 Solution

Use $\sin \theta \cos \theta =\frac{1}{2}\sin 2\theta$ with $\theta =\pi x/l$. The initial shape becomesThe sine functions are orthogonal on $[0,l]$, so this decomposition directly identifies the two nonzero spatial normal mode coefficients. Since the string is released from rest, every time-dependent sine coefficient vanishes and the subsequent motion isThus the signed mode amplitudes are $A_1=A$, $A_2=A/2$ and $A_n=0$ for $n\ge 3$; their magnitude amplitudes are $|A|$ and $|A|/2$. The factor of two in the second angular frequency is fixed by the same boundary conditions as its spatial wavelength.