ch16-p033

# ch16-p033 - 2 y m cos( 1 2 2 ) 2 = 2 cos 1 (2.00/9.00) =...

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33. (a) The amplitude of the second wave is 9.00 mm m y = , as stated in the problem. (b) The figure indicates that λ = 40 cm = 0.40 m, which implies that the angular wave number is k = 2 π /0.40 = 16 rad/m. (c) The figure (along with information in the problem) indicates that the speed of each wave is v = dx / t = (56.0 cm)/(8.0 ms) = 70 m/s. This, in turn, implies that the angular frequency is ω = kv =1100 rad/s = 1.1 × 10 3 rad/s. (d) The figure depicts two traveling waves (both going in the – x direction) of equal amplitude y m . The amplitude of their resultant wave, as shown in the figure, is y m = 4.00 mm. Eq. 16-52 applies: y m =
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Unformatted text preview: 2 y m cos( 1 2 2 ) 2 = 2 cos 1 (2.00/9.00) = 2.69 rad. (e) In making the plus-or-minus sign choice in y = y m sin( k x t + ), we recall the discussion in section 16-5, where it shown that sinusoidal waves traveling in the x direction are of the form y = y m sin( k x + t + ). Here, should be thought of as the phase difference between the two waves (that is, 1 = 0 for wave 1 and 2 = 2.69 rad for wave 2). In summary, the waves have the forms (with SI units understood): y 1 = (0.00900)sin(16 x +1100 t ) and y 2 = (0.00900)sin(16 x + 1100 t + 2.7 ) ....
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