A doubt on Waves

Here is a question,

A string of mass $m$ and length $l$, both fixed at both ends is vibrating in its fundamental mode. The maximum amplitude is $a$ and the tension in the string is $T$. Find the energy of vibrations of the string in terms of $a$, $T$ and $l$.

This is how I answered,

Let the equation of wave be $y=f\left( x,t \right) =a\sin { \left( \omega t-kx \right) }$

Differentiating w.r.t $t$,

$\frac { \delta y }{ \delta t } ={ v }_{ p }=a\omega \cos { \left( \omega t-kx \right) }$

$\therefore$ Kinetic Energy, $K.E=\frac { 1 }{ 2 } m{ a }^{ 2 }{ \omega }^{ 2 }\cos ^{ 2 }{ \left( \omega t-kx \right) }$

$\therefore$Average Kinetic Energy, $\left< K.E \right> =\frac { \int _{ 0 }^{ T }{ \frac { 1 }{ 2 } m{ a }^{ 2 }{ \omega }^{ 2 }\cos ^{ 2 }{ \left( \omega t-kx \right) } dt } }{ \int _{ 0 }^{ T }{ dt } } =\frac { 1 }{ 4 } m{ a }^{ 2 }{ \omega }^{ 2 }$

Potential Energy, $P.E=\frac { 1 }{ 2 } m{ \omega }^{ 2 }{ a }^{ 2 }\sin ^{ 2 }{ \left( \omega t-kx \right) }$

$\therefore$ Average Potential Energy,$\left< P.E \right> =\frac { \int _{ 0 }^{ T }{ \frac { 1 }{ 2 } m{ a }^{ 2 }{ \omega }^{ 2 }\sin ^{ 2 }{ \left( \omega t-kx \right) } dt } }{ \int _{ 0 }^{ T }{ dt } } =\frac { 1 }{ 4 } m{ a }^{ 2 }{ \omega }^{ 2 }$

Therefore Total energy=$\left< K.E \right>+\left< P.E \right>=\frac { 1 }{ 2 } m{ a }^{ 2 }{ \omega }^{ 2 }$

Now,

${ \omega }^{ 2 }={ v }^{ 2 }{ k }^{ 2 }={ v }^{ 2 }\frac { 4{ \pi }^{ 2 } }{ { \lambda }^{ 2 } } =\frac { T }{ \frac { m }{ L } } \frac { 4{ \pi }^{ 2 } }{ { \lambda }^{ 2 } } =\frac { 2T{ \pi }^{ 2 } }{ m\lambda } =\frac { T{ \pi }^{ 2 } }{ mL }$

$\therefore$ Total Energy$=\boxed{\frac { T{ a }^{ 2 }{ \pi }^{ 2 } }{ 2L } }$

But the given answer is $\boxed{\frac { T{ a }^{ 2 }{ \pi }^{ 2 } }{ 4L} }$

Can someone explain this?