Hw 10 Solutions

# D the string speed is given by u x t y t ω y m sin

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(d) The string speed is given by u ( x, t ) = y / t = – ω y m sin( kx )sin( ω t ). For the given coordinate and time, 53. (a) The angular frequency is ω = 8.00 π /2 = 4.00 π rad/s, so the frequency is f = ω /2 π = (4.00 π rad/s)/2 π = 2.00 Hz. (b) The angular wave number is k = 2.00 π /2 = 1.00 π m –1 , so the wavelength is λ = 2 π / k = 2 π /(1.00 π m –1 ) = 2.00 m. (c) The wave speed is

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(d) We need to add two cosine functions. First convert them to sine functions using cos α = sin ( α + π /2), then apply Letting α = kx and β = ω t , we find Nodes occur where cos( kx ) = 0 or kx = n π + π /2, where n is an integer (including zero). Since k = 1.0 π m –1 , this means . Thus, the smallest value of x which corresponds to a node is x = 0.500 m ( n =0). (e) The second smallest value of x which corresponds to a node is x = 1.50 m ( n =1). (f) The third smallest value of x which corresponds to a node is x = 2.50 m ( n =2). (g) The displacement is a maximum where cos( kx ) = ± 1. This means kx = n π , where n is an integer. Thus, x = n (1.00 m). The smallest value of x which corresponds to an anti- node (maximum) is x = 0 ( n =0). (h) The second smallest value of x which corresponds to an anti-node (maximum) is ( n =1). (i) The third smallest value of x which corresponds to an anti-node (maximum) is ( n =2). 79. We use Eq. 16-2, Eq. 16-5, Eq. 16-9, Eq. 16-13, and take the derivative to obtain the transverse speed u . (a) The amplitude is y m = 2.0 mm. (b) Since ω = 600 rad/s, the frequency is found to be f = 600/2 π 95 Hz.
(c) Since k = 20 rad/m, the velocity of the wave is v = ω / k = 600/20 = 30 m/s in the + x direction. (d) The wavelength is λ = 2 π / k 0.31 m, or 31 cm. (e) We obtain so that the maximum transverse speed is u m = (600)(2.0) = 1200 mm/s, or 1.2 m/s. 16. We use to obtain 19. (a) We read the amplitude from the graph. It is about 5.0 cm.

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