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Unformatted text preview: 8. (a) The amplitude is ym = 6.0 cm.
(b) We find λ from 2π/λ = 0.020π: λ = 1.0×102 cm.
(c) Solving 2πf = ω = 4.0π, we obtain f = 2.0 Hz.
(d) The wave speed is v = λf = (100 cm) (2.0 Hz) = 2.0×102 cm/s.
(e) The wave propagates in the –x direction, since the argument of the trig function is kx
+ ωt instead of kx – ωt (as in Eq. 16-2).
(f) The maximum transverse speed (found from the time derivative of y) is ( ) umax = 2π fym = 4.0 π s −1 ( 6.0 cm ) = 75cm s.
(g) y(3.5 cm, 0.26 s) = (6.0 cm) sin[0.020π(3.5) + 4.0π(0.26)] = –2.0 cm. ...
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- Spring '08