This preview shows pages 1–2. 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: PHYS-1200 PHYSICS II SPRING 2006 HOMEWORK SOLUTIONS CLASS 14. 16.06, 16.19, 16.89; Q17.01, Q33.02 16.06. y = (6.0 cm) sin [(0.020 π cm-1 ) x + (4.0 π s-1 ) t ], and we know, y = y m sin ( kx ± ϖ t ). a) By inspection, y m = 6.0 cm ¯¯¯¯¯¯¯¯¯¯ Hz . 2 2 s . 4 2 c) cm 100 cm 020 . 2 2 b) 1- 1- = = = = = = f f k π π π ϖ λ π π π λ d) v = f λ = (2.0 Hz)(100 cm) v = 200 cm/s e) The + sign in the expression tells us that the wave is going in the – x direction f) Taking the derivative of the expression for the displacement of the wave with respect to time gives the speed of a particle in the string. Then, u y t y kx t y m m = = ± ∂ ∂ ϖ ϖ ϖ cos( ). The maximum value is . Then, u m = ϖ y m = (4.0 π s-1 )(6.0 cm) u m = 24 π cm/s = 75 cm/s ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ g) y = (6.0 cm) sin [(0.020 π cm-1 )(3.5 cm) + (4.0 π s-1 )(0.26 s)] y = – 2.0 cm 16.19. a) The amplitude can be read from the graph. y m = 5.0 cm ¯¯¯¯¯¯¯¯¯¯ b) λ can be read off the graph also. can be read off the graph also....
View Full Document