55. Using Eq. 17-37, we have
y
=
0.60 cos
π
6
sin
5πx
− 200πt +
π
6
with length in meters and time in seconds (see Eq. 17-42 for comparison).
(a) The amplitude is seen to be
0.60 cos
π
6
= 0.3
√
3 = 0.52 m
.
(b) Since k = 5π and ω = 200π, then (using Eq. 17-11)
v =
ω
k
= 40 m/s .
(c) k = 2π/λ leads to λ = 0.40 m.