A string of mass  $m$ is fixed at both its ends. The fundamental mode of string is excited  and it has an angular frequency $\omega$ and the maximum displacement amplitude A. Then :

Let the displacement of the string be given by 
$y(x,t)=A\sin \frac{\pi x}{L}\cos (\omega t+\delta )$ 
Where  $\delta$  is a phase factor.  So the transverse velocity is given by 
 $y(x,t)=\frac{\partial y}{\partial t}=-\omega A\sin \frac{\pi x}{L}\sin (\omega t+\delta )$
The maximum kinetic energy is equal to the string’s total energy of oscillation.  Note  that all points of the  string achieve their maximum kinetic energy at the same instant of time, where $y=0forallx.$  Since   $dm=\mu dxwhere\mu =\left(\frac{m}{L}\right)$
Is the mass per unit length of the uniform string.  The maximum kinetic energy.
$E_K=\max .of[\frac{1}{2}\int {\left(\frac{\partial y}{\partial t}\right)}^{2}dm]$ 
$=\max .of\left[\frac{1}{2}\underset{0}{\overset{L}{\int }}{\left(\frac{\partial y}{\partial t}\right)}^{2}dx\right]$ 
The maximum value of ${\left(\frac{\partial y}{\partial t}\right)}^2,$  occurs when 
${\sin }^2(\omega t+\delta )=1$ 
Hence   $E_K=\frac{\mu }{2}{A}^2{\omega }^2\underset{0}{\overset{L}{\int }}{\sin }^2\frac{\pi x}{L}dx$
The integral $\underset{0}{\overset{L}{\int }}{\sin }^2\frac{\pi x}{L}dx$   over the half-cycle has 
The average value of  $\frac{L}{2}$