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

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Unformatted text preview: 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 = = = 2. (a) The angular wave number is 1 2 2 3.49m . 1.80m k- = = = (b) The speed of the wave is ( 29 ( 29 1.80m 110rad s 31.5m s. 2 2 v f = = = = 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 + = sin- 1 (2.0/6.0) and kx + 600 t 2 + = 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 = - y m cos ( k x- t + ) (see Eq. 16-21 or Eq. 16-28) gives 703 CHAPTER 16 u = - y m cos (- t + ) as the function being plotted in the graph. We note that it has a positive slope (referring to its t-derivative) at t = 0: = = - y m sin (- t + ) > 0 at t = 0 . This implies that sin > 0 and consequently that 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 . Therefore, u = - u max cos (- t + ) | t = 0 = cos- 1 ( 29 = 0.6435 rad (bear in mind that cos = cos(- ) ), and we must choose = - 0.64 rad (since this is about - 37 and is in fourth quadrant). Of course, this answer added to 2n is still a valid answer (where n is any integer), so that, for example, = - 0.64 + 2 = 5 .64 rad is also an acceptable result. 5. Using v = f , we find the length of one cycle of the wave is = 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 radians, so that /3 rad corresponds to one-sixth of a cycle. The corresponding length, therefore, is /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 rad. Thus, the phase difference is (1/2)2 = rad. 6. (a) The amplitude is y m = 6.0 cm. (b) We find from 2 / = 0.020 : = 1.010 2 cm. (c) Solving 2 f = = 4.0 , we obtain f = 2.0 Hz. (d) The wave speed is v = f = (100 cm) (2.0 Hz) = 2.010 2 cm/s. (e) The wave propagates in the x direction, since the argument of the trig function is kx + t instead of kx t (as in Eq. 16-2). (f) The maximum transverse speed (found from the time derivative of y ) is ( 29 ( 29 1 max 2 4.0 s 6.0cm 75cm s....
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CH-16 - Chapter 16 1. (a) The motion from maximum...

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