ⎟ , with f air = 250.0 Hz v = so w = When the sculpture is in μ vair Fair ⎝ vair ⎠ air, Fair = w = mg = ρaVg When the sculpture is submerged in water, Fw = w − B = ( ρa − ρ w )Vg vw ρa − ρ w 1000 kg/m3 = and f w = (250.0 Hz) − = 198 Hz vair ρa 2700 kg/m3 15.81 (b) The sculpture has a large mass and therefore very little displacement EVALUATE: We have neglected the buoyant force on the wire itself IDENTIFY: When the rock is submerged in the liquid, the buoyant force on it reduces the tension in the wire supporting it This in turn changes the frequency of the fundamental frequency of the vibrations of the wire The buoyant force depends on the density of the liquid (the target variable) The vertical forces on the rock balance in both cases, and the buoyant force is equal to the weight of the liquid displaced by the rock (Archimedes’s principle) F SET UP: The wave speed is v = and v = f λ B = ρliqVrock g ΣFy = μ EXECUTE: λ = L = 6.00 m In air, v = f λ = (42.0 Hz)(6.00 m) = 252 m/s v = μ= F v = 164.0 N (252 m/s) F μ so = 0.002583 kg/m In the liquid, v = f λ = (28.0 Hz)(6.00 m) = 168 m/s F = μ v = (0.002583 kg/m)(168 m/s) = 72.90 N F + B − mg = B = mg − F = 164.0 N − 72.9 N = 91.10 N For the rock, V = m ρ = (164.0 N/9.8 m/s ) 3200 kg/m3 = 5.230 × 10−3 m3 91.10 N B = = 1.78 × 103 kg/m3 Vrock g (5.230 × 10−3 m3 )(9.8 m/s2 ) EVALUATE: This liquid has a density 1.78 times that of water, which is rather dense but not impossible IDENTIFY: Compute the wavelength from the length of the string Use Eq (15.1) to calculate the wave speed and then apply Eq (15.13) to relate this to the tension (a) SET UP: The tension F is related to the wave speed by v = F/μ (Eq (15.13)), so use the information B = ρliqVrock g and ρliq = 15.82 given to calculate v EXECUTE: λ /2 = L λ = L = 2(0.600 m) = 1.20 m Figure 15.82 v = f λ = (65.4 Hz)(1.20 m) = 78.5 m/s μ = m/L = 14.4 × 10−3 kg/0.600 m = 0.024 kg/m Then F = μ v = (0.024 kg/m)(78.5 m/s) = 148 N (b) SET UP: F = μ v and v = f λ give F = μ f 2λ μ is a property of the string so is constant λ is determined by the length of the string so stays constant μ , λ constant implies F/f = μλ = constant, so F1/f12 = F2 /f 22 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Mechanical Waves 15-31 15.83 ⎛ f ⎞ ⎛ 73.4 Hz ⎞ EXECUTE: F2 = F1 ⎜ ⎟ = (148 N) ⎜ ⎟ = 186 N ⎝ 65.4 Hz ⎠ ⎝ f1 ⎠ F − F 186 N − 148 N The percent change in F is = = 0.26 = 26% F1 148 N EVALUATE: The wave speed and tension we calculated are similar in magnitude to values in the examples Since the frequency is proportional to F , a 26% increase in tension is required to produce a 13% increase in the frequency IDENTIFY: Stress is F/A, where F is the tension in the string and A is its cross-sectional area SET UP: A = π r For a string fixed at each end, f1 = v F F = = L L μ mL EXECUTE: (a) The cross-section area of the string would be A = (900 N)/(7.0 × 108 Pa) = 1.29 × 10−6 m , corresponding to a radius of 0.640 mm The length is the volume divided by the area, and the volume is V = m/ρ , so L= V m/ρ (4.00 × 10−3 kg) = = = 0.40 m A A (7.8 × 10 kg/m3 )(1.29 × 10−6 m ) 900 N = 375 Hz, or 380 Hz to two (4.00 × 10−3 kg)(0.40 m) (b) For the maximum tension of 900 N, f1 = 15.84 figures EVALUATE: The string could be shorter and thicker A shorter string of the same mass would have a higher fundamental frequency IDENTIFY: Apply ΣFy = to segments of the cable The forces are the weight of the diver, the weight of the segment of the cable, the tension in the cable and the buoyant force on the segment of the cable and on the diver SET UP: The