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. 17.

Transcription

1 MECHANICAL WAVES 5 5 IDENTIFY: v= fλ T = / f is the time for one complete vibration SET UP: The frequenc of the note one octave higher is 568 Hz 344 m/s EXECUTE: (a) λ = v 0439 m f = 784 Hz = T = = 8 ms f 344 m/s (b) λ = v 09 m f = 568 Hz = EVALUATE: When f is doubled, λ is halved 5 IDENTIFY: The distance between adjacent dots is λ v= fλ The long-wavelength sound has the lowest frequenc, 00 Hz, and the short-wavelength sound has the highest frequenc, 00 khz SET UP: For sound in air, v = 344 m/s 344 m/s EXECUTE: (a) Red dots: λ = v 7 m f = 00 Hz = 344 m/s Blue dots: λ = = 007 m = 7 cm Hz (b) In each case the separation easil can be measured with a meterstick 480 m/s (c) Red dots: λ = v 740 m f = 00 Hz = 480 m/s Blue dots: λ = = m = 740 cm In each case the separation easil can be measured with a Hz 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 frequenc, the wavelength increases 53 IDENTIFY: v= fλ = λ / T SET UP: 0 h = 3600 s The crest to crest distance is λ m 0 m/s 800 km EXECUTE: v = = v = = 800 km/h 3600 s 0 h EVALUATE: Since the wave speed is ver high, the wave strikes with ver little warning 54 IDENTIFY: fλ = v SET UP: 0 mm = 0000 m v 500 m s 6 EXECUTE: f = = = 5 0 Hz λ 0000 m EVALUATE: The frequenc is much higher than the upper range of human hearing 55 IDENTIFY: v= fλ T = / f SET UP: 9 nm = 0 m 8 c m/s 4 5 EXECUTE: (a) λ = 400 nm : f = = = Hz T = / f = 33 0 s 9 λ m m/s 4 5 λ = 700 nm : f = = 49 0 Hz T = 33 0 s The frequencies of visible light lie between m Hz and Hz The periods lie between 33 0 s and 33 0 s (b) T is ver short and cannot be measured with a stopwatch EVALUATE: Longer wavelength corresponds to smaller frequenc and larger period 5-

2 5- Chapter 5 56 IDENTIFY: Compare xt (, ) given in the problem to the general form of Eq(54) f = / T and v= fλ SET UP: The comparison gives EXECUTE: (a) 650 mm (b) 80 cm (c) f = = 78 Hz s (d) v = (080 m)(78 Hz) = 778 m s A = 650 mm, λ = 80 cm and T = 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 57 IDENTIFY: Use Eq(5) to calculate v T = / f and k is defined b Eq(55) The general form of the wave function is given b Eq(58), which is the equation for the transverse displacement SET UP: v = 800 m/s, A = m, λ = 030 m EXECUTE: (a) v= fλ so f = v/ λ = (800 m/s)/(030 m) = 50 Hz T = / f = / 50 Hz = s k = π / λ = π rad/030 m = 96 rad/m (b) For a wave traveling in the x-direction, x (, t) = Acos π ( x/ λ + tt / ) (Eq(58)) At x = 0, (0, t) = Acos π ( t/ T), so = A at t = 0 This equation describes the wave specified in the problem Substitute in numerical values: x (, t) = (00700 m)cos( π ( x/ 030 m + t/ s)) Or, x t x t (, ) = (00700 m)cos((96 m ) + (57 rad/s) ) (c) From part (b), = (00700 m)cos( π ( x/ 030 m + t/ s)) Plug in x = 0360 m and t = 050 s: = (00700 m) cos( π (0360 m/030 m s/00400 s)) = (00700 m) cos[ π (4875 rad)] = m =+ 495 cm (d) In part (c) t = 050 s = A means cos( π ( x/ λ + t/ T)) = cosθ = for θ = 0, π, 4 π, = n( π ) or n = 0,,, So = A when π ( x/ λ+ t/ T) = n( π) or x / λ + xt / = n t = T( n x/ λ) = (00400 s)( n 0360 m/030 m) = (00400 s)( n 5) For n = 4, t = 050 s (before the instant in part (c)) For n = 5, t = 0550 s (the first occurrence of = A after the instant in part (c)) Thus the elapsed time is 0550 s 0500 s = s EVALUATE: Part (d) sas = A at 05 s and next at 055 s; the difference between these two times is 0040 s, which is the period At t = 050 s the particle at x = 0360 m is at = 495 cm and traveling upward It takes T / 4 = 0000 s for it to travel from = 0 to = A, so our answer of s is reasonable 58 IDENTIFY: The general form of the wave function for a wave traveling in the x-direction is given b Eq(58) The time for one complete ccle to pass a point is the period T and the number that pass per second is the frequenc f The speed of a crest is the wave speed v and the maximum speed of a particle in the medium v = ωa is max SET UP: Comparison to Eq(58) gives A = 375 cm, k = 0450 rad/cm and ω = 540 rad/s π rad π rad EXECUTE: (a) T = = = 6 s In one ccle a wave crest travels a distance ω 540 rad/s π rad π rad λ = = = 040 m k 0450 rad/cm (b) k = 0450 rad/cm f = / T = 086 Hz = 086 waves/second (c) v= fλ = (086 Hz)(040 m) = 0 m/s vmax = ωa= (540 rad/s)(375 cm) = 00 m/s EVALUATE: The transverse velocit of the particles in the medium (water) is not the same as the velocit of the wave

3 Mechanical Waves IDENTIFY: Evaluate the partial derivatives and see if Eq(5) is satisfied SET UP: cos( kx + ωt) = k sin( kx + ωt) cos( kx + ωt) = ωsin( kx + ωt) sin( kx + ωt) = k cos( kx + ωt) sin( kx + ωt) = ωsin( kx + ωt) EXECUTE: (a) = Ak cos( kx + ωt) = Aω cos( kx+ ωt) Eq(5) is satisfied, if v= ω / k (b) = Ak sin( kx + ωt) = Aω sin( kx+ ωt) Eq(5) is satisfied, if v= ω / k (c) = kasin( kx) = k Acos( kx) = ω Asin( ω t) = ω Acos( ωt) Eq(5) is not satisfied (d) v = = ωacos( kx+ ωt) a = = Aω sin( kx+ ωt) EVALUATE: The functions cos( kx + ωt) and sin( kx + ωt) differ onl in phase 50 IDENTIFY: v and a are given b Eqs(59) and (50) SET UP: The sign of v determines the direction of motion of a particle on the string If v = 0 and a 0 the speed of the particle is increasing If v 0, the particle is speeding up if v and a have the same sign and slowing down if the have opposite signs EXECUTE: (a) The graphs are given in Figure 50 (b) (i) v = ωasin(0) = 0 and the particle is instantaneousl at rest a = ω A cos(0) = ω A and the particle is speeding up (ii) v = ωasin( π 4) = ωa, and the particle is moving up a = ω A cos( π 4) = ω A, and the particle is slowing down ( v and a have opposite sign) (iii) v = ωasin( π ) = ωa and the particle is moving up a = ω Acos( π ) = 0 and the particle is instantaneousl not accelerating (iv) v = ωasin(3π 4) = ωa, and the particle is moving up a = ω Acos(3π 4) = ω A, and the particle is speeding up (v) v = ωasin( π) = 0 and the particle is instantaneousl at rest a = ω Acos( π) = ω A and the particle is speeding up (vi) v = ωasin(5π 4) = ωa and the particle is moving down a = ω Acos(5π 4) = ω A and the particle is slowing down ( v and a have opposite sign) (vii) v = ωasin(3π ) = ωa and the particle is moving down a = ω Acos(3π ) = 0 and the particle is instantaneousl not accelerating (viii) v = ωasin(7π 4) = ωa, and the particle is moving down a = ω Acos(7π 4) = ω A and the particle is speeding up ( v and a have the same sign) EVALUATE: At t = 0 the wave is represented b Figure 50a in the textbook: point (i) in the problem corresponds to the origin, and points (ii)-(viii) correspond to the points in the figure labeled -7 Our results agree with what is shown in the figure Figure 50

