8, Nov. 2010
PHYS 2101
HW#11 WileyPlus Problem Solutions
HW #11 Solutions or at least most of them (USING BOOK VALUES 9th Edition in problems; problem # indicated by []) 1) [16-CQ4]
2 2 = = 3.49m1. 1.80m (1.80 m ) (110 rad s ) (b) The speed of the wave is v = f = = = 31.5m s. 2 2
2) [16-3] . (a) The angular wave number is k = 3) [16-5] (a) The motion from maximum displacement to zero is one-fourth of a cycle. One-fourth of a period is 0.170 s, so the period is T = 4(0.170 s) = 0.680 s.
1 1 = = 1.47 Hz. T 0.680s 1.40m (c) A sinusoidal wave travels one wavelength in one period: v = = = 2.06m s. T 0.680s
(b) The frequency is the reciprocal of the period: f = 4) [16-9] (a) The amplitude ym is half of the 6.00 mm vertical range shown in the figure, that is, ym = 3.0 mm. (b) The speed of the wave is v = d/t = 15 m/s, where d = 0.060 m and t = 0.0040 s. The angular wave 2 number is k = 2/ where = 0.40 m. Thus, k = = 16 rad/m . (c) The angular frequency is found from = k v = (16 rad/m)(15 m/s) = 2.4102 rad/s. (d) We choose the minus sign (between kx and t) in the argument of the sine function because the wave is shown traveling to the right (in the +x direction, see Section 16-5). Therefore, with SI units understood, 2 we obtain y = ym sin(kx kvt) 0.0030 sin(16 x 2.4 10 t) . 5) [16-10] (a) The amplitude is ym = 6.0 cm. (b) We find from 2/ = 0.020: = 1.0102 cm. (c) Solving 2f = = 4.0, we obtain f = 2.0 Hz. (d) The wave speed is v = f = (100 cm) (2.0 Hz) = 2.0102 cm/s. (e) The wave propagates in the x direction, since the argument of the trig function is kx + t instead of kx t (as in Eq. 16-2). (f) The maximum transverse speed (found from the time derivative of y) is umax = 2 fym = 4.0 s 1 ( 6.0 cm ) = 75cm s.
(
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(g) y(3.5 cm, 0.26 s) = (6.0 cm) sin[0.020(3.5) + 4.0(0.26)] = 2.0 cm. 6) [16-11] . From Eq. 16-10, a general expression for a sinusoidal wave traveling along the +x direction is y ( x, t ) = ym sin(kx t + ) . (a) The figure shows that at x = 0, y (0, t ) = ym sin( t + ) is a positive sine function, that is,
y (0, t ) = + ym sin t. Therefore, the phase constant must be = . At t = 0, we then have y ( x, 0) = ym sin(kx + ) = ym sin kx
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Nov. 8, 2010
PHYS 2101
HW#11 WileyPlus Problem Solutions
which is a negative sine function. A plot of y(x, 0) is depicted on the right (b) From the figure we see that the amplitude is ym = 4.0 cm. (c) The angular wave number is given by k = 2/ = /10 = 0.31 rad/cm. (d) The angular frequency is = 2/T = /5 = 0.63 rad/s. (e) As found in part (a), the phase is = . (f) The sign is minus since the wave is traveling in the +x direction. (g) Since the frequency is f = 1/T = 0.10 s, the speed of the wave is v = f = 2.0 cm/s. (h) From the results above, the wave may be expressed as
Taking the derivative of y with respect to t, we find y x t u ( x, t ) = = 4.0 cos which yields u(0, 5.0) = 2.5 cm/s. t t 10 5 7) [16-12] With length in centimeters and time in seconds, we have u = Squaring this and adding it to the square of 15y, we have u2 + (15y)2 = (225 )2 [sin2 (x 15 t) + cos2 (x 15 t)] so that u = (225 ) 2 (15 y ) 2 = 15 152 y 2 . Therefore, where y = 12, u must be 135. Consequently, the speed there is 424 cm/s = 4.24 m/s. 8) [16-14] (a) Comparing with Eq. 16-2, we see that k = 20/m and = 600/s. Therefore, the speed of the wave is (see Eq. 16-13) v = /k = 30 m/s. 15 (b) From Eq. 1626, we find 2 = = 2 = 0.017 kg m = 17 g m. v 30 9) [16-23] . (a) We read the amplitude from the graph. It is about 5.0 cm. (b) We read the wavelength from the graph. The curve crosses y = 0 at about x = 15 cm and again with the same slope at about x = 55 cm, so = (55 cm 15 cm) = 40 cm = 0.40 m. (c) The wave speed is v = / , where is the tension in the string and is the linear mass density of du = 225 sin (x 15t) . dt
