This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View
Full
Document
This note was uploaded on 04/09/2008 for the course PHYS 131 taught by Professor Drty during the Spring '08 term at Lafayette.
 Spring '08
 drty
 mechanics, Mass

Click to edit the document details