4 5-4 Chapter 5 5 IDENTIFY and SET UP: Read A and T from the graph Appl Eq(54) to determine λ and then use Eq(5) to calculate v EXECUTE: (a) The maximum is 4 mm (read from graph) (b) For either x the time for one full ccle is 0040 s; this is the period (c) Since = 0 for x = 0 and t = 0 and since the wave is traveling in the + x-direction then x (, t) = Asin[ π ( tt / x/ λ)] (The phase is different from the wave described b Eq(54); for that wave = A for x = 0, t = 0) From the graph, if the wave is traveling in the + x-direction and if x = 0 and x = 0090 m are within one wavelength the peak at t = 00 s for x = 0 moves so that it occurs at t = 0035 s (read from graph so is approximate) for x = 0090 m The peak for x = 0 is the first peak past t = 0 so corresponds to the first maximum in sin[ π ( t/ T x/ λ)] and hence occurs at π ( t/ T x/ λ) = π / If this same peak moves to t = 0035 s at x = 0090 m, then π ( t / T x / λ) = π / λ Solve for : t / T x / λ = /4 x/ λ = t/ T / 4 = 0035 s/ 0040 s 05 = 065 λ = x / 065 = 0090 m / 065 = 04 m Then v= fλ = λ / T = 04 m / 0040 s = 35 m /s (d) If the wave is traveling in the x-direction, then x (, t) = Asin( π ( tt / + x/ λ)) and the peak at t = 0050 s for x = 0 corresponds to the peak at t = 0035 s for x = 0090 m This peak at x = 0 is the second peak past the origin so corresponds to π ( t/ T + x/ λ) = 5 π / If this same peak moves to t = 0035 s for x = 0090 m, then π( t/ T + x/ λ) = 5 π / t/ T + x/ λ = 5/4 x/ λ = 5/ 4 t/ T = 5/ s / 0040 s = 0375 λ = x / 0375 = 0090 m / 0375 = 04 m Then v= fλ = λ / T = 04 m / 0040 s = 60 m /s EVALUATE: No Wouldn t know which point in the wave at x = 0 moved to which point at x = 0090 m 5 IDENTIFY: v = v= fλ = λ / T t π πv π SET UP: Acos ( x vt) =+ A sin ( x vt) λ λ λ x t EXECUTE: (a) Acos π π λ π λ =+ Acos x t =+ Acos ( x vt) where = λ f = v has been used λ T λ T λ T πv π (b) v = = Asin ( x vt) λ λ (c) The speed is the greatest when the sine is, and that speed is π va λ This will be equal to v if A = λ π, less than v if A < λ π and greater than v if A > λ π 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 53 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: x (cm) (cm) The graph is shown in Figure 53a (b) (i) t = 0400 s: x (cm) (cm) The graph is shown in Figure 53b (ii) t = 0800 s: x (cm) (cm) The graph is shown in Figure 53c (iii) The graphs show that the wave is traveling in + x-direction

5 Mechanical Waves 5-5 EVALUATE: We know that Eq(53) is for a wave traveling in the + x-direction, and xt (, ) is derived from this This is consistent with the direction of propagation we deduced from our graph Figure IDENTIFY: The frequenc and wavelength determine the wave speed and the wave speed depends on the tension SET UP: F v = μ = ml / v= fλ μ 00 kg EXECUTE: F = μv = μ( fλ) = ([400 Hz][0750 m]) = 43 N 50 m EVALUATE: If the frequenc is held fixed, increasing the tension will increase the wavelength 55 IDENTIFY and SET UP: Use Eq(53) to calculate the wave speed Then use Eq(5) to calculate the wavelength EXECUTE: (a) The tension F in the rope is the weight of the hanging mass: F = mg = (50 kg)(980 m/s ) = 47 N v= F/ μ = 47 N/(00550 kg/m) = 63 m/s (b) v= fλ so λ = v/ f = (63 m/s)/0 Hz = 036 m (c) EVALUATE: v= F/ μ, where F = mg Doubling m increases v b a factor of λ = vf / f remains 0 Hz and v increases b a factor of, so λ increases b a factor of 56 IDENTIFY: For transverse waves on a string, v= F/ μ The general form of the equation for waves traveling in the + x-direction is xt (, ) = Acos( kx ωt) For waves traveling in the x-direction it is xt (, ) = Acos( kx+ ωt) v= ω / k

