# The initial conditions for the harmonic travelling waves

I'm given the following problem:

For a harmonic travelling wave travelling in the positive direction, the initial transverse speed is 3 m/s and the initial $y$ displacement or transverse displacement is 0.1m. If the frequency of oscillation is 10 Hz, find amplitude and $\phi$ constant.

It's not a difficult problem. However, I do not understand initial speed and initial y displacement conditions.
Namely, I don't understand the following:

The general wave formula is:
y(x,t) = Asin(kx - wt + phi) and if initial "y" displacement is 0.10, then
y(0, 0) = 0.1 and then 0.1 = Asin(phi) and
u(0, 0) = 3 and 3 = -Acos(phi).

How do we know that x in these formulas will be zero?

And could someone explain me, why when we derive y(x,y) to get u(x,t), we must keep x as a constant ?

$y$ the displacement of a particle from its equilibrium position, at a position $x$ front the origin/source, at a time $t$.
The word "initial" is used twice in the formulation of the problem and without any further information you can interpret that as $x=0$ and $t=0$.
So you have been given two piece of information about a particle at the origin $x=0$ at a time $t=0 which is undergoing simple harmonic motion so that you can find the two unknowns: the amplitude and the phase. When you differentiate$y$with respect to$t$you are evaluating the velocity of a particle at a fixed position$x\$.