A string of length '$L$' is fixed at both ends.  It is vibrating in its $3^{rd}$ overtone with  maximum amplitude $\text{'}a\text{'}.$ The amplitude at a distance $L/3$  from one end is $\sqrt{n}\frac{a}{2}$. find n. 

For a string vibrating in its nth overtone $(n+1)thharmonic)$
$(n+1)\frac{\lambda }{2}=L\Rightarrow \lambda =\frac{2L}{n+1}$    
     
 $kx=\frac{2\pi x}{\lambda }=\frac{\pi (n+1)x}{L}$
 $y=2A\sin \left(\frac{(n+1)\pi x}{L}\right)\cos \omega t$
Here  $2A=aandn=3.$  
For  $x=\frac{L}{3},y=\left[a\sin (\frac{4\pi }{L}\times \frac{L}{3})\right]\cos \omega t$
$=\alpha \sin \frac{4\pi }{3}\cos \omega t=-a\left(\frac{\sqrt{3}}{2}\right)\cos \omega t$ 
$atx=L/3;$  the amplitude is   $\sqrt{3}a/2$