6 5-6 Chapter 5 SET UP: Comparison to the general equation gives has mass 08 kg and μ = m/ L= kg/m A = 850 mm, 7 rad/m ω 730 rad/s d 50 m EXECUTE: (a) v = = = 59 m/s t = = = s k 7 rad/m v 59 m/s (b) W = F = μv = (00850 kg/m)(59 m/s) = 5 N k = and ω = 730 rad/s The string π rad π rad (c) λ = = = m The number of wavelengths along the length of the string is k 7 rad/m 50 m m = (d) For a wave traveling in the opposite direction, xt (, ) = (850 mm)cos([7 rad/m] x+ [730 rad/s] t) EVALUATE: We have assumed that the tension in the string is constant and equal to W In realit the tension will var along the length of the string because of the weight of the string and the wave speed will therefore var along the string The tension at the lower end of the string will be W = 5 N and at the upper end it is W + 5 N = 8 N, an increase of 6% 57 IDENTIFY: For transverse waves on a string, v= F/ μ v= fλ SET UP: The wire has μ = m/ L= (0065 kg)/(0750 m) = 000 kg/m EXECUTE: (a) v= fλ = (875 Hz)(333 0 m) = 9 m/s The tension is F = μv = (000 kg/m)(9 m/s) = 86 N (b) v = 9 m/s EVALUATE: If λ is kept fixed, the wave speed and the frequenc increase when the tension is increased 58 IDENTIFY: Appl F = 0 to determine the tension at different points of the rope v= F/ μ SET UP: From Example 53, m samples = 00 kg, m rope = 00 kg and μ = 0050 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 885m s as found in Example 53 (b) The tension at the middle of the rope is (0 kg)(9 80m s ) = 058 N and the wave speed is 907m s (c) The tension at the top of the rope is (0 kg)(980 m/s ) = 56 m/s and the speed is 99 m/s (See Challenge Problem (58) for the effects of varing 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 59 IDENTIFY: v= F/ μ v= fλ The general form for xt (, ) is given in Eq(54), where T = / f Eq(50) sas that the maximum transverse acceleration is amax = ω A= ( π f) A SET UP: μ = kg/m EXECUTE: (a) v= F μ = (500 N) (00500 kg m) = 00 m s (b) λ = v f = (00 m s) (400 Hz) = 050 m (c) xt (, ) = Acos( kx ωt) k = π λ = 800 π rad m; ω = π f = 800 π rad s xt (, ) = (300 cm)cos[ π(800 rad m) x (800 π rad s) t] (d) v =+ Aω sin( kx ωt) and a = Aω cos( kx ωt) a, max = Aω = A( π f) = 890 m s (e) a is,max much larger than g, so it is a reasonable approximation to ignore gravit EVALUATE: xt (, ) in part (c) gives (0,0) = A, which does correspond to the oscillator having maximum upward displacement at t = 0 50 IDENTIFY: Appl Eq(55) SET UP: ω = π f μ = m/ L EXECUTE: (a) P μfω A av = P av = = 080 m kg (50 N)( (00 Hz)) (6 0 3 m) π 03 W or 0 W to two figures (b) Pav is proportional to A, so halving the amplitude quarters the average power, to 0056 W EVALUATE: The average power is also proportional to the square of the frequenc

7 P I 5 IDENTIFY: For a point source, I = and 4π r I SET UP: EXECUTE: (a) r (b) (c) I I 3 3 r 6 μw = 0 W r = r I 00 W/m = = (300 m) = 95 km r I 6 0 W/m r =, with I = 0 μw/m and r 3 = r P= I πr = π = 5 (4 ) (00 W/m )(4 )(300 m) 0 W r I3 = I = I/4= 05 μw/m r3 Mechanical Waves 5-7 EVALUATE: These are approximate calculations, that assume the sound is emitted uniforml in all directions and that ignore the effects of reflection, for example reflections from the ground 5 IDENTIFY: Appl Eq(56) SET UP: EXECUTE: I = 0 W/m r = 75 m Set r I 0 W m I = 0 W/m and solve for r I 0 W m r = = (75 m) = 5 m, so it is possible to move r r = 75 m 5 m = 50 m closer to the source EVALUATE: I increases as the distance r of the observer from the source decreases 53 IDENTIFY and SET UP: Appl Eq(56) to relate I and r Power is related to intensit at a distance r b P= I(4 π r ) Energ is power times time EXECUTE: (a) Ir = Ir I = I ( r / r ) = (006 W/m )(43 m/3 m) = 0050 W/m (b) P= π = π = 4 r I 4 (43 m) (006 W/m ) 604 W 4 Energ = Pt = (604 W)(3600 s) = 0 J EVALUATE: We could have used r = 3 m and I = 0050 W/m in P= 4π r I and would have obtained the same P Intensit becomes less as r increases because the radiated power spreads over a sphere of larger area 54 IDENTIFY: The tension and mass per unit length of the rope determine the wave speed Compare xt (, ) given in the problem to the general form given in Eq(58) v= ω / k The average power is given b Eq (55) SET UP: Comparison with Eq(58) gives A = 33 mm, k = 698 rad/m and ω = 74 rad/s EXECUTE: (a) A = 30 mm 74 rad s (b) f = ω 8 Hz π = π = (c) λ = π = π = 090 m k 698 rad m 74 rad s (d) v = ω = = 06 m s k 698 rad m (e) The wave is traveling in the (f ) The linear mass densit is F (g) 3 = μv = (504 0 kg m)(063 m s) = 83 N x direction because the phase of x (, t ) has the form kx + ωt μ 3 3 = (338 0 kg) (35 m) = kg/m, so the tension is P = μfω A = = 3 3 av (50 0 kg m)(83 N)(74 rad s) (30 0 m) 039 W EVALUATE: In part (d) we could also calculate the wave speed as v= fλ and we would obtain the same result 55 IDENTIFY: P= 4π r I SET UP: From Example 55, EXECUTE: EVALUATE: I = = 050 W/m at r 50 m P= π = π = 4 r I 4 (50 m) (050 W/m ) 707 W I = 000 W/m at 750 m and sinusoidal waves emitted b the source 4 π (750 m) (000 W/m ) = 707 W P is the average power of the

8 5-8 Chapter 5 56 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, respectivel, is shown in Figure 56a (b) The wave form for the given times, respectivel, is shown in Figure 56b 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 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, respectivel, is shown in Figure 57a (b) The wave form for the given times, respectivel, is shown in Figure 57b 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 IDENTIFY: Appl 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 58 EVALUATE: The pulses interfere when the overlap but resume their original shape after the have completel passed through each other Figure IDENTIFY: Appl 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 59

9 Mechanical Waves 5-9 EVALUATE: The pulses interfere when the overlap but resume their original shape after the have completel passed through each other Figure 59

10 5-0 Chapter IDENTIFY: Appl 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 530 EVALUATE: The pulses interfere when the overlap but resume their original shape after the have completel passed through each other Figure IDENTIFY: Appl 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 53 EVALUATE: The pulses interfere when the overlap but resume their original shape after the have completel passed through each other Figure 53

