 Phyllis Fleming Physics Physics 104
 Answers - Wave Motion

1.  (a) From Figures 1a and 1b above, we find the wavelength to be 4.00 cm.

(b) The crest at x = 1.00 cm in Fig. 1a moves to x = 3.00 cm in Fig. 1b in 10 s. Since the wave moves one wavelength in a period T and (3.00 cm - 1.00 cm) is one-half wavelength, the period is 20 s. The frequency f = 1/T = (1/20) s-1.

(c) v = f λ = (1/20) s-1 (4.00 cm) = 0.20 cm/s.

(d) At t = 10 s, there is a crest at x = 3.00 cm. There will be two more crests at 3.00 cm +λ and 3.00 cm + 2 λ or at x = 7.00 cm and x = 11.00 cm.

(e) At t = 0 there is a crest at x =1.00 cm. There will be another crest at T =
20 s.

(f) The rope has zero displacement at any x twice each period.

(g) There is one crest at any point in one period. In 80 s there will be 80/20 =
4 crests.

 2 (a) In Fig. 1a above, for y = 0  at  t = 0  and  x = 0, you see a trough approaching from the left. Similarly in Fig. 1b above, for y = 0  at  t = 10 s  and  x = 0, you see a crest approaching from the left. This is shown below in a plot of y as a function of  t  for  x = 0 (Fig. for #2a): (b) In Fig. 1a above, for  y = 0  at  t = 0  and  x = 2.00 cm, you see a crest approaching from the left. A plot of y as a function of t for x = 2 cm is shown in Fig. for #2b below: (c) In Fig. 1a, for  y = 0  at  t = 0  and  x = 4 cm, you see a trough approaching from the left. In Fig. 1b, for  y = 0  at  t = 10 s  and  x = 4 cm, you see a crest approaching from the left. A plot of y as a function of t for x = 4.0 cm is shown in fig. for #2c below: 3 The question is answered simply by looking at Fig. for #2a through #2c above.There is no question that Fig. for #2a is equivalent to #2c. Points are in phase when they are an integral number of λ's apart. Points at x = 0 and x = 2 cm are λ/2 apart and are 1800 out of phase. Points at x = 0 and x = 4 cm are λapart and are in phase.
 4 For a wave traveling to the right, we may write y(x,t) = ym sin (2 πx/ λ - 2πt/T) where ym is the amplitude of the wave, λ its wavelength, and T the period. For this case ym = 10 cm, λ = 4.0 cm, T = 20 s, and y(x,t) = 10 cm sin [(2 πx/4.0)cm-1 - (2 πt/20)s-1]          = 10 cm sin [( πx/2)cm-1 - (0.1 πt)s-1]
 5 (a) For x = 0, y(0,t) = 10 cm sin (-0.1πt s-1)          = -10 cm sin (0.1πt s-1)      as shown in Fig. for #2a above. (b) For x = 2.0 cm, y(2.0 cm, t) = 10 cm sin [(2π/2) - (0.1πt)s-1]                   = 10 cm{ sin π cos 0.1πt s-1 - cos π sin 0.1πt s-1}                   = 10 cm{(0) cos 0.1πt s-1 - (-1) sin 0.1πt s-1}                   = 10 cm sin 0.1πt s-1        as shown in Fig. for #2b above. (c) For x = 4.0 cm, y(4.0 cm, t) = 10 cm sin [(4π/2) - (0.1πt)s-1]                   = 10 cm{ sin 2π cos 0.1πt s-1 - cos 2πsin 0.1πt s-1}                   = 10 cm{(0) cos 0.1πt s-1 - (1) sin 0.1πt s-1}                   = -10 cm sin 0.1πt s-1       as shown in Fig. for #2c above.
 6 For x = 0, y(0,t) = -10 cm sin (0.1 πt s-1): (a) v(0, t) = dx/dt = -1.0 π cm/s cos (0.1πt s-1) (b) a(0,t) = dv/dt = +0.1π2 cm/s2 sin (0.1 πt s-1) = -(0.1π2 s-2)y(0,t)
 7 Dashed wave: λ = 4.0 cm: solid: λ’ = 2.0 cm. Dashed wave: ym = 1.0 cm; solid: ym’ = 0.5 cm (a) Two waves of the same amplitude, but different wavelengths (Fig. for #7a). (b) Two waves of the same wavelength, but different amplitudes (Fig. for #7b).
 8 Since y has a maximum displacement of 10 cm at x = 0, we use a cosine function for the wave of amplitude xm = 10 cm. With a frequency f of 10 s-1 and a velocity v = 40 cm/s,  λ = v/f = 40 cm/s/10 s-1 = 4.0 cm. (a) A plot of y as a function of x for t = 0 is shown in Fig. for #8a above. Notice the maximum displacement of 10 cm for x = 0 and the wavelength of the wave as 4.0 cm. (b) For f = 10 s-1, T = 1/f = 0.10 s. A plot of y as a function of x for t = 0.05 s = T/2 is shown in Fig. for #8b above. Notice that the crest at x = 0 for t = 0 has moved to x = 2.0 cm = λ/2 in time T/2. In general for a wave moving to the right, y(x,t) = ym cos(2 πx/  λ - 2πt/T) For this wave, y(x,t) = 10 cm cos [( πx/2)cm-1 - (20 πt)s-1]
 9 (a) In general, y(x,t) = ym sin (2 πx/  λ - 2πt/t) for a wave to the right. For our case, y(x,t) = 2.0 cm sin (0.50 π cm-1x - 200 π s-1t). By comparison, ym = 2.0 cm; 2π/  λ = 0.50 π cm-1, or λ = 4.0 cm; 2 π/T = 200 π s-1, or T = 0.01 s; frequency f = 1/T = 100 s-1; v = λf = (4.0 cm)(100 s-1) = 40 cm/s; direction of the wave is to the right because of - sign in wave equation. (b) There is a crest at x =1.00 cm at t = 0 (Fig. for #9a above). This crest moves to the right to x = 2.00 cm or a distance of 1.00 cm = λ/4 in time 0.0025s = T/4 (Fig. for #9b above). The crests moves to the right to x = 3.00 cm or a distance of 2.00 cm = λ/2 in time 0.0050 s = T/2 (Fig. for #9c above). (c) y(x, t) = 2.0 cm sin(0.5 π cm-1x-200 πs-1t).   (dy/dt)x =1cm = -400 π cm/s cos(0.5 π -200 πs-1t).   (dy/dt)x=1cm,t=0 = -400 π cm/s cos 0.5π = 0.   (dy/dt)x=1cm,t = 0.0025s = -400 π cm/s cos(0.5 π - 0.5 π) = -400 π cm/s.

