The following fig. show a snapshot of a vibrating string at t=0. The particle P is observed moving up with velocity $20\sqrt{3}cm/s.$ The tangent at P makes an angle  ${60}^0$ with the x-axis The equation of the wave is (Given amplitude = 4 mm)

 $v_p=-v\frac{dy}{dx},\frac{dy}{dx}=\tan 60=\sqrt{3}$
$v_p=-v\sqrt{3}\Rightarrow P$ is moving along upward direction 
 wave must be travelling along negative x direction
The equation of the wave is  $y=A\sin (wt+kx+f)$
 $$\begin{gathered} A=4\times {10}^{-3}m,v=\frac{20\sqrt{3}}{\sqrt{3}}=20cm{s}^{-1} \\ \lambda =4\times {10}^{-2}m=4cm;f=\frac{0.2}{4\times {10}^{-2}}5Hz \\ \Rightarrow w=2\pi f=10\pi \\ K=\frac{2\pi }{0}=\frac{2\pi }{4}=\frac{\pi }{2}c{m}^{-1} \end{gathered}$$
 at  $t=0,x=0,y=2\sqrt{2}\times {10}^{-3}m=\frac{2\sqrt{2}}{10}cm$
 $$\begin{gathered} \Rightarrow \frac{2\sqrt{2}}{10}\times 4\times {10}^{-1}\sin \left(f\right) \\ \Rightarrow \frac{2\sqrt{2}}{10}\times 4\times {10}^{-1}\sin \left(f\right)\frac{1}{\sqrt{2}}\Rightarrow f\frac{\pi }{4}or\frac{3\pi }{4} \end{gathered}$$
 Taking at x=1.5,t=0,y=0 $\Rightarrow f=\frac{\pi }{4}$
 $\Rightarrow y=0.4\sin (10pt-\frac{\pi x}{2}+\frac{\pi }{4})$