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58. We use Eq. 172, Eq. 175, Eq. 179, Eq. 1712, and take the derivative to obtain the transverse speed
u
.
(a) The amplitude is
y
m
=2
.
0 mm.
(b) Since
ω
= 600 rad/s, the frequency is found to be
f
= 600
/
2
π
≈
95 Hz.
(c) Since
k
= 20 rad/m, the velocity of the wave is
v
=
ω/k
= 600
/
20 = 30 m/s in the
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Unformatted text preview: + x direction. (d) The wavelength is Î» = 2 Ï€/k â‰ˆ . 31 m, or 31 cm. (e) We obtain u = dy dt = âˆ’ Ï‰y m cos( kx âˆ’ Ï‰t ) = â‡’ u m = Ï‰y m so that the maximum transverse speed is u m = (600)(2 . 0) = 1200 mm/s, or 1 . 2 m/s....
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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
 SPRUNGER
 Physics

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