10. A wave moves one wavelength in one period. From Figs. 2a and 2b above, we see that the wavelength of the waves = 8.0 cm . In one-half period the waves moves 4.0 cm. Crest C, on the wave that is moving to the right, at x = 0 in Fig. 2a, moves to x = 4.0 cm in T/2. Crest C’, on the wave that is moving to the left, at x = 8.0 cm in Fig. 2b, moves to x = 4.0 cm. At t = T/2, there will be a double crest at x = 4.0 cm with a displacement y = 4.00 cm.  The individual waves to the right and to the left, respectively, are shown immediately above in Fig for #10 (b) at t = 3T/4. Now C has moved to the right 3λ/4 = 6.0 cm to x = 6.0 cm and C’ has moved 6.0 cm to the left to x = 2.0 cm.
C at x = 6.0 cm is superimposed with a trough at this point from the wave moving to the left.  C’ at x = 2.0 cm is superimposed with a trough at this point from the wave moving to the right.  The entire string has zero displacement at
t = 3T/4.

 11 (a) Points on a rope along which a traveling wave exists vibrate up and down with simple harmonic motion. (b) All point on the rope except the nodal points on a rope along which a standing wave exists vibrate up and down with simple harmonic motion. (c) All points on a rope along which a traveling wave exits vibrate with the same amplitude. (d) The amplitude of a rope along which a standing wave exists varies from a maximum at the antinodes to zero at the nodes. (e) The entire rope along which a traveling wave exits never has zero displacement. (f) The entire rope along which a standing wave exists has zero displacement twice each period.
 12 The plots are shown in Fig. for #12 below for a wave traveling to the right. The wavelength of the wave is 4.0 cm. The crest at x = 0 at -T/4 moves to x = 1.0 cm at t = 0; to x = 2.0 cm at t = T/4, and to x = 3.0 cm at t = T/2. 13.

