al-ameri (aoa434) Hwk07 Stokes (19102) This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. This is ONLINE HOMEWORK No 7. It is due before 7.00 am on MONDAY 16 March, A
al-ameri (aoa434) Hwk02 Stokes (19102) This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. This is Online Homework No. 2 (OH-02). All answers should be submitted onlin
al-ameri (aoa434) Hwk09 Stokes (19102) This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. This is Online Homework No9. It is due before 6.00 am on Tuesday 7 April, Ab
111. We find the difference in the two applications of the Doppler formula:
f 2 f1 = 37 Hz = f
340 m/s + 25 m/s 340 m/s 25 m/s =f 340 m/s 15 m/s 340 m/s 15 m/s 340 m/s 15 m/s
which leads to f = 4.8 102 Hz .
49. The harmonics are integer multiples of the fundamental, which implies that the difference between any successive pair of the harmonic frequencies is equal to the fundamental frequency. Thus, f1 = (390 Hz 325 Hz) = 65 Hz. This further implies that the
50. Since the rope is fixed at both ends, then the phrase second-harmonic standing wave pattern describes the oscillation shown in Figure 1623(b), where (see Eq. 1665) =L and f= v . L
(a) Comparing the given function with Eq. 16-60, we obtain k = /2 and =
51. (a) The amplitude of each of the traveling waves is half the maximum displacement of the string when the standing wave is present, or 0.25 cm. (b) Each traveling wave has an angular frequency of = 40 rad/s and an angular wave number of k = /3 cm1. The
52. The nodes are located from vanishing of the spatial factor sin 5x = 0 for which the solutions are 123 5x = 0, , 2,3, x = 0, , , , 555 (a) The smallest value of x which corresponds to a node is x = 0. (b) The second smallest value of x which correspond
53. (a) The waves have the same amplitude, the same angular frequency, and the same angular wave number, but they travel in opposite directions. We take them to be y1 = ym sin(kx t), y2 = ym sin(kx + t). The amplitude ym is half the maximum displacement o
54. From the x = 0 plot (and the requirement of an anti-node at x = 0), we infer a standing wave function of the form y ( x, t ) = (0.04) cos(kx) sin(t ), where = 2 / T = rad/s , with length in meters and time in seconds. The parameter k is determined by
55. (a) The angular frequency is = 8.00/2 = 4.00 rad/s, so the frequency is f = /2 = (4.00 rad/s)/2 = 2.00 Hz. (b) The angular wave number is k = 2.00/2 = 1.00 m1, so the wavelength is = 2/k = 2/(1.00 m1) = 2.00 m. (c) The wave speed is
v = f = (2.00 m) (
56. Reference to point A as an anti-node suggests that this is a standing wave pattern and thus that the waves are traveling in opposite directions. Thus, we expect one of them to be of the form y = ym sin(kx + t) and the other to be of the form y = ym si
57. Recalling the discussion in section 16-12, we observe that this problem presents us with a standing wave condition with amplitude 12 cm. The angular wave number and frequency are noted by comparing the given waves with the form y = ym sin(k x t). The
58. With the string fixed on both ends, using Eq. 16-66 and Eq. 16-26, the resonant frequencies can be written as
f=
nv n n mg = = , 2L 2L 2L
n = 1, 2,3,
(a) The mass that allows the oscillator to set up the 4th harmonic ( n = 4 ) on the string is
m=
4 L
59. (a) The frequency of the wave is the same for both sections of the wire. The wave speed and wavelength, however, are both different in different sections. Suppose there are n1 loops in the aluminum section of the wire. Then,
L1 = n11/2 = n1v1/2f,
wher
60. With the string fixed on both ends, using Eq. 16-66 and Eq. 16-26, the resonant frequencies can be written as
f=
nv n n mg = = , 2L 2L 2L
n = 1, 2,3,
The mass that allows the oscillator to set up the nth harmonic on the string is
m= 4 L2 f 2 . n2 g
T
48. Using Eq. 16-26, we find the wave speed to be v=
65.2 106 N = = 4412 m/s. 3.35 kg/ m
The corresponding resonant frequencies are fn = nv n = , 2L 2L n = 1, 2,3,
(a) The wavelength of the wave with the lowest (fundamental) resonant frequency f1 is 1 =
47. (a) The resonant wavelengths are given by = 2L/n, where L is the length of the string and n is an integer, and the resonant frequencies are given by f = v/ = nv/2L, where v is the wave speed. Suppose the lower frequency is associated with the integer
46. Use Eq. 1666 (for the resonant frequencies) and Eq. 1626 (v = / ) to find fn: fn = which gives f3 = (3/2L) i . (a) When f = 4i, we get the new frequency f3 = 3 f = 2 f3 . 2L
v 2 L = = 3. f3 3
nv n = 2L 2L
(b) And we get the new wavelength 3 =
71. (a) Hookes law and the work done by a spring is discussed in the chapter. Taking absolute values, and writing that law in terms of differences F and x , we analyze the first two pictures as follows: | F | = k | x| 240 N 110 N = k (60 mm 40 mm) which y
112. (a) We proceed by dividing the (velocity) equation involving the new (fundamental) frequency f by the equation when the frequency f is 440 Hz to obtain f / = / f f = f
where we are making an assumption that the mass-per-unit-length of the string does
72. (a) Using Eq. 7-8 and SI units, we find W = F d = (2 4 (8 + c = 16 4c i j) i j) which, if equal zero, implies c = 16/4 = 4 m. (b) If W > 0 then 16 > 4c, which implies c < 4 m. (c) If W < 0 then 16 < 4c, which implies c > 4 m.
Create assignment, 20188, Homework 6, May 01 at 10:07 am This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. Please remember to carry out
73. A convenient approach is provided by Eq. 7-48. P = F v = (1800 kg + 4500 kg)(9.8 m/s2)(3.80 m/s) = 235 kW. Note that we have set the applied force equal to the weight in order to maintain constant velocity (zero acceleration).
74. (a) The component of the force of gravity exerted on the ice block (of mass m) along the incline is mg sin , where = sin 1 0.91 15 gives the angle of inclination for the . inclined plane. Since the ice block slides down with uniform velocity, the work
75. (a) The plot of the function (with SI units understood) is shown below.
Estimating the area under the curve allows for a range of answers. Estimates from 11 J to 14 J are typical. (b) Evaluating the work analytically (using Eq. 7-32), we have W=
2 0
1
76. (a) Eq. 7-10 (along with Eq. 7-1 and Eq. 7-7) leads to vf = (2 m F cos )1/2= (cos )1/2, where we have substituted F = 2.0 N, m = 4.0 kg and d = 1.0 m. (b) With vi = 1, those same steps lead to vf = (1 + cos )1/2. (c) Replacing with 180 , and still usi
which (by taking two derivatives) we find the acceleration to be a = 0.20 m/s2. The (constant) force is therefore F = ma = 0.40 N, with a corresponding work given by W = 2 Fx = 50 t(t 10). It also follows from the x expression that vo = 1.0 m/s. This mean
78. The problem indicates that SI units are understood, so the result (of Eq. 7-23) is in Joules. Done numerically, using features available on many modern calculators, the result is roughly 0.47 J. For the interested student it might be worthwhile to quo
79. (a) To estimate the area under the curve between x = 1 m and x = 3 m (which should yield the value for the work done), one can try counting squares (or half-squares or thirds of squares) between the curve and the axis. Estimates between 5 J and 8 J ar