 Solution

Fixed endpoints require $k_n=n\pi /L$, with $n=1,2,\dots$. Writing $c_s=\sqrt{T/\rho }$,For equal amplitudes $A$, the two requested profiles areThe first is asymmetric, with an interior zero at $x=2L/3$; the second has its interior node at $L/2$. For two-mode string recurrence, orthogonality means the whole initial profile recurs only when both cosine coefficients equal one. Since ${\omega }_2=2{\omega }_1$,

The weak-stiffness string frequency shift follows by expanding the supplied dispersion relation:Higher normal modes acquire larger fractional frequency shifts. Let $\eta =\alpha k_1^2/c_s^2\ll 1$ and retain ${\omega }_1^{(0)}=c_s{k}_1$ for the original ideal frequency. At the originally defined time $t_1=\pi /(2{\omega }_1^{(0)})$,Thus the $t=0$ sketch is unchanged, whereas the original quarter-period sketch gains a small negative fundamental component. If $t_1$ is redefined using the new fundamental frequency, its fundamental component is exactly zero and the second-mode change there is only second order. Both conventions agree that generic later shapes change because of relative phase drift.

The new frequency ratio isAn exact recurrence now requires ${\omega }_{1}t=2\pi q$ and ${\omega }_{2}t=2\pi p$ for integers $p,q$. If $r=p/q$ in lowest terms, the first recurrence is $2\pi q/{\omega }_1$; for irrational $r$ there is no exact finite recurrence. Near-recurrences remain possible. Positive stiffness alone does not prove that the ratio is irrational.