 The plot of y vs t for x = 0 is shown to the right in Fig for #13. If you look above at the graphs for Fig. for #12, you see that the point at x = 0 has the following values of y: y = 5.0 cm at t =-T/4, y = 0 at t = 0, y = 5.0 cm at t = T/4, and y = 0 at t = T/2. 14 The plots are shown in the graphs below in Fig. for #14 for a wave traveling to the left. The crest at x = 6 cm at t = -T/4 moves  λ/4 = 1 cm to x = 5 cm at t = 0; to x = 4 cm at x = T/4, and to x = 3 cm at x = T/2. 15.

If you look back at the graphs for Fig. for #12 and Fig. for #14, above, and add the displacement y for various values of x for t = -T/4, t = 0, t = T/4, and t = T/2, you will get the graphs shown below in Fig. for #15.  At times t = -T/4 and
t = T/4, the entire rope has zero displacement.  16 The point on the rope at x = 1.0 cm is an antinodal point.  If you look above at the Fig. for #15 you see the point at x = 1.0 cm has y = 10 cm at t = 0, y = 0 at t = T/4, and y = -10 cm at T/2. These correspond to the values of y as a function of time shown in the Fig. for #16 below. The amplitude of an antinodal point varies from 0 to a maximum, in this case, 10 cm. 17 (a) The unit of force is N or kg-m/s2. The unit of mass per unit length µ is kg/m, the unit of velocity v is m/s. The unit of µ/F is (kg/m)/(kg-m/s2) = (s/m)2 ≠ m/s. The unit of F/µ is (kg-m/s2)/(kg/m) = (m/s)2 ≠ m/s. We could stop here and write down the answer, but let's carry on. The unit of (µ/F)1/2 = s/m ≠ m/s. The unit of (F/µ)1/2 = m/s! (b) The distance between nodes is λ/2. Thus the number of nodes is inverserly proportional to λ.   λf = v. The velocity v is directly proportional to the square root of the tension F. If the tension is increased by a factor of four, the square root of the tension and the velocity is increased by a factor of two. For a fixed frequency f, the wavelength is directly proportional to the velocity v. If the velocity is doubled, the wavelength is doubled. If the wavelength is doubled, the number of nodes, which is inversely proportional to the wavelength, is halved.
 18 In general for a wave to the left, y(x,t) = ym sin (2 πx/ λ - 2πt/T). When we compare y(x,t) = 2.0 cm sin (2πx cm-1 - 600πt s-1), you see that 2 π/λ = 2 π cm-1, or λ= 1.0 cm.  Also 2 π/T = 600 π s-1,  or T = (1/300) s and f= 1/T = 300 s-1. (a) v = λf = 1.0 cm(300 s-1) = 300 cm/s. (b) v = (F/µ)1/2, or v2 = F/µ.     µ = F/v2 = 18 N/(3 X 102 m/s)2 = 2 x 10-4 (kg-m/s2/m2/s2) = 2 x 10-4 kg/m.
 19 To find the equation of a wave, we must find the amplitude A, wavelength λ, and frequency f of the wave. We are given ym = 1.0 cm and f = 100 s-1,  the tension in the string F = 1.6 N,  the mass per unit length µ = 4.0 x 10-3 kg/m. We know that  λ = v/f and v = (F/µ)1/2 = (1.6 N/4.0 x10-3 kg/m)1/2 = 20 m/s. Then λ = v/f = 20 m/s/100 s-1 = 0.20 m = 20 cm. The wave is going toward smaller values of x, or to the left. For a wave to the left,  y(x,t) = ym sin (2πx/  λ + 2 πft). For our case,  y(x,t) = 1.0 cm sin (0.10 πx cm-1 + 200 πt s-1).
 20 For a string fixed at both ends, the allowed frequencies are fn = nv/2L, where n is an integer, v the velocity, and L the length of the string.  For f = 440 s-1 and L = 50 cm,  nv = 2fnL = 2(440 s-1)50 cm = 4.4 x 104 cm/s.  For a frequency of 528 s-1 = nv/2L’ = 4.4 x 104 cm/s/2L’ or L’ = 2.2 x 104 cm/528 = 41.7 cm.
 21 The lowest frequency occurs for n = 1 in fn = nv/2L. 30 s-1 = v/2(60 cm);  v = 3.6 x 103 cm/s.
