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Unformatted text preview: 6. Setting x = 0 in u = −ω ym cos(k x − ω t + φ) (see Eq. 1621 or Eq. 1628) gives
u = −ω ym cos(−ω t+φ)
as the function being plotted in the graph. We note that it has a positive “slope”
(referring to its tderivative) at t = 0:
du
dt = d (−ω ym cos(−ω t+
dt φ)) = − ym ω² sin(−ω t + φ) > 0 at t = 0. This implies that – sinφ > 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) umax = 5 m/s. We
note that umax = ym ω. Therefore,
u = − umax cos(− ω t + φ) t = 0 φ = cos−1( 4 ) = ± 0.6435 rad
5 (bear in mind that cosθ = 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 2nπ is still a valid
answer (where n is any integer), so that, for example, φ = −0.64 + 2π = 5.64 rad is also
an acceptable result. ...
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 Spring '08
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 Physics

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