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Chapter-16 - Chapter 16 1(a The angular wave number is k =...

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693 Chapter 16 1. (a) The angular wave number is 1 2 2 3.49m . 1.80m k - π π = = = λ (b) The speed of the wave is ( ( 1.80m 110rad s 31.5m s. 2 2 v f ϖ λ = λ = = = π π 2. The distance d between the beetle and the scorpion is related to the transverse speed t v and longitudinal speed v l as t t d v t v t = = l l where t t and t l are the arrival times of the wave in the transverse and longitudinal directions, respectively. With 50 m/s t v = and 150 m/s v = l , we have 150 m/s 3.0 50 m/s t t t v t v = = = l l . Thus, if 3 3 3.0 2.0 4.0 10 s 2.0 10 s , t t t t t t t t - - ∆ = - = - = = × = × l l l l l then 3 (150 m/s)(2.0 10 s) 0.30 m 30 cm. d v t - = = × = = l l 3. (a) The motion from maximum displacement to zero is one-fourth of a cycle so 0.170 s is one-fourth of a period. The period is T = 4(0.170 s) = 0.680 s. (b) The frequency is the reciprocal of the period: 1 1 1.47Hz. 0.680s f T = = = (c) A sinusoidal wave travels one wavelength in one period: 1.40m 2.06m s. 0.680s v T = = = λ 4. (a) The speed of the wave is the distance divided by the required time. Thus,
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CHAPTER 16 694 853 seats 21.87 seats/s 22 seats/s 39 s v = = . (b) The width w is equal to the distance the wave has moved during the average time required by a spectator to stand and then sit. Thus, (21.87 seats/s)(1.8 s) 39 seats w vt = = . 5. Let y 1 = 2.0 mm (corresponding to time t 1 ) and y 2 = –2.0 mm (corresponding to time t 2 ). Then we find kx + 600 t 1 + φ = sin - 1 (2.0/6.0) and kx + 600 t 2 + φ = sin - 1 (–2.0/6.0) . Subtracting equations gives 600( t 1 t 2 ) = sin - 1 (2.0/6.0) – sin - 1 (–2.0/6.0). Thus we find t 1 t 2 = 0.011 s (or 1.1 ms). 6. Setting x = 0 in u = - ϖ y m cos ( k x - ϖ t + φ ) (see Eq. 16-21 or Eq. 16-28) gives u = - ϖ y m cos ( - ϖ t+ φ ) as the function being plotted in the graph. We note that it has a positive “slope” (referring to its t -derivative) at t = 0: d u d t = d (-ϖ y m cos (-ϖ t+ φ 29 d t = - y m ϖ ² sin ( - ϖ t + φ ) > 0 at t = 0 . This implies that – sin φ > 0 and consequently that φ is in either the third or fourth quadrant. The graph shows (at t = 0) u = - 4 m/s, and (at some later t ) u max = 5 m/s. We note that u max = y m ϖ . Therefore, u = - u max cos ( - ϖ t + φ ) | t = 0 φ = cos - 1 ( 4 5 29 = ± 0.6435 rad ( bear in mind that cos θ = cos( - θ ) ) , and we must choose φ = - 0.64 rad (since this is about - 37° and is in fourth quadrant). Of course, this answer added to 2n π is still a valid answer (where n is any integer), so that, for example, φ = - 0.64 + 2 π = 5 .64 rad is also an acceptable result.
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695 7. Using v = f λ , we find the length of one cycle of the wave is λ = 350/500 = 0.700 m = 700 mm. From f = 1/ T , we find the time for one cycle of oscillation is T = 1/500 = 2.00 × 10 –3 s = 2.00 ms. (a) A cycle is equivalent to 2 π radians, so that π /3 rad corresponds to one-sixth of a cycle. The corresponding length, therefore, is λ /6 = 700/6 = 117 mm. (b) The interval 1.00 ms is half of T and thus corresponds to half of one cycle, or half of 2 π rad. Thus, the phase difference is (1/2)2 π = π rad. 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.
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