Past exam of the mathematical methods II course of the University of Oxford / 2014 resit / 10

Interpret “held” and then “released” as zero initial velocity. In the small-slope wave equation on a string model, the elastic energy is $U=(T/2){\int }_0^L{y}_x^{2}dx$ and the initial kinetic energy is zero. The given profile hasThe two cosine functions are orthogonal on $[0,L]$ and each has squared integral $L/2$. ThusThe implicit leading displacement amplitude is one in the source's chosen length units. If that amplitude is written explicitly as $A$, all energies here and below acquire a factor $A^2$.

For a short string element $dx$, small slope makes its tension magnitude approximately constant and its vertical component approximately $T{y}_x$. The net vertical force is $T[y_x(x+dx,t)-y_x(x,t)]\simeq T{y}_{xx}dx$. Its mass is $\rho dx$. Newton's second law givesThis assumes uniform equilibrium tension, small slopes, negligible bending stiffness and dissipation. The symbol $\rho$ is linear mass density here, not volume density.

Fixed endpoints select the spatial normal modes $\sin (n\pi x/L)$, with angular frequencies ${\omega }_n=n\pi c/L$. Zero initial velocity selects cosine time dependence. Matching the two nonzero initial amplitudes givesEach term separately satisfies the wave equation and fixed-end boundary conditions. At $t=0$ the prescribed profile is recovered, and differentiating in time gives zero initial velocity.

To recover the original signed displacement profile, orthogonality requires both cosine time factors to equal one. Since ${\omega }_3=3{\omega }_1$, the smallest positive time isAt $L/c$ both factors are $-1$, giving the negative of the initial profile, rather than the same profile. More generally, simultaneous recurrence requires commensurate excited-mode frequencies and undiminished amplitudes. The ideal nondispersive, fixed-end string has integer-multiple frequencies, whereas damping or a dispersive correction can destroy exact recurrence.

For a mode $y_n=A_n\sin (n\pi x/L)\cos ({\omega }_{n}t)$, integrate the kinetic-energy density:Here $A_1=1$, $A_3=-1/3$ and ${\omega }_3=3{\omega }_1$, soAll other modes have zero energy. The equal modal energies for inverse-harmonic string amplitudes give each excited mode total energy $T{\pi }^2/(4L)$, even though their displacement amplitudes differ. The corresponding potential energy replaces sine squared by cosine squared, so their sum recovers the constant total energy from part (a).