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# CH-16 - Chapter 16 1(a The motion from maximum displacement...

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Chapter 16 1. (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 2. (a) The angular wave number is 1 2 2 3.49m . 1.80m k - π π = = = λ (b) The speed of the wave is ( 29 ( 29 1.80m 110rad s 31.5m s. 2 2 v f ϖ λ = λ = = = π π 3. 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). 4. Setting x = 0 in u = - ϖ y m cos ( k x - ϖ t + φ ) (see Eq. 16-21 or Eq. 16-28) gives 703

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