11 53 IDENTIFY: net = + The string never moves at values of x for which sin kx = 0 SET UP: sin( A± B) = sin Acos B± cos Asin B EXECUTE: (a) net = Asin( kx+ ωt) + Asin( kx ωt) Mechanical Waves 5- net = A[sin( kx)cos( ωt) + cos( kx)sin( ωt) + sin( kx)cos( ωt) cos( kx)sin( ωt)] = Asin( kx)cos( ωt) nπ nπ nλ (b) sin kx = 0 for kx = nπ, n = 0,,, x = = = k π / λ EVALUATE: Using = Asin( kx± ωt) instead of = Acos( kx± ωt) corresponds to a particular choice of phase and corresponds to = 0 at x = 0, for all t 533 IDENTIFY and SET UP: Nodes occur where sin kx = 0 and antinodes are where sin kx =± EXECUTE: Eq(58): = ( A sin kx)sinωt SW (a) At a node = 0 for all t This requires that sin kx = 0 and this occurs for kx = nπ, n = 0,,, nπ x= nπ / k = = (33 m) n, n= 0,,, 0750 π rad/m (b) At an antinode sin n = 0,,, kx =± so will have maximum amplitude This occurs when = ( + ) kx n π π x= ( n+ ) π / k = ( n+ ) = (33 m) ( n+ ), n= 0,,, 0750 π rad/m EVALUATE: λ = π / k = 66 m Adjacent nodes are separated b λ /, adjacent antinodes are separated b λ /, and the node to antinode distance is λ / IDENTIFY: Appl Eqs(58) and (5) At an antinode, t () = ASW sinωt k and ω for the standing wave have the same values as for the two traveling waves SET UP: A SW = 0850 cm The antinode to antinode distance is λ /, so λ = 300 cm v = / t EXECUTE: (a) The node to node distance is λ / = 50 cm (b) λ is the same as for the standing wave, so λ = 300 cm A= A SW = 045 cm λ 0300 m v= fλ = = = 33 m/s T s (c) v = = ASWω sin kxcosωt At an antinode sin kx =, so v SW cos = A ω ωt vmax = ASWω π rad π rad ω = = = 838 rad/s v max = ( m)(838 rad/s) = 007 m/s v min = 0 T s (d) The distance from a node to an adjacent antinode is λ / 4 = 750 cm EVALUATE: The maximum transverse speed for a point at an antinode of the standing wave is twice the maximum transverse speed for each traveling wave, since ASW = A 535 IDENTIFY: Evaluate / and / and see if Eq(5) is satisfied for v= ω / k SET UP: sin kx = k coskx coskx = ksinkx sinωt = ωcosωt cosωt = ωsinωt EXECUTE: (a) = k [ Asw sinωt] sin kx, = ω [ Asw sinωt] sin kx, so for xt (, ) to be a solution of ω ω Eq(5), k =, and v = v k (b) A standing wave is built up b the superposition of traveling waves, to which the relationship v= λ k applies EVALUATE: xt (, ) = ( ASW sin kx)sinωt is a solution of the wave equation because it is a sum of solutions to the wave equation 536 IDENTIFY and SET UP: cos( kx ± ωt) = coskxcosωt sin kxsinωt EXECUTE: + = A[ cos( kx+ ωt) + cos( kx ωt)] + = A[ coskxcosωt + sin kxsinωt + coskxcosωt + sin kxsin ωt] = Asin kxsinωt EVALUATE: The derivation shows that the standing wave of Eq(58) results from the combination of two waves with the same A, k, and ω that are traveling in opposite directions,

12 5- Chapter IDENTIFY: Evaluate / and / and show that Eq(5) is satisfied SET UP: ( + ) = + and ( + ) = + EXECUTE: = + and = + The functions and are given as being solutions to the wave equation, so = + = + = + = and so = + is a solution of v v v v Eq (5) EVALUATE: The wave equation is a linear equation, as it is linear in the derivatives, and differentiation is a linear operation L v 538 IDENTIFY: For a string fixed at both ends, λ n = and fn = n n L SET UP: For the fundamental, n = For the second overtone, n = 3 For the fourth harmonic, n = 4 EXECUTE: (a) λ = L = 300 m f (480 m s) = v = = 60 Hz L (50 m) (b) 3= 00 m 3 f = 3 f = 480 Hz (c) 4= 075 m 4 f3 = 4f = 640 Hz EVALUATE: As n increases, λ decreases and f increases 539 IDENTIFY: Use Eq(5) for v and Eq(53) for the tension F v = / t and a = v / t (a) SET UP: The fundamental standing wave is sketched in Figure 539 Figure 539 EXECUTE: v= fλ = (600 Hz)(60 m) = 960 m/s (b) The tension is related to the wave speed b Eq(53): v= F/ μ so F = μv μ = m/ L= kg/0800 m = kg/m F = μv = = (00500 kg/m)(960 m/s) 46 N (c) ω = π f = 377 rad/s and x (, t) = ASW sinkxsinωt v = ωa sin kxcos ωt; max SW SW = ω a A kx t SW sin sin ω ( v ) = ωa = (377 rad/s)(0300 cm) = 3 m/s ( a ) = ω A = (377 rad/s) (0300 cm) = 46 m/s max SW f = 600 Hz From the sketch, λ /= L so λ = L = 60 m EVALUATE: The transverse velocit is different from the wave velocit The wave velocit and tension are similar in magnitude to the values in the Examples in the text Note that the transverse acceleration is quite large 540 IDENTIFY: The fundamental frequenc depends on the wave speed, and that in turn depends on the tension SET UP: F v v = where μ = m/ L f = The nth harmonic has frequenc fn = nf μ L (800 N)(0400 m) = = = = 37 m/s / kg F FL EXECUTE: (a) v 3 m L m f v 37 m/s = = = 409 Hz L (0400 m) 0,000 Hz (b) n = = 44 The 4 th harmonic is the highest that could be heard f EVALUATE: In part (b) we use the fact that a standing wave on the wire produces a sound wave in air of the same frequenc

13 Mechanical Waves IDENTIFY: Compare x (, t ) given in the problem to Eq(58) From the frequenc and wavelength for the third harmonic find these values for the eighth harmonic (a) SET UP: The third harmonic standing wave pattern is sketched in Figure 54 Figure 54 EXECUTE: (b) Eq (58) gives the general equation for a standing wave on a string: x (, t) = ( A sin kx)sinωt SW ASW = A, so A= A SW / = (560 cm)/ = 80 cm (c) The sketch in part (a) shows that L = 3( λ / ) k = π / λ, λ = π /k Comparison of x (, t ) given in the problem to Eq (58) gives λ = π /(00340 rad/cm) = 848 cm L = 3( λ / ) = 77 cm (d) λ = 85 cm, from part (c) ω = 500 rad/s so f = ω / π = 796 Hz period T = / f = 06 s v= fλ = 470 cm/s (e) v = d/ dt = ωa sinkxcosωt SW v, max = ωasw = (500 rad/s)(560 cm) = 80 cm/s (f ) f3 = 796 Hz = 3 f, so f = 65 Hz is the fundamental f8 = 8f = Hz; ω8 = π f8 = 33 rad/s λ = v/ f = (470 cm/s)/( Hz) = 693 cm and k = π / λ = rad/cm x (, t) = (560 cm)sin([00906 rad/cm] x)sin([33 rad/s] t) k = rad/cm So, EVALUATE: The wavelength and frequenc of the standing wave equals the wavelength and frequenc of the two traveling waves that combine to form the standing wave In the 8th harmonic the frequenc and wave number are larger than in the 3rd harmonic 54 IDENTIFY: Compare the xt (, ) specified in the problem to the general form of Eq(58) SET UP: The comparison gives A SW = 444 mm, k = 35 rad/m and ω = 754 rad/s EXECUTE: (a) A A SW = = (444 mm) = mm (b) λ = π = π = 093 m k 35 rad m 754 rad s (c) f = ω = = 0 Hz π π 754 rad s (d) v = ω = = 3 m s k 35 rad m (e) If the wave traveling in the + x direction is written as ( x, t ) = A cos( kx ωt ), then the wave traveling in the x-direction is ( x, t) Acos( kx ωt) = +, where A = mm from part (a), k = 35 rad m and ω = 754 rad s (f ) The harmonic cannot be determined because the length of the string is not specified EVALUATE: The two traveling waves that produce the standing wave are identical except for their direction of propagation 543 (a) IDENTIFY and SET UP: Use the angular frequenc and wave number for the traveling waves in Eq(58) for the standing wave EXECUTE: The traveling wave is x (, t) = (30 mm) cos([698 rad/m] x) + [74 rad/s] t) A = 30 mm so A SW = 460 mm; k = 698 rad/m and ω = 74 rad/s The general equation for a standing wave is x (, t) = ( A sin kx)sin ωt, so SW x (, t) = (460 mm)sin([698 rad/m] x)sin([74 rad/s] t) (b) IDENTIFY and SET UP: Compare the wavelength to the length of the rope in order to identif the harmonic EXECUTE: L = 35 m (from Exercise 54) λ = π / k = 0900 m L = 3( λ / ), so this is the 3rd harmonic

