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Unformatted text preview: 6. (a) The amplitude is ym = 6.0 cm.
(b) We ﬁnd λ from 2π/λ = 0.020π : λ = 100 cm.
(c) Solving 2πf = ω = 4.0π , we obtainf = 2.0 Hz.
(d) The wavespeed is v = λf = (100 cm)(2.0 Hz) = 200 cm/s.
(e) The wave propagates in the negative x direction, since the argument of the trig function is kx + ωt
instead of kx − ωt (as in Eq. 17-2).
(f) The maximum transverse speed (found from the time derivative of y ) is
umax = 2πf ym = 4.0π s−1 (6.0 cm) = 75 cm/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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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
- Fall '08