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Problem_Set_11_Solutions

# Problem_Set_11_Solutions - 10 With length in centimeters...

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10. With length in centimeters and time in seconds, we have u = du dt = 225 π sin ( π x 15 π t ) . Squaring this and adding it to the square of 15 π y , we have u 2 + (15 π y ) 2 = (225 π ) 2 [sin 2 ( π x 15 π t ) + cos 2 ( π x 15 π t )] so that u = (225 π ) 2 - (15 π y ) 2 = 15 π 15 2 - y 2 . Therefore, where y = 12, u must be ± 135 π . Consequently, the speed there is 424 cm/s = 4.24 m/s.

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24. (a) The tension in each string is given by τ = Mg /2. Thus, the wave speed in string 1 is 2 1 1 1 (500g)(9.80m/s ) 28.6m/s. 2 2(3.00g/m) Mg v τ μ μ = = = = (b) And the wave speed in string 2 is 2 2 2 (500g)(9.80m/s ) 22.1m/s. 2 2(5.00g/m) Mg v μ = = = (c) Let 1 1 1 2 2 2 /(2 ) /(2 ) v M g v M g = = = μ μ and M 1 + M 2 = M . We solve for M 1 and obtain 1 2 1 500g 187.5g 188g. 1 / 1 5.00/3.00 M M μ μ = = = + + (d) And we solve for the second mass: M 2 = M M 1 = (500 g – 187.5 g) 313 g.
30. The wave 1 1 ( , ) (4.00 mm) [(30 m ) (6.0 s ) ] y x t h x t = + is of the form ( ) h kx t ω with angular wave number 1 30 m k = and angular frequency 6.0 rad/s ω = . Thus, the speed of the wave is 1 / (6.0 rad/s)/(30 m ) 0.20 m/s. v k ω = = =

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38. We see that y 1 and y 3 cancel (they are 180º) out of phase, and y 2 cancels with y 4 because their phase difference is also equal to π rad (180º). There is no resultant wave in this case.
and = v L f L λ = . (a) Comparing the given function with Eq. 16-60, we obtain k = π /2 and ω = 12 π rad/s. Since k = 2 π / λ then 2 4.0m 4.0m. 2 L π π = ¡ λ = ¡ = λ (b) Since ω = 2 π f then 2 12 rad/s, f π = π which yields 6.0Hz 24m/s. f v f = ¡ = λ = (c) Using Eq. 16–26, we have 200 N 24 m/s /(4.0 m) v m τ μ = ¡ = which leads to m = 1.4 kg.

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