Past exam of the mathematical methods II course of the University of Oxford / 2014 / 11

For an element $dx$, the vertical components of the approximately constant tension are $T{y}_x$ at its two ends. Their difference is $T{y}_{xx}dx$ to first order in the small slope. Its inertial response is $\rho dx{y}_{tt}$. Thus Newton's second law givesHere $\rho$ is mass per unit length. Small slopes, negligible gravity, uniform tension and negligible bending stiffness or dissipation are the assumptions in this wave equation on a string.

Use separation of variables, $y(x,t)=X(x)Q(t)$. The wave equation implies $X^{\prime \prime }/X=\ddot{Q}/(c^{2}Q)=-k^2$, giving $X^{\prime \prime }+k^{2}X=0$ and $\ddot{Q}+c^2{k}^{2}Q=0$. The fixed-end boundary conditions $X(0)=X(l)=0$ selectEach such product is a standing wave. A general released string is a superposition of these normal modes, not necessarily a single separated standing wave. Absence of damping keeps their amplitudes constant.

Use the product identity $\sin z\cos z=\frac{1}{2}\sin 2z$ to decompose the initial shape:The Fourier sine series is already explicit. Zero initial velocity gives cosine time factors, soand $y(x,t)=A\sin (\pi x/l)\cos (\pi ct/l)+(A/2)\sin (2\pi x/l)\cos (2\pi ct/l)$. Orthogonality of the sine modes makes this decomposition unique in the usual square-integrable fixed-end setting.

For one real mode $y_n=A_n\sin (k_{n}x)\cos ({\omega }_{n}t+{\varphi }_n)$, the kinetic and small-slope potential energy densities integrate toSince $\rho {\omega }_n^2=T{k}_n^2$, their sum is constant: $E_n=Tl{A}_n^2{k}_n^2/4$. The initial condition has ${\varphi }_n=0$ and $A_1=A$, $A_2=A/2$, henceAlthough the second mode's displacement amplitude is halved, its wavenumber and angular frequency are doubled, giving the same total energy. Orthogonality eliminates cross terms when the full string's energies are integrated.