One end of a taut string of length 3 m along the x-axis is fixed at $X=0$ The speed of the wave is $100\mathrm{m}/\mathrm{s}$ Stationary waves are set in the string such that its other end $x = 3 m$ vibrates freely along the y direction. The possible functions of these stationary waves are:

One end of the string is a node and the other end is an antinode. These boundary conditions will be satisfied if

$(2n-1)\frac{\lambda }{4}=3\mathrm{m}\Rightarrow \lambda =\frac{12}{2n-1}\mathrm{m}$

Hence

$\begin{array}{l}k=\frac{2\pi }{\lambda }=(2n-1)\frac{\pi }{6}\text{ and }\\ \omega =vk=(2n-1)\frac{\pi }{6}\times 100\mathrm{m}/\mathrm{s}=(2n-1)\frac{50\pi }{3}\mathrm{rad}/\mathrm{s}\\ \ln \left(1\right) k=\frac{\pi }{6},n=1\text{ and }\omega =\frac{50\pi }{3}\Rightarrow \text{ correct }\end{array}$

$\begin{array}{l}\ln \left(2\right) ,n=2 and k=2\left(\frac{\pi }{6}\right)\text{ and }\omega =2\left(\frac{50\pi }{3}\right)\Rightarrow \text{ incorrect }\\ \ln \left(3\right) , n=5 and k=5\left(\frac{\pi }{6}\right)\text{ and }\omega =5\left(\frac{50\pi }{3}\right)\Rightarrow \text{ correct }\\ \ln \left(4\right), n=15 and k=15\left(\frac{\pi }{6}\right)\text{ and }\omega =15\left(\frac{50\pi }{3}\right)\Rightarrow \text{ correct }\end{array}$