P17_059.PDF

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Chapter 17 - 17.59
59. (a) Recalling the discussion in
§
17-5, we see that the speed of the wave given by a function with
5 t (where x is in centimeters and t is in seconds) must be 5 cm/s.
(b) In part (c), we show several “snapshots” of the wave: the one on the left is as shown in Figure
17-47 (at t = 0), the middle one is at t =1 . 0 s, and the rightmost one is at t =2 . 0s. Itisclear
that the wave is traveling to the right (the + x direction).
(c) The third picture in the sequence below shows the pulse at 2 s. The horizontal scale (and, presum-
ably, the vertical one also) is in centimeters.
−
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
2
4
6
8
10
12
14
4) / 5=1 . 2 s. The particle (say,
of the string that carries the pulse) at that location reaches a maximum displacement h =2cm
at t =(10
−
−
3) / 5=1 . 4 s. Finally, the the trailing edge of the pulse departs from x =10cmat
t =(10
−
1) / 5=1 . 8 s. Thus, we find for h ( t )at x = 10 cm (with the horizontal axis, t , in seconds):
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
1.2
1.3
1.4
1.5
1.6
1.7
1.8
argument x
(d) The leading edge of the pulse reaches x =10cmat t =(10
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