position in cm, time in sec
\(\displaystyle y = 3\sin(2x-5t)\)
transverse velocity in cm/sec
\(\displaystyle \frac{\partial y}{\partial t} = v_y = -15\cos(2x-5t)\)
it's important to understand that as the wave passes through the string, any point on the string is displaced only in the y-direction ... displacement of the string in the x-direction is always 0. so, when the question asks for the displacement of the string, it's asking for \(\displaystyle \Delta y\) of that point. And, although not stated as such in your problem statement, I suspect that the question wants the displacement of the point on the string from its equilibrium position because it did not state specifically between any two times or any two speeds.
for any point on the string (that is, any x-value), that point has maximum speed when that point is at equilibrium. its speed is 0 when the same point is at maximum displacement (3 cm in this case) from equilibrium. so, as the point on the string travels from equilibrium to its maximum displacement the point on the string slows down ... and as it travels from maximum displacement back to equilibrium, it speeds up in the opposite direction.
now, the point on the string defined by its x-value, x = 1.0472 ... was that a given value for the problem, or did you come up with it?
reason I ask is this ... at x = 1.0472 and t = 0, the transverse velocity of the string is 7.5 cm/sec, which makes me think you figured that out yourself.
that's fine, but now calculate the position of the point x = 1.0472 at t = 0 ...
\(\displaystyle y = 3\sin(2.0944 - 0) \approx 2.6 \, cm\)
and that value is the displacement of that point on the string from equilibrium.