14 5-4 Chapter 5 (c) For this 3rd harmonic, f = ω / π = 8 Hz f = 3 f so f = (8 Hz)/3 = 393 Hz 3 EVALUATE: The wavelength and frequenc of the standing wave equals the wavelength and frequenc of the two traveling waves that combine to form the standing wave The nth harmonic has n node-to-node segments and the node-to-node distance is λ /, so the relation between L and λ for the nth harmonic is L= n( λ/) L 544 IDENTIFY: v= F/ μ v= fλ The standing waves have wavelengths λ n = and frequencies fn = nf The n standing wave on the string and the sound wave it produces have the same frequenc SET UP: For the fundamental n = and for the second overtone n = 3 The string has μ 3 = m/ L= (875 0 kg)/(0750 m) = 7 0 kg/m EXECUTE: (a) λ = L /3 = (0750 m)/3 = 0500 m The sound wave has frequenc v 344 m/s 4 f = = = 03 0 Hz For waves on the string, λ m 4 3 v= fλ = (03 0 Hz)(0500 m) = 55 0 m/s The tension in the string is F (b) 3 5 = μv = (7 0 kg/m)(55 0 m/s) = 30 0 N f /3 (03 0 Hz)/ Hz f = = = EVALUATE: The waves on the string have a much longer wavelength than the sound waves in the air because the speed of the waves on the string is much greater than the speed of sound in air 545 IDENTIFY and SET UP: Use the information given about the A 4 note to find the wave speed, that depends on the linear mass densit of the string and the tension The wave speed isn t affected b the placement of the fingers on the bridge Then find the wavelength for the D 5 note and relate this to the length of the vibrating portion of the string EXECUTE: (a) f = 440 Hz when a length L = 0600 m vibrates; use this information to calculate the speed v of waves on the string For the fundamental λ / = L so λ = L = (0600 m) = 0 m Then v= fλ = (440 Hz)(0 m) = 58 m/s Now find the length L = x of the string that makes f = 587 Hz v 58 m/s λ = = = 0900 m f 587 Hz L = λ/ = 0450 m, so x = 0450 m = 450 cm (b) No retuning means same wave speed as in part (a) Find the length of vibrating string needed to produce f = 39 Hz v 58 m/s λ = = = 35 m f 39 Hz L = λ / = 0675 m; string is shorter than this No, not possible EVALUATE: Shortening the length of this vibrating string increases the frequenc of the fundamental xt (, ) = ( A sin kx)sinωt v = / t a = / t 546 IDENTIFY: SW SET UP: vmax = ( ASW sin kx) ω amax = ( ASW sin kx) ω EXECUTE: (a) (i) x = λ is a node, and there is no motion (ii) x = λ is an antinode, and vmax = A( π f) = π fa, 4 amax = ( π f) vmax = 4π f A (iii) cos π = and this factor multiplies the results of (ii), so v 4 max = π fa, = π f A amax (b) The amplitude is Asin kx, or (i) 0, (ii) A, (iii) A (c) The time between the extremes of the motion is the same for an point on the string (although the period of the zero motion at a node might be considered indeterminate) and is / f EVALUATE: An point in a standing wave moves in SHM All points move with the same frequenc but have different amplitude 547 v IDENTIFY: For the fundamental, f = v= L F/ μ A standing wave on a string with frequenc f produces a sound wave that also has frequenc f SET UP: f = 45 Hz L = 0635 m EXECUTE: (a) v fl = = (45 Hz)(0635 m) = 3 m s

15 Mechanical Waves 5-5 (b) The frequenc of the fundamental mode is proportional to the speed and hence to the square root of the tension; (45 Hz) 0 = 46 Hz (c) The frequenc will be the same, 45 Hz The wavelength will be λ air = vair f = (344 m s) (45 Η z) = 40 m, which is larger than the wavelength of standing wave on the string b a factor of the ratio of the speeds EVALUATE: Increasing the tension increases the wave speed and this in turn increases the frequencies of the standing waves The wavelength of each normal mode depends onl on the length of the string and doesn't change when the tension changes 548 IDENTIFY: The ends of the stick are free, so the must be displacement antinodes The first harmonic has one node, at the center of the stick, and each successive harmonic adds one node SET UP: The node to node and antinode to antinode distance is λ / EXECUTE: The standing wave patterns for the first three harmonics are shown in Figure 548 st harmonic: L= λ λ = L= 40 m nd harmonic: L= λ λ = L= 0 m 3 rd 3 L harmonic: L = λ3 λ3 = = 33 m 3 EVALUATE: The higher the harmonic the shorter the wavelength Figure IDENTIFY and SET UP: Calculate v, ω, and k from Eqs(5), (55), and (56) Then appl Eq(57) to obtain x (, t ) 3 A = 50 0 m, λ = 80 m, v = 360 m/s EXECUTE: (a) v= fλ so f = v/ λ = (360 m/s)/80 m = 00 Hz ω = π f = π(00 Hz) = 6 rad/s k = π / λ = π rad/80 m = 349 rad/m (b) For a wave traveling to the right, x (, t) = Acos( kx ωt) This equation gives that the x = 0 end of the string has maximum upward displacement at t = 0 3 Put in the numbers: x (, t) = (50 0 m)cos((349 rad/m) x (6 rad/s) t) (c) The left hand end is located at x = 0 Put this value into the equation of part (b): 3 (0, t) =+ (50 0 m)cos((6 rad/s) t) (d) Put x = 35 m into the equation of part (b): 3 (35 m, t) = (50 0 m) cos((349 rad/m)(35 m) (6 rad/s) t) 3 (35 m, t) = (50 0 m) cos(47 rad) (6 rad/s) t 3 47 rad = 3 π / and cos( θ ) = cos( θ ), so (35 m, t) = (50 0 m) cos((6 rad/s) t 3 π / rad) (e) = Acos( kx ωt) ((part b)) The transverse velocit is given b v = = A cos( kx ωt) =+ Aωsin( kx ωt) 3 The maximum v is Aω = (50 0 m)(6 rad/s) = 035 m/s 3 (f ) x (, t) = (50 0 m)cos((349 rad/m) x (6 rad/s) t) t = 0065 s and x = 35 m gives 3 3 = (50 0 m) cos((349 rad/m)(35 m) (6 rad/s)(0065 s)) = 50 0 m v =+ Aωsin( kx ωt) =+ (035 m/s)sin((349 rad/m) x (6 rad/s) t) t = 0065 s and x = 35 m gives v = (035 m/s)sin((349 rad/m)(35 m) (6 rad/s)(0065 s)) = 00 EVALUATE: The results of part (f ) illustrate that v = 0 when = ± A, as we saw from SHM in Chapter IDENTIFY: Compare xt (, ) given in the problem to the general form given in Eq(58) SET UP: The comparison gives A = 0750 cm, k = 0400 π rad/cm and ω = 50 π rad/s

