$y= a \sin \left(\omega t-k x+\phi_{0}\right)$

As at $t=0$ at $x=0$ and $y=0$,

$\Rightarrow \quad \phi_{0}=0$

So, equation of wave over string is

$y= a \sin (\omega t-k x)$

Here,

$\omega=2 \pi f=2 \pi \times 500$

$=1000 \pi\left(\frac{ rad }{ s }\right)$

Angular wave number,

$k=\frac{\omega}{v}=\frac{1000 \pi}{100}=10 \pi\left( m ^{-1}\right)$

So, wave equation is

$y=a \sin (1000 \pi \cdot t-10 \pi \cdot x)$

Given at $t=0$ at $x=0.25 m$ and $y=0.02 m$.

So, $\quad 0.02=a \sin (-10 \pi \times 0.25)$

$=-a \sin \left(\frac{5}{2} \pi\right)$

$[\therefore \sin (-\theta)=-\sin \theta]$

$\Rightarrow \quad 0.02=-a \sin \left(2 \pi+\frac{\pi}{2}\right)$

$\Rightarrow \quad 0.02=-a \sin \left(\frac{\pi}{2}\right)$ or $\alpha =-0.02$

Here note that $a$ is amplitude and its positive and negative values are same. When we are getting a negative value this means particle is displaced below mean position.

So, we have

$y=-0.02 \sin (1000 \pi t-10 \pi x)$

Now, at $t=5 \times 10^{-4} s$ and $x=0.2 m$, value of displacement of particle is

$y =-0.02 \sin \left(1000 \pi \times 5 \times 10^{-4}\right.$

$=-0.02 \sin \left(\frac{\pi}{2}-2 \pi\right)$

$=0.02 \sin \left(2 \pi-\frac{\pi}{2}\right)$

${\left[\therefore \sin \left(\frac{\pi}{2}-2 \pi\right)=-\sin \left(2 \pi-\frac{\pi}{2}\right)\right.}$

$\quad \text { Also, } \sin (2 \pi-\theta)=-\sin \theta]$

$=-0.02\,m$