Answer:
Explanation:
A general wave function is given by:

A: amplitude of the wave = 0.075m
k: wave number
w: angular frequency
a) You use the following expressions for the calculation of k, w, T and λ:



b) Hence, the wave function is:

c) for x=3m you have:

d) the speed of the medium:

you can see the velocity of the medium for example for x = 0:

Explanation:
25 ft I believe that this is the best answer