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Unformatted text preview: Chapter 16 – Student Solutions Manual 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 = (0.0 60 0 kg)/(2.00 m) = 0.0300 kg/m. Thus, 500 N 129m s. 0.0300kg m v = = 17. (a) The amplitude of the wave is y m =0.120 mm. (b) The wave speed is given by v = τ μ , where τ is the tension in the string and μ is the linear mass density of the string, so the wavelength is λ = v / f = τ μ / f and the angular wave number is ( ) 1 2 0.50kg 2 2 100Hz 141m . 10N k f m − π = = π = π = λ μ τ (c) The frequency is f = 100 Hz, so the angular frequency is ω = 2 π f = 2 π (100 Hz) = 628 rad/s. (d) We may write the string displacement in the form y = y m sin( kx + ω t ). The plus sign is used since the wave is traveling in the negative x direction. In summary, the wave can be expressed as ( ) ( ) ( ) 1 1 0.120mm sin 141m + 628s . y x t − − ⎡ ⎤ = ⎣ ⎦ 21. (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 λ = (55 cm – 15 cm) = 40 cm = 0.40 m. (c) The wave speed is / , v = τ μ where τ is the tension in the string and μ is the linear mass density of the string. Thus, 3 3.6 N 12 m/s. 25 10 kg/m v − = = × (d) The frequency is f = v / λ = (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 u m = ω y m = 2 π fy m = 2 π (30 Hz) (5.0 cm) = 940 cm/s = 9.4 m/s. (f) The angular wave number is k = 2 π / λ = 2 π /(0.40 m) = 16 m –1 . (g) The angular frequency is ω = 2 π f = 2 π (30 Hz) = 1.9×10 2 rad/s (h) According to the graph, the displacement at x = 0 and t = 0 is 4.0 × 10 –2 m. The formula for the displacement gives y (0, 0) = y m sin φ . We wish to select φ so that 5.0 × 10 –2 sin φ = 4.0 × 10 –2 . 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 φ = 0.93 rad....
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