buoyant force on an object of volume V that is completely submerged in water is B = ρ waterVg EXECUTE: (a) The tension is the difference between the diver’s weight and the buoyant force, F = (m − ρ waterV ) g = (120 kg − (1000 kg/m3 )(0.0800 m3 ))(9.80 m/s ) = 392 N (b) The increase in tension will be the weight of the cable between the diver and the point at x, minus the buoyant force This increase in tension is then ( μ x − ρ ( Ax)) g = (1.10 kg/m − (1000 kg/m3 )π (1.00 × 10−2 m) )(9.80 m/s2 ) x = (7.70 N/m) x The tension as a function of x is then F ( x) = (392 N) + (7.70 N/m) x (c) Denote the tension as F ( x ) = F0 + ax, where F0 = 392 N and a = 7.70 N/m Then the speed of transverse waves as a function of x is v = surface is found from t = ∫ dt = ∫ dx = ( F0 + ax )/μ and the time t needed for a wave to reach the dt μ dx =∫ dx dx/dt F0 + ax Let the length of the cable be L, so t = μ L ∫0 dx F0 + ax = μ a F0 + ax L = μ ( F0 + aL − F0 ) a 1.10 kg/m ( 392 N + (7.70 N/m)(100 m) − 392 N ) = 3.89 s 7.70 N/m EVALUATE: If the weight of the cable and the buoyant force on the cable are neglected, then the tension would L 392 N F = = 18.9 m/s and t = = 5.29 s have the constant value calculated in part (a) Then v = v μ 1.10 kg/m t= The weight of the cable increases the tension along the cable and the time is reduced from this value © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 15-32 15.85 Chapter 15 IDENTIFY: Carry out the analysis specified in the problem SET UP: The kinetic energy of a very short segment Δx is ΔK = 12 (Δm)v 2y v y = ∂y/∂t The work done by the tension is F times the increase in length of the segment Let the potential energy be zero when the segment is unstretched 2 ΔK (1/2)Δmv y ⎛ ∂y ⎞ = = μ⎜ ⎟ ⎝ ∂t ⎠ Δx Δm/μ ∂y (b) = ω A sin(kx − ω t ) and so uk = μω A2 sin (kx − ω t ) ∂t ∂y (c) The piece has width Δx and height Δx , and so the length of the piece is ∂x EXECUTE: (a) uk = 1/2 1/ 2 ⎛ ⎛ ⎛ ∂y ⎞ ⎞ ⎡ ⎛ ∂y ⎞ ⎤ ⎛ ∂y ⎞ ⎞ ⎜ ( Δx) + ⎜ Δx ⎟ ⎟ = Δx ⎜1 + ⎜ ⎟ ⎟ ≈ Δx ⎢1 + ⎜ ⎟ ⎥ , where the relation given in the hint has ⎜ ⎜ ⎝ ∂x ⎠ ⎟ ⎝ ∂x ⎠ ⎟⎠ ⎝ ⎝ ⎠ ⎣⎢ ⎝ ∂x ⎠ ⎦⎥ been used Δx ⎡1 + 12 (∂y/∂x) ⎤ − Δx ⎛ ∂y ⎞2 ⎦ (d) up = F ⎣ = F⎜ ⎟ Δx ⎝ ∂x ⎠ ∂y = − kA sin(kx − ω t ), and so up = Fk A2 sin ( kx − ωt ) (e) ∂x (f) Comparison with the result of part (c) with k = ω /v = ω μ /F shows that for a sinusoidal wave uk = up (g) The graph is given in Figure 15.85 In this graph, uk and up coincide, as shown in part (f) At y = 0, the string is stretched the most, and is moving the fastest, so uk and up are maximized At the extremes of y, the string is unstretched and is not moving, so uk and up are both at their minimum of zero P v EVALUATE: The energy density travels with the wave, and the rate at which the energy is transported is the product of the density per unit length and the speed (h) uk + up = Fk A2 sin ( kx − ω t ) = Fk (ω /v) A2sin (kx − ω t ) = Figure 15.85 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher ... Figure 15.13a (b) (i) t = 0.400 s: 0.00 1.50 3.00 4.50 6.00 7.50 9.00 10.50 12.00 x(cm) 0.203 0.300 0.221 0. 0131 −0.221 −0. 0131 −0.203 −0.300 −0.221 y(cm) The graph is shown in Figure 15.13b (ii)... 0.146 kg/m v 22.9 m/s L 2(5.00 m) = = 1.67 m f = = = 13. 7 Hz The sound waves have frequency 13. 7 Hz and 6 λ 1.67 m 344 m/s wavelength λ = = 25.0 m 13. 7 Hz EVALUATE: The frequency of the sound wave... f8 = f1 = 21.2 Hz; ω8 = 2π f8 = 133 rad/s λ = v/f = (1470 cm/s)/(21.2 Hz) = 69.3 cm and k = 2π /λ = 0.0906 rad/cm y ( x,t ) = (5.60 cm)sin([0.0906 rad/cm]x )sin( [133 rad/s]t ) 15.44 EVALUATE: The