= Solution
For a single <normal mode> $y_n=A_n\sin(k_nx)\cos(\omega_nt+\phi_n)$, the <kinetic energy> is the integral of $\tfrac12\rho y_t^2$. Using $\int_0^l\sin^2(k_nx)\,dx=l/2$ gives
$$
\boxed{K_n(t)=\frac{\rho l\omega_n^2A_n^2}{4}\sin^2(\omega_nt+\phi_n)}.
$$
The extra string length is $\int_0^l[\sqrt{1+y_x^2}-1]dx\simeq\tfrac12\int_0^l y_x^2dx$. Multiplying by <tension> gives the quadratic <potential energy>, so $\int_0^l\cos^2(k_nx)dx=l/2$ implies
$$
\boxed{U_n(t)=\frac{Tl k_n^2A_n^2}{4}\cos^2(\omega_nt+\phi_n)}.
$$
Since $\rho\omega_n^2=Tk_n^2$, the sum is time independent:
$$
\boxed{E_n=K_n+U_n=\frac{Tl k_n^2A_n^2}{4}=\frac{n^2\pi^2TA_n^2}{4l}}.
$$
For the released shape, $A_1=A$ and $A_2=A/2$, so
$$
\boxed{E_1=E_2=\frac{\pi^2TA^2}{4l},\qquad E_{\rm total}=\frac{\pi^2TA^2}{2l}}.
$$
The second mode's smaller displacement is offset by its doubled spatial <wavenumber> and <angular frequency>. <orthogonality> of distinct sine and cosine spatial modes eliminates cross terms in the integrated energies, so the total is their sum. Initially the <kinetic energy> is zero and all the <energy> is stored in the string's deformation; during the motion the total remains constant.
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