16 5-6 Chapter 5 EXECUTE: (a) A = 0750 cm, λ = = 500 cm, f = 5 Hz, T = = s and 0400 rad/cm f v= λ f = 65 m/s (b) The sketches of the shape of the rope at each time are given in Figure 550 (c) To sta with a wavefront as t increases, x decreases and so the wave is moving in the (d) From Eq (53), the tension is F = μv = (050 kg m) (65 m s) = 95 N (e) P = μfω A = av 54 W x -direction EVALUATE: The argument of the cosine is ( kx + ωt) for a wave traveling in the x-direction, and that is the case here Figure IDENTIFY: The speed in each segment is v= F μ The time to travel through a segment is t = L/ v μ 4μ μ SET UP: The travel times for each segment are t = L, t = L, and t3 = L F F 4F μ μ μ 7 μ EXECUTE: Adding the travel times gives ttotal = L + L + L = L F F F F (b) No The speed in a segment depends onl 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

17 Mechanical Waves IDENTIFY: Appl τ z = 0 to find the tension in each wire Use v= F/ μ to calculate the wave speed for each wire and then t = L/ vis the time for each pulse to reach the ceiling, where L = 5 m m 50 N SET UP: The wires have μ = = = 004 kg/m The free-bod diagram for the beam is L (980 m/s )(5 m) given in Figure 55 Take the axis to be at the end of the beam where wire A is attached EXECUTE: τ z = 0 gives TL B = wl ( /3) and TB = w/3 = 583 N TA + TB = 750 N, so T A = 67 N v A TA 67 N 5 m 583 N = = = 756 m/s t A = = 0065 s v B = = 535 m/s μ 004 kg/m 756 m/s 004 kg/m 5 m t B = = 0034 s Δ t = tb ta = 69 ms 535 m/s EVALUATE: The wave pulse travels faster in wire A, since that wire has the greater tension Figure IDENTIFY and SET UP: The transverse speed of a point of the rope is v = / t where x (, t ) is given b Eq(57) EXECUTE: (a) x (, t) = Acos( kx ωt) v = d/ dt =+ Aω sin( kx ωt) v = Aω = π fa, max v F f = and v =, so λ ( m/ L) FL f = λ M π A FL v, max = λ M (b) To double v, max increase F b a factor of 4 EVALUATE: Increasing the tension increases the wave speed v which in turn increases the oscillation frequenc With the amplitude held fixed, increasing the number of oscillations per second increases the transverse velocit 554 IDENTIFY: The maximum vertical acceleration must be at least g SET UP: amax = ω A gλμ min EXECUTE: g = ω A min and thus Amin = g/ ω Using ω = π f = πv/ λ and v= F/ μ, this becomes A = 4π F EVALUATE: When the amplitude of the motion increases, the maximum acceleration of a point on the rope increases 555 IDENTIFY and SET UP: Use Eq(5) and ω = π f to replace v b ω in Eq(53) Compare this equation to ω = k / m from Chapter 3 to deduce k EXECUTE: (a) ω = π f, f = v/ λ, and v= F/ μ These equations combine to give ω = π f = π( v/ λ) = ( π/ λ) F/ μ But also ω = k / m Equating these expressions for ω gives k = m( π / λ) ( F/ μ) But m= μ Δ x so k =Δ x( πλ / ) 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 inversel proportional to the wavelength λ The tension supplies the restoring force and the /λ factor represents the dependence of the restoring force on the curvature of the string

18 5-8 Chapter IDENTIFY: Appl τ z = 0 to one post and calculate the tension in the wire v= F/ μ for waves on the wire v= fλ The standing wave on the wire and the sound it produces have the same frequenc For standing waves on L the wire, λ n = n SET UP: For the 7 th overtone, n = 8 The wire has μ = m/ L= (073 kg)/(500 m) = 046 kg/m The free-bod diagram for one of the posts is given in Figure 556 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 L EXECUTE: τ z = 0 gives w cos570 T( Lsin570 ) = 0 w 35 N T = = = 763 N For tan570 tan570 F 763 N waves on the wire, v = = = 9 m/s For the 7 th overtone standing wave on the wire, μ 046 kg/m L (500 m) v 9 m/s λ = = = 5 m f = = = 83 Hz The sound waves have frequenc 83 Hz and 8 8 λ 5 m 344 m/s wavelength λ = = 88 m 83 Hz EVALUATE: The frequenc of the sound wave is at 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 Figure IDENTIFY: The magnitude of the transverse velocit is related to the slope of the t versus x curve The transverse acceleration is related to the curvature of the graph, to the rate at which the slope is changing SET UP: If increases as t increases, v is positive a has the same sign as v if the transverse speed is increasing EXECUTE: (a) and (b) (): The curve appears to be horizontal, and v = 0 As the wave moves, the point will begin to move downward, and a < 0 (): As the wave moves in the + x-direction, the particle will move upward so v > 0 The portion of the curve to the left of the point is steeper, so a > 0 (3) The point is moving down, and will increase its speed as the wave moves; v < 0, a < 0 (4) The curve appears to be horizontal, and v = 0 As the wave moves, the point will move awa from the x-axis, and a > 0 (5) The point is moving downward, and will increase its speed as the wave moves; v < 0, a< 0 (6) The particle is moving upward, but the curve that represents the wave appears to have no curvature, so v > 0 and a = 0 (c) The accelerations, which are related to the curvatures, will not change The transverse velocities will all change sign EVALUATE: At points, 3, and 5 the graph has negative curvature and a < 0 At points and 4 the graph has positive curvature and a > IDENTIFY: The time it takes the wave to travel a given distance is determined b the wave speed v A point on the string travels a distance 4A in time T SET UP: v= fλ T = / f d d 800 m EXECUTE: (a) The wave travels a horizontal distance d in a time t = 0333 s v = λ f = 0600 m 400 Hz = ( )( )

