Problem_Set_11_Solutions

Problem_Set_11_Solutions - 10 With length in centimeters...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
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 ) 2 [sin 2 ( x 15 t ) + cos 2 ( x 15 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.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
24. (a) The tension in each string is given by τ = Mg /2. Thus, the wave speed in string 1 is 2 1 11 (500g)(9.80m/s ) 28.6m/s. 2 2(3.00g/m) Mg v μμ == = = (b) And the wave speed in string 2 is 2 2 2 22.1m/s. 2 2(5.00g/m) Mg v μ = (c) Let 111 2 22 /(2 ) /(2 ) vM g g = and M 1 + M 2 = M . We solve for M 1 and obtain 1 21 500g 187.5g 188g. 1 / 1 5.00/3.00 M M = ++ (d) And we solve for the second mass: M 2 = M M 1 = (500 g – 187.5 g) 313 g.
Background image of page 2
30. The wave 11 ( , ) (4.00 mm) [(30 m ) (6.0 s ) ] yxt h x t −− =+ is of the form ( ) hkx t ω with angular wave number 1 30 m k = and angular frequency 6.0 rad/s = . Thus, the speed of the wave is 1 / (6.0 rad/s)/(30 m ) 0.20 m/s. vk == =
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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.
Background image of page 4
and = v Lf 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 ππ = ¡ ¡ = λ (b) Since = 2 π f then 2 12 rad/s, f π= π which yields 6.0Hz 24m/s. fv f = ¡ = (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. (d) With 33 ( 2 4 m / s ) 9.0Hz 22 ( 4 . 0 m ) v f L == = The period is T = 1/ f = 0.11 s.
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 14

Problem_Set_11_Solutions - 10 With length in centimeters...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online