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.