x t x t y( x, t ) = 4.0sin + = 4.0sin . 10 5 10 5
3.6 N = 12 m/s. 25 103 kg/m (d) The frequency is f = v/ = (12 m/s)/(0.40 m) = 30 Hz and the period is T = 1/f = 1/(30 Hz) = 0.033 s. (e) The maximum string speed is um = ym = 2fym = 2(30 Hz) (5.0 cm) = 940 cm/s = 9.4 m/s. (f) The angular wave number is k = 2/ = 2/(0.40 m) = 16 m1. (g) The angular frequency is = 2f = 2(30 Hz) = 1.9102 rad/s. (h) According to the graph, the displacement at x = 0 and t = 0 is 4.0 102 m. The formula for the displacement gives y(0, 0) = ym sin . We wish to select so that 5.0 102 sin = 4.0 102.
the string. Thus, v =
2
Nov. 8, 2010
PHYS 2101
HW#11 WileyPlus Problem Solutions
The solution is either 0.93 rad or 2.21 rad. In the first case the function has a positive slope at x = 0 and matches the graph. In the second case it has negative slope and does not match the graph. We select = 0.93 rad. (i) The string displacement has the form y (x, t) = ym sin(kx + t + ). A plus sign appears in the argument of the trigonometric function because the wave is moving in the negative x direction. Using the results obtained above, the expression for the displacement is y ( x, t ) = 5.0 102 m sin (16 m 1 ) x + (190s 1 )t + 0.93 .
(
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10) [16-32] . (a) Let the phase difference be . Then from Eq. 16-52, 2ym cos(/2) = 1.50ym, which gives 1.50 ym = 2 cos 1 = 82.8. 2 ym (b) Converting to radians, we have = 1.45 rad. (c) In terms of wavelength (the length of each cycle, where each cycle corresponds to 2 rad), this is equivalent to 1.45 rad/2 = 0.230 wavelength. 11) [16-41] . (a) The wave speed is given by v = , where is the tension in the string and is the linear mass density of the string. Since the mass density is the mass per unit length, = M/L, where M is L (96.0 N) (8.40 m) the mass of the string and L is its length. Thus v = = = 82.0 m/s. M 0.120 kg (b) The longest possible wavelength for a standing wave is related to the length of the string by L = /2, so = 2L = 2(8.40 m) = 16.8 m. (c) The frequency is f = v/ = (82.0 m/s)/(16.8 m) = 4.88 Hz. 12) [16-49] . (a) Equation 16-26 gives the speed of the wa
v=
150 N = = 144.34 m/s 1.44 102 m/s. 3 7.20 10 kg/m
(b) From the figure, we find the wavelength of the standing wave to be = (2/3)(90.0 cm) = 60.0 cm. v 1.44 102 m/s (c) The frequency is f = = = 241Hz. 0.600 m 13) [16-52] Since the rope is fixed at both ends, then the phrase second-harmonic standing wave pattern describes the oscillation shown in Figure 16-20(b), where (see Eq. 16-65) = L
and
(a) Comparing the given function with Eq. 16-60, we obtain k = /2 and = 12 rad/s. Since k = 2/,
v f= . L
2 = = 4.0m L = 4.0m. 2 (b) Since = 2f, then 2 f = 12 rad/s, which yields f = 6.0 Hz
then (c) Using Eq. 16-26, we have
v = f = 24 m/s.
v=
200 N which leads to m = 1.4 kg. 24 m/s = m /(4.0 m)
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Nov. 8, 2010
PHYS 2101
HW#11 WileyPlus Problem Solutions
(d) With
f=
3v 3(24 m/s) = = 9.0 Hz the period is T = 1/f = 0.11 s. 2 L 2(4.0 m)
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