19 Mechanical Waves 5-9 (b) A point on the string will travel a vertical distance of 4A each ccle Although the transverse velocit v ( x, t ) is not constant, a distance of h = 800 m corresponds to a whole number of ccles, 3 = (4 ) = (800 m) [4(500 0 m)] = 400, so the amount of time is t = nt = n f = (400) (400 Hz) = 00 s n h A 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 559 IDENTIFY: ( x, ) + z ( x, ) = A The trajector is a circle of radius A SET UP: v = / t, v = z/ t a = v / t, a = v / t z EXECUTE: At t = 0, (0,0) = A, z(0,0) = 0 At t = π ω, (0, π ω) = 0, z(0, π ω) = A z At t = πω, (0, πω) = A, z(0, πω) = 0 At t = 3π ω, (0,3π ω) = 0, z(0,3π ω) = A The trajector and these points are sketched in Figure 559 (b) v = t =+ Aω sin( kx ωt), vz = z t = Aωcos( kx ωt) # ˆ ˆ [sin( ) ˆ cos( ) ˆ v= vj + vzk = Aω kx ω j kx ωt k] v= v + vz = Aω so the speed is constant # r = ˆj + zkˆ # # r v = v + zvz = A ωsin ( kx ωt)cos( kx ωt) A ωcos( kx ωt) sin( kx ωt) = 0 r # v # =0, so v # is tangent to the circular path (c) a = v t = Aω cos( kx ωt), az = vz t = Aω sin( kx ωt) # # # # r a = a + zaz = A ω [cos ( kx ωt) + sin ( kx ωt)] = A ω r = A, r a = ra r # a # = ra cos φ so φ = 80 and a # # # is opposite in direction to r ; a is radiall inward For these x (, t) and zx (, t ), + z = A, so the path is again circular, but the particle rotates in the opposite sense compared to part (a) + x-direction The displacement is transverse, so v # and a # lie in the z- EVALUATE: The wave propagates in the plane z Figure 559 L 560 IDENTIFY: The wavelengths of the standing waves on the wire are given b λ n = When the ball is changed n Fl0 the wavelength changes because the length of the wire changes; Δ l = AY 0 SET UP: For the third harmonic, n = 3 For copper, Y = 0 Pa The wire has cross-sectional area A= πr = π(05 0 m) = 84 0 m 3 7 (0 m) EXECUTE: (a) λ 3 = = 0800 m 3 ( ΔFl ) 0 (b) The increase in length when the 000 N ball is replaced b the 5000 N ball is given b Δ l =, where AY Δ F = 4000 N is the increase in the force applied to the end of the wire (4000 N)(0 m) 3 Δ l = = m The change in wavelength is Δ λ = Δ l = 35 mm (84 0 m )( 0 Pa) EVALUATE: The change in tension changes the wave speed and that in turn changes the frequenc of the standing wave, but the problem asks onl about the wavelength

20 5-0 Chapter 5 56 IDENTIFY: Follow the procedure specified in part (b) u u SET UP: If u = x vt, then = v and = EXECUTE: (a) As time goes on, someone moving with the wave would need to move in such a wa that the wave appears to have the same shape If this motion can be described b x = vt + b, with b a constant, then xt (, ) = fb ( ), and the waveform is the same to such an observer (b) d f = du and d f du = v, (c) This is of the form xt (, ) = fu ( ), with u= x vtand to determine the speed v= C B so xt (, ) = f( x vt) is a solution to the wave equation with wave speed v f( u) De B ( x Ct/ B) = The result of part (b) ma be used EVALUATE: The wave in part (c) moves in the + x-direction The speed of the wave is independent of the constant D 56 IDENTIFY: The phase angle determines the value of for x = 0, t = 0 but does not affect the shape of the xt (, ) versus x or t graph cos( kx ωt + φ) SET UP: = ωsin( kx ωt + φ) EXECUTE: (a) The graphs for each φ are sketched in Figure 56 (b) = ωasin( kx ωt + φ) (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 identifφ uniquel, the quadrant in whichφ lies must be known In phsical terms, the signs of both the position and velocit, and the magnitude of either, are necessar to determine φ (within additive multiples of π ) EVALUATE: The phase φ = 0 corresponds to = Aat x = 0, t = 0 Figure IDENTIFY and SET UP: Use Eq(53) to replace μ, and then Eq(56) to replace v EXECUTE: (a) Eq(55): v= F/ μ sas F / v P = μfω A av μ = so = ( ) ω = ω P F / v F A F A / v av ω = π f so ω / v= π f / v= π / λ = k and P = FkωA as was to be shown av, (b) IDENTIFY: For the ω dependence, use Eq(55) since it involves just ω, not k: SET UP: P, μ, A all constant so av Fω is constant, and EXECUTE: /4 /4 /4 Fω ω = ω ( F / F ) = ω ( F /4 F) = ω (4) = ω / ω must be changed b a factor of / (decreased) F ω = IDENTIFY: For the k dependence, use the equation derived in part (a), P = FkωA av SET UP: If P av and A are constant then Fkω must be constant, and Fkω = Fk ω P μfω A av =

21 F ω F ω EXECUTE: k = k = k k k k/ 8 F ω 4F = = = ω 4 6 / Mechanical Waves 5- k must be changed b a factor of / 8 (decreased) EVALUATE: Power is the transverse force times the transverse velocit To keep P av constant the transverse velocit must be decreased when F is increased, and this is done b decreasing ω 564 IDENTIFY: The wave moves in the + x direction with speed v, so to obtain xt (, ) replace x with x vt in the expression for x (,0) SET UP: Pxt (, ) is given b Eq(5) EXECUTE: (a) The wave pulse is sketched in Figure 564 (b) 0 for ( x vt) < L h( L+ x vt) / L for L< ( x vt) < 0 xt (, ) = h( L x+ vt) / L for 0 < ( x vt) < L 0 for ( x vt) > L (c) From Eq(5): F(0)(0) = 0 for ( x vt) < L xt (, ) xt (, ) FhL ( / )( hvl / ) = FvhL ( / ) for L < ( x vt) < 0 Pxt (, ) = F = x t F( h/ L)( hv/ L) = Fv( h/ L) for 0 < ( x vt) < L F(0)(0) = 0 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 velocit v is constant in magnitude and has opposite sign on either side of the peak of the pulse Figure IDENTIFY and SET UP: The average power is given b Eq(55) Rewrite this expression in terms of v and λ in place of F and ω EXECUTE: (a) P = μfω A av v= F/ μ so F = v/ μ ω = π f = π( v/ λ) Using these two expressions to replace λ P av A = 707 cm 3 = 4π v μ (b) EVALUATE: P ~ av v 3 so doubling v increases av 3 F and ω gives Pav = μπ v A / λ ; μ = P b a factor of 8 3 (600 0 kg)/(800 m) P av = 8(500 W) = 4000 W 566 IDENTIFY: Draw the graphs specified in part (a) SET UP: When xt (, ) is a maximum, the slope / is zero The slope has maximum magnitude when xt (, ) = 0 EXECUTE: (a) The graph is sketched in Figure 566a (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

