Chapter 16 Solutions_8

Chapter 16 Solutions_8 - Chapter 16 3(a The motion from...

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Unformatted text preview: Chapter 16 3. (a) The motion from maximum displacement to zero is one-fourth of a cycle so 0.170 s is onefourth of a period. The period is T = 4(0.170 s) = 0.680 s. (b) The frequency is the reciprocal of the period: f= 1 1 = = 1.47 Hz. T 0.680s (c) A sinusoidal wave travels one wavelength in one period: v= λ T = 1.40 m = 2.06 m s. 0.680s 7. Using v = fλ, we find the length of one cycle of the wave is λ = 350/500 = 0.700 m = 700 mm. From f = 1/T, we find the time for one cycle of oscillation is T = 1/500 = 2.00 × 10–3 s = 2.00 ms. (a) A cycle is equivalent to 2π radians, so that π/3 rad corresponds to one‐sixth of a cycle. The corresponding length, therefore, is λ/6 = 700/6 = 117 mm. (b) The interval 1.00 ms is half of T and thus corresponds to half of one cycle, or half of 2π rad. Thus, the phase difference is (1/2)2π = π rad. 8. (a) The amplitude is ym = 6.0 cm. (b) We find λ from 2π/λ = 0.020π: λ = 1.0×102 cm. (c) Solving 2πf = ω = 4.0π, we obtain f = 2.0 Hz. (d) The wave speed is v = λf = (100 cm) (2.0 Hz) = 2.0×102 cm/s. (e) The wave propagates in the –x direction, since the argument of the trig function is kx + ωt instead of kx – ωt (as in Eq. 16‐2). (f) The maximum transverse speed (found from the time derivative of y) is umax = 2π fym = 4.0 π s −1 ( 6.0 cm ) = 75cm s. (g) y(3.5 cm, 0.26 s) = (6.0 cm) sin[0.020π(3.5) + 4.0π(0.26)] = –2.0 cm. 15. The wave speed v is given by v = τ μ , where τ is the tension in the rope and μ is the linear mass density of the rope. The linear mass density is the mass per unit length of rope: ( ) μ = m/L = (0.0600 kg)/(2.00 m) = 0.0300 kg/m. Thus, v= 500 N = 129 m s. 0.0300 kg m 20. (a) Comparing with Eq. 16‐2, we see that k = 20/m and ω = 600/s. Therefore, the speed of the wave is (see Eq. 16‐13) v = ω/k = 30 m/s. (b) From Eq. 16–26, we find μ= τ v 2 = 15 = 0.017 kg m = 17 g m. 302 26. Using Eq. 16–33 for the average power and Eq. 16–26 for the speed of the wave, we solve for f = ω/2π: f= 1 2πym 2 Pavg = 1 2π(7.70 × 10−3 m) 2(85.0 W) = 198 Hz. (36.0 N) (0.260 kg / 2.70 m ) μ τ /μ 28. Comparing y ( x, t ) = (3.00 mm)sin[(4.00 m −1 ) x − (7.00 s −1 )t ] to the general expression y ( x, t ) = ym sin(kx − ω t ) , we see that k = 4.00 m −1 and ω = 7.00 rad/s . The speed of the wave is v = ω / k = (7.00 rad/s)/(4.00 m −1 ) = 1.75 m/s. 32. (a) Let the phase difference be φ. Then from Eq. 16–52, 2ym cos(φ/2) = 1.50ym, which gives φ = 2 cos −1 ⎜ ⎛ 1.50 ym ⎞ ⎟ = 82.8°. ⎝ 2 ym ⎠ (b) Converting to radians, we have φ = 1.45 rad. (c) In terms of wavelength (the length of each cycle, where each cycle corresponds to 2π rad), this is equivalent to 1.45 rad/2π = 0.230 wavelength. 41. Possible wavelengths are given by λ = 2L/n, where L is the length of the wire and n is an integer. The corresponding frequencies are given by f = v/λ = nv/2L, where v is the wave speed. The wave speed is given by v = τ μ = τ L / M , where τ is the tension in the wire, μ is the linear mass density of the wire, and M is the mass of the wire. μ = M/L was used to obtain the last form. Thus fn = n τL n = 2L M 2 τ LM = n 250 N = n (7.91 Hz). 2 (10.0 m) (0.100 kg) (a) The lowest frequency is f1 = 7.91 Hz. (b) The second lowest frequency is f 2 = 2(7.91 Hz) = 15.8 Hz. (c) The third lowest frequency is f3 = 3(7.91 Hz) = 23.7 Hz. 50. Since the rope is fixed at both ends, then the phrase “second‐harmonic standing wave pattern” describes the oscillation shown in Figure 16–23(b), where (see Eq. 16–65) λ=L and f= v . L (a) Comparing the given function with Eq. 16‐60, we obtain k = π/2 and ω = 12π rad/s. Since k = 2π/λ then 2π π = ⇒ λ = 4.0 m λ2 (b) Since ω = 2πf then 2 πf = 12π rad/s, which yields ⇒ L = 4.0 m. f = 6.0 Hz ⇒ (c) Using Eq. 16–26, we have v = f λ = 24 m/s. v= which leads to m = 1.4 kg. (d) With τ 200 N ⇒ 24 m/s = m /(4.0 m) μ f= The period is T = 1/f = 0.11 s. 3v 3(24 m/s) = = 9.0 Hz 2 L 2(4.0 m) 63. We compare the resultant wave given with the standard expression (Eq. 16–52) to obtain k = 20 m −1 = 2π / λ, 2 ym cos ( 1 φ ) = 3.0 mm , and 1 φ = 0.820 rad . 2 2 (a) Therefore, λ = 2π/k = 0.31 m. (b) The phase difference is φ = 1.64 rad. (c) And the amplitude is ym = 2.2 mm. 70. We write the expression for the displacement in the form y (x, t) = ym sin(kx – ωt). (a) The amplitude is ym = 2.0 cm = 0.020 m, as given in the problem. (b) The angular wave number k is k = 2π/λ = 2π/(0.10 m) = 63 m–1 (c) The angular frequency is ω = 2πf = 2π(400 Hz) = 2510 rad/s = 2.5×103 rad/s. (d) A minus sign is used before the ωt term in the argument of the sine function because the wave is traveling in the positive x direction. Using the results above, the wave may be written as y ( x, t ) = ( 2.00 cm ) sin ( 62.8 m −1 ) x − ( 2510s −1 ) t . (e) The (transverse) speed of a point on the cord is given by taking the derivative of y: ( ) u ( x, t ) = ∂y = −ω ym cos ( kx − ω t ) ∂t which leads to a maximum speed of um = ωym = (2510 rad/s)(0.020 m) = 50 m/s. (f) The speed of the wave is v= λ ω 2510 rad s == = 40 m s. T k 62.8 rad/m ...
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This note was uploaded on 06/11/2010 for the course PHYSICS 207 taught by Professor Ellis during the Spring '10 term at SUNY Buffalo.

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