Answer
See the detailed answer below.
Work Step by Step
a) We know that the wave speed in a stretched string is given by
$$v=\sqrt{\dfrac{T_S}{\mu}}$$
Plugging the known;
$$v=\sqrt{\dfrac{20}{2\times 10^{-3}}}=\color{red}{\bf 100}\;\rm m/s$$
The wavelength is given by
$$v=\lambda d$$
$$\lambda=\dfrac{v}{f}=\dfrac{100}{100}=\color{red}{\bf 1.0}\;\rm m$$
b) The amplitude is the maximum displacement, so
$$A=\color{red}{\bf 1.0}\;\rm mm$$
And the phase constant is found at $x=0$, and $t=0$, when $D=-A$
$$D_{(x,t)}=A\sin\left[ \dfrac{2\pi x}{\lambda}-\dfrac{2\pi t}{T}+\phi_0 \right]$$
And at $t=0$, $x=0$,
$$D_{(0,0)}=-A= A\sin\phi_0 $$
Hence,
$$\phi_0\sin^{-1}(-1)=\color{red}{\bf -\dfrac{\pi}{2}}\;\rm rad$$
c)
We know that the displacement equation for a sinusoidal wave
$$D_{(x,t)}=A\sin\left[ \dfrac{2\pi x}{\lambda}- 2\pi f t +\phi_0 \right]$$
Plugging the known;
$$\boxed{D_{(x,t)}=(1.0{\;\rm mm})\sin\left[ 2\pi x - 200\pi t -\frac{\pi}{2}\right]}$$
d) At $t=15$ ms, and $x=0.5$ m, then
$$D_{(x,t)}= 1.0 \sin\left[ 2\pi(0.5) - 200\pi (15\times 10^{-3}) -\frac{\pi}{2}\right]$$
$$D_{(x,t)}= \color{red}{\bf -1.0}\;\rm mm$$
