Answer
$y(x,t) = (0.16~m)~sin[(1.57~m^{-1}) x]~sin [(31.4~s^{-1}) t]$
Work Step by Step
We found that the wave speed is $20~m/s$ and the amplitude is $y_m = 0.16~m$
We can find $k$:
$k = \frac{\omega}{v}$
$k = \frac{2\pi f}{v}$
$k = \frac{(2\pi)(5.0~Hz)}{20~m/s}$
$k = 1.57~m^{-1}$
Note that: $~~\omega = 2\pi f = (2\pi)(5.0 ~Hz) = 31.4~s^{-1}$
We can write the equation for the standing wave:
$y(x,t) = y_m~sin(kx)~cos(\omega t+\phi)$
$y(x,t) = (0.16~m)~sin[(1.57~m^{-1}) x]~cos [(31.4~s^{-1}) t+\phi]$
At $t = 0$, the transverse displacement of the point $x = 1.0~m$ is $y = 0$
We can find $\phi$:
$y(1.0~m,0) = (0.16~m)~sin[(1.57~m^{-1}) (1.0~m)]~cos [(31.4~s^{-1}) (0)+\phi] = 0$
$cos [0+\phi] = 0$
$\phi = \frac{\pi}{2}, \frac{3\pi}{2}$
If $t$ increases slightly more than $t = 0$, then $y \gt 0$ since the point $x = 1.0~m$ is moving in the positive direction of the y axis. Therefore, $\phi = \frac{3\pi}{2}$
We can write the equation for the standing wave:
$y(x,t) = (0.16~m)~sin[(1.57~m^{-1}) x]~cos [(31.4~s^{-1}) t+\phi]$
$y(x,t) = (0.16~m)~sin[(1.57~m^{-1}) x]~cos [(31.4~s^{-1}) t+\frac{3\pi}{2}]$
$y(x,t) = (0.16~m)~sin[(1.57~m^{-1}) x]~sin [(31.4~s^{-1}) t]$
