Answer
$$D_{(x,t)}=({0.01\;\rm mm})\sin\left[ \pi x-400\pi t +\frac{\pi}{2} \right]$$
Work Step by Step
We know that the displacement equation for a sinusoidal wave
$$D_{(x,t)}=A\sin\left[ \dfrac{2\pi }{\lambda}x- 2\pi f t +\phi_0 \right]$$
And we know that $v=400$ m/s, $f=200$ Hz, $A=0.01$ mm, and $\phi_0=\dfrac{\pi}{2}$
So we have to find $\lambda$ which is given by
$$\lambda=\dfrac{v}{f}=\dfrac{400}{200}=2\;\rm m$$
Thus,
$$\boxed{D_{(x,t)}=({0.01\;\rm mm})\sin\left[ \pi x-400\pi t +\frac{\pi}{2} \right]}$$
It is moving in the positive $x$-direction so the original signs remain.
