A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, $\left.y(x,t)=\left(0.01\mathrm{m}\right)\sin \left[\left(62.8{\mathrm{m}}^{-1}\right)x\right]\cos \left[628{\mathrm{s}}^{-1}\right)t\right]$ Assuming $\pi =3.14$ the correct statements(s) is (are)

Comparing the given equation with the standing wave equation

$y(x,t)=2a\sin \left(kx\right)\cos \left(\omega t\right)$

we get 

$\begin{array}{r}k=62.8\Rightarrow \frac{2\pi }{\lambda }=62.8\\ \Rightarrow \lambda =\frac{2\times 3.14}{62.8}=0.1\mathrm{m}\end{array}$

and $\omega =628\Rightarrow 2\pi v=628\Rightarrow v=\frac{628}{2\times 3.14}$

$=100\mathrm{Hz}$

Since the string is vibrating it the fifth harmonic, the number of nodes= 6. (see figure) 

$\begin{array}{l}\frac{5\lambda }{2}=L(\mathrm{L}=\text{ length of string })\\ \Rightarrow L=\frac{5\times 0.1}{2}=0.25\mathrm{m}\end{array}$

There is an antinode at the mid-point of the string where $y$ is maximum. From the given equation

$y_{max}=0.01\mathrm{m}$

Fundamental frequenty $v_0=\frac{v}{2L}=\frac{v\lambda }{2L}=\frac{100\times 0.1}{2\times 0.25}$

$=20\mathrm{Hz}$