22 5- Chapter 5 (c) The energ flow is alwas in the same direction (d) In this case, = kasin( kx +ω t) and Eq(5) becomes Pxt (,) = FkωA sin( kx+ ωt) The power is now negative (energ flows in the x-direction ), but the qualitative relations of part (b) are unchanged The graph is sketched in Figure 566b EVALUATE: cosθ and sinθ are 80 out of phase, so for fixed t, maximum corresponds to zero P and = 0 corresponds to maximum P Figure IDENTIFY and SET UP: v= F/ μ The coefficient of linear expansion α is defined b Δ L = L0 α Δ T This can FA / be combined with Y = to give Δ F = YαAΔ T for the change in tension when the temperature changes b Δ L/ L0 Δ T Combine the two equations and solve for α EXECUTE: v = F/ μ, v = F/ μ and F = μv The length and hence μ sta the same but the tension decreases b Δ F = YαA Δ T v = ( F +Δ F)/ μ = ( F Yα A Δ T)/ μ v = F/ μ YαA Δ T/ μ = v YαA Δ T/ μ And μ = m/ L so A/ μ = AL/ m= V/ m= / ρ (A is the cross-sectional area of the wire, V is the volume of a v v length L) Thus v v = α( Y Δ T/ ρ) and α = (/ Y ρ) ΔT EVALUATE: When T increases the tension decreases and v decreases 568 IDENTIFY: The time between positions and 5 is equal to T / v= fλ The velocit of points on the string is given b Eq(59) 60 s SET UP: Four flashes occur from position to position 5, so the elapsed time is 4 = 0048 s The figure 5000 in the problem shows that λ = L = 0500 m At point P the amplitude of the standing wave is 5 cm EXECUTE: (a) T / = 0048 s and T = 0096 s f = / T = 04 Hz λ = 0500 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 st overtone and nd harmonic (c) v= fλ = (04 Hz)(0500 m) = 50 m/s (d) In position, 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 = ωa= π fa= π(04 Hz)(005 m) = 0980 m/s F FL FL (00 N)(0500 m) (e) v = = and m = = = 85 g μ m v (50 m/s) EVALUATE: The standing wave is produced b 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 569 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 /, so the distance between an two nodes is n( λ / ) for n =,, 3, This must equal 450 cm, since there are nodes at the post and clothespin Use this in Eq(5) to get an expression for the possible frequencies f EXECUTE: 450 cm = n( λ / ), λ = v/ f, so f = nv [ /(900 cm)] = (0800 Hz) n, n =,, 3, EVALUATE: Higher frequencies have smaller wavelengths, so more node-to-node segments fit between the post and clothespin

23 Mechanical Waves IDENTIFY: The displacement of the string at an point is xt (, ) = ( ASW sin kx)sin ωt For the fundamental mode λ = L, so at the midpoint of the string sin kx = sin( π λ)( L ) =, and = A sin ωt SW The transverse velocit is v = / and the transverse acceleration is a = v / t SET UP: Taking derivatives gives v cos = = ωasw ωt, with maximum value v,max = ωasw, and v sin a = = ω ASW ωt, with maximum value a, max = ω ASW EXECUTE: ω = a = =, and then 3 3,max v,max (840 0 m s ) (380 m s) 0 rad s A = v ω = = 3 3 SW,max (380 m s) ( 0 rad s) 7 0 m 3 (b) v= λf = ( L)( ω π) = Lω π = (0386 m)( 0 rad s) π = 7 m s EVALUATE: The maximum transverse velocit and acceleration will have different (smaller) values at other points on the string 57 IDENTIFY: To show this relationship is valid, take the second time derivative SET UP: sinωt = cosωt cosωt = ωsinωt EXECUTE: (a) xt (, ) = [( A SW sin kx)sin ωt] = ω [( ASW sin kx)cos ωt] xt (, ) = ω [( Asw sin kx)sin ωt] = ω ( x, t) This equation shows that a = ω This is characteristic of simple harmonic motion; each particle of the string moves in simple harmonic motion (b) Yes, the traveling wave is also a solution of this equation When a string carries a traveling wave each point on the string moves in simple harmonic motion EVALUATE: A standing wave is the superposition of two traveling waves, so it is not surprising that for both tpes of waves the particles on the string move in SHM 57 IDENTIFY and SET UP: Carr out the analsis specified in the problem EXECUTE: (a) The wave moving to the left is inverted and reflected; the reflection means that the wave moving to the left is the same function of x, and the inversion means that the function is f ( x) (b) The wave that is the sum is f ( x) f ( x) (an inherentl odd function), and for an f, f(0) f( 0) = 0 (c) The wave is reflected but not inverted (see the discussion in part (a) above), so the wave moving to the left in Figure 5 in the textbook is + f ( x) ( ) ( ) ( ) ( ) ( ) (d) ( ( ) ( )) x df x df x d = f x + f x = + = + x = df df dx dx dx dx dx d( x) dx dx dx x = x At x = 0, the terms are the same and the derivative is zero EVALUATE: Our results verif the behavior shown in Figures 50 and 5 in the textbook 573 IDENTIFY: Carr out the derivation as done in the text for Eq(58) The transverse velocit is v = / t and the transverse acceleration is a = v / t (a) SET UP: For reflection from a free end of a string the reflected wave is not inverted, so x (, t) = ( x, t) + ( x, t), where ( x, t) Acos( kx ωt) = + (traveling to the left) ( x, t) Acos( kx ωt) = (traveling to the right) Thus x (, t) = A[cos( kx+ ωt) + cos( kx ωt)] EXECUTE: Appl the trig identit cos( a± b) = cosacosb sin asinb with a = kx and b= ωt: cos( kx + ωt) = coskxcosωt sin kxsinωt and cos( kx ωt) = coskxcosωt + sin kxsin ωt Then x (, t) = (Acos kx)cosωt (the other two terms cancel) (b) For x = 0, coskx = and x (, t) = Acos ωt The amplitude of the simple harmonic motion at x = 0 is A, which is the maximum for this standing wave, so x = 0 is an antinode (c) max = A from part (b) cosωt v = = [(Acos kx) cos ωt] = Acoskx = Aωcos kxsin ωt