One end of a taut string of length 3m along the x-axis is fixed at $x=0$  . The speed of the waves in the string is $100 m{s}^{-1}$  .  The other end of the sting is vibrating in the y direction so that stationary waves are set up in the sting. The possible waveform(s) of these stationary waves is (are)

As at one end sting is fixed where node is formed and at other end where it is vibrating antinodes is formed so length of sting is writing as 
 $$\begin{gathered} \frac{n\lambda }{2}+\frac{\lambda }{4}=L \Rightarrow \lambda \left(\frac{2n+1}{4}\right)=L \\ \Rightarrow \frac{2\pi }{k}\left(\frac{2n+1}{4}\right)=3 \Rightarrow k=\frac{2\pi (2n+1)}{12},n=0,1,\mathrm{2.....} \\ \Rightarrow k=\frac{\pi }{6},\frac{3\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{9\pi }{2}\mathrm{.....} \end{gathered}$$
 
 For above value of k, equations in options (A), (B),(C) and (D), it matches with 
 $k=\frac{\pi }{6},\frac{5\pi }{6},\frac{5\pi }{2}$
Oscillation frequency of sting is given as 
 $$\begin{gathered} \Rightarrow f\frac{v}{\lambda }=\frac{v}{\frac{4L}{2n+1}}=\frac{v(2n+1)}{4L} \\ \Rightarrow \omega =2\pi f=\frac{2\pi (2n+1)}{24}=\frac{\pi \left(100\right)}{2\times 3}(2n+1) \\ \Rightarrow \omega =\frac{100\pi (2n+1)}{6} \omega =\frac{50\pi }{3},50\pi ,\frac{250\pi }{3}\mathrm{......} \end{gathered}$$
 
 For $n=0 and 2$  it matches with options (A) and (C). Thus  all these three options can be the correct equations under different resonating conditions of the sting.