 22 (a) To dislodge bird 2 without disturbing bird 1, she needs an antinode at the position of bird 2 and a node at the position of bird 1. With a node at both ends of the wire, the longest wavelength is in the Fig. for #22 above. From the figure, 3λ/2 = 36 cm,  λ = 24 cm. (b) f = v/  λ = 48 m/s/24 m = 2 s-1. (c) For a standing wave,  y(x,t) = ym sin 2 πx/  λ cos 2 πft.        dy/dt = -2π ym sin 2 πx/  λ sin 2 πft.        a = d2y/dt2 = -(2πf)2 ym sin 2 πx/  λ cos 2 πft. The maximum value of the sine and cosine is 1. If the maximum absolute value of the acceleration bird 2 can withstand is 48 m/s2, then 48 m/s2 = (2 πf)2ymax, where ymax  is the maximum amplitude he can withstand. ymax = 48 m/s2/(4 πs-1)2 = 0.30m.  Any amplitude slightly greater than this is the smallest amplitude needed to dislodge the bird.
 23 General equation for a standing wave is y(x,t) = ym sin 2πx/ λ cos 2πft Compare this with y(x,t) = 1.0 cm (sin px/3 cm-1) cos 20pt s-1 to find that (a) A = 1.0 cm, (b) 2π λ = π/3 cm-1 or λ = 6 cm, 2 πf = 20 π s-1 or f = 10 s-1      and v = λf = 6 cm(10 s-1) = 60 cm/s. (c) We need two waves of the same wavelength and frequency,      traveling in opposite directions with an amplitude = 0.5 cm.      y1(x,t) = 0.5 cm sin ( πx/3 cm-1 - 20 πt s-1) and y2(x,t) =      0.5 cm sin ( πx/3 cm-1 + 20 πt s-1). (d) the distance between nodes = λ/2 = 3.0 cm. (e) v(x,t) = dx/dt = -20 π cm/s sin πx/3 cm-1 sin 20 πt s-1.      v(1.5 cm, 1/4 s) = -20 π cm/s (sin π/2) (sin 5 π)                                = -20 π(1)(0) = 0.
 24 (a) Distance between nodes = λ/2 = 0.250 m.  λ = 0.500 m. (b) For an antinodal point, y(t) = 10.0 cos 2 πt/T.      For t = 1.0 s, 5.0 cm =10.0 cos 2 π s/T.      cos-1 0.50 = π/6 = 2 π s/T.      T = 12 s. (c) # of half wavelengths in 1.25 m = 1.25 m/0.25 m = 5. There is a node at      both ends. The rope is fixed at both ends. (d) Now y(x,t) = A sin 2πx/λ cos 2 πft = 10 cm sin 4 πx cm-1 cos πt s-1/6 (e) λn = 2L/n.  For n = 1, λ1 = 2L = 2(1.25 m) = 2.50 m.  For n = 2, λ2 = L =      1.25 m.  For n = 3,  λ3 = 2.50 m/3 = 0.833 m.  The figures are shown below      in Fig. for #24. 25 For a standing wave, y(x,t) = ym sin 2 πx/  λ cos 2 πft. The end at which a string is fixed must have a node or y = 0 at that end.  For a string fixed at x = 0, this expression gives y(0,t) = 0.  For a string also fixed at L, y(L,t) = 0 = ym sin 2 πL/  λ cos2 πft.  For this to be 0, sin 2 πL/  λ = 0 or 2 πL/  λ = n π and λn = 2L/n, where n = 1, 2, 3, . . .   For a string fixed at x = 0, you still use the sine, but now with the string no longer fixed at L you want an antinode at x = L, or dy/dx = 0 at x = L.  dy/dx = 2 πym/  λ cos 2 πx/  λcos 2 πft. This will be zero at all times at x = L if cos 2πL/  λ = 0 or 2 πL/  λ = (n - 1/2) π and  λn = 2L/(n - 1/2) with n = 1, 2, . .
 26 Consider an observer at the wall. For her, the velocity of the source is vs = u and her own velocity is zero. The frequency heard by the observer at the wall is, νwall = νo[v/(v-u)]        (Equation 1) The reflected sound now comes from a stationary source and is heard by the moving observer. The frequency heard by the moving observer is ν = νwall [(v + u)/(v)]     (Equation 2) Substituting νwall from Eq. 1 into Eq. 2: ν = νo[(v + u)/(v - u)] The beat frequency is ν - νo = νo[(v + u)/v - u) - 1] = νo [2u/(v - u)] = 100 Hz[2(10 m/s)/(340 - 10 m/s)] = = 6.1 Hz

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