This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Shoreline CC PHYS 122 Homework #4 Solution Chap 16, Problems 1, 8, 9 ,13, 22 •1 A wave has an angular frequency of 110 rad/s and a wavelength of 1.80 m. Calculate (a) the angular wave number and (b) the speed of the wave. 2. (a) The angular wave number is 1 2 2 3.49m . 1.80m k- π π = = = λ (b) The speed of the wave is ( 29 ( 29 1.80m 110rad s 31.5m s. 2 2 v f ϖ λ = λ = = = π π ••8 The equation of a transverse wave traveling along a very long string is y = 6.0 sin(0.020 π x + 4.0 π t ), where x and y are expressed in centimeters and t is in seconds. Determine (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the speed, (e) the direction of propagation of the wave, and (f) the maximum transverse speed of a particle in the string. (g) What is the transverse displacement at x = 3.5 cm when t = 0.26 s? 8. (a) The amplitude is y m = 6.0 cm. (b) We find λ from 2 π / λ = 0.020 π : λ = 1.0×10 2 cm. (c) Solving 2 π f = ϖ = 4.0 π , we obtain f = 2.0 Hz. (d) The wave speed is v = λ f = (100 cm) (2.0 Hz) = 2.0×10 2 cm/s. (e) The wave propagates in the – x direction, since the argument of the trig function is kx + ϖ t instead of kx – ϖ t (as in Eq. 16-2). (f) The maximum transverse speed (found from the time derivative of y ) is ( 29 ( 29 1 max 2 4.0 s 6.0cm 75cm s. m u fy- = π = π = (g) y (3.5 cm, 0.26 s) = (6.0 cm) sin[0.020 π (3.5) + 4.0 π (0.26)] = –2.0 cm. Shoreline CC PHYS 122 ••9 A transverse sinusoidal wave is moving along a string in the positive direction of an...
View Full Document
- Winter '08
- Work, 20 cm, 1.3 m, 4.0 G, 1 2 k, 31.5 m