A uniform string of length $l$  is fixed at both ends such that tension T is produced in it.  The string is excited to vibrate with maximum displacement amplitude  $a_0$. The maximum kinetic energy of the string for its first overtone frequency is  $2\left(\frac{a_0^2{\pi }^{2}T}{x\mathcal{l}}\right)$. The value of x is _____.

Since, maximum kinetic energy of a particle performing SHM is equal to its oscillation energy, therefore, maximum kinetic energy of the string in its fundamental tone.
 $=\frac{{a_0}^2{\pi }^{2}T}{4\mathcal{l}}$
First Overtone
Frequency,  $n_1=\frac{v}{{\lambda }_1}=\frac{1}{\lambda }\sqrt{\frac{T}{m}}$
Considering an elemental length dx of string at a distance x from left end, its mass = mdx
Its oscillation energy $=\frac{1}{2}\left(mdx\right)a_0^2.{\sin }^2\left(\frac{2\pi x}{\mathcal{l}}\right){\left(2\pi n_1\right)}^2$
 $=\frac{2{\pi }^2{a}_0^{2}T}{t_2}{\sin }^2\left(\frac{2\pi x}{\mathcal{l}}\right).dx$
  Total oscillation energy of the string $=\frac{2{\pi }^2{a}_0^{2}T}{t_2}\underset{0}{\overset{\mathcal{l}}{\int }}{\sin }^2\left(\frac{2\pi x}{\mathcal{l}}\right).dx$
or maximum kinetic energy of string in its first overtone  $\frac{a_0^2{\pi }^{2}T}{\mathcal{l}}$