(N/A) The equation of a travelling wave propagating along the positive $x$-direction is given by $y(x, t) = a \sin(\omega t - kx + \phi)$.
Given $y(0, 0) = 0$,we have $a \sin(\phi) = 0$,so $\phi = 0$.
Linear mass density $\mu = 8.0 \times 10^{-3} \; kg \; m^{-1}$.
Frequency $v = 256 \; Hz$.
Amplitude $a = 5.0 \; cm = 0.05 \; m$.
Tension $T = mg = 90 \times 9.8 = 882 \; N$.
Wave velocity $v = \sqrt{T/\mu} = \sqrt{882 / (8.0 \times 10^{-3})} = \sqrt{110250} \approx 332 \; m/s$.
Angular frequency $\omega = 2\pi v = 2 \times 3.14159 \times 256 \approx 1608 \; rad/s$.
Propagation constant $k = \omega / v = 1608 / 332 \approx 4.84 \; m^{-1}$.
Substituting these values,the displacement equation is $y(x, t) = 0.05 \sin(1608t - 4.84x) \; m$.