 Solution

For a small-slope string element, the transverse net tension is $T[y_x(x+\delta x)-y_x(x)]\simeq T{y}_{xx}\delta x$. Its mass is $\rho \delta x$, so Newton's second law givesIts extra length is $[\sqrt{1+y_x^2}-1]\delta x\simeq y_x^2\delta x/2$. Work against the approximately constant tension gives the elastic energy of a small-slope string, while transverse motion gives its kinetic energy:For a single travelling wave $y=f(x\pm ct)$, $y_t^2=c^2{y}_x^2$, so $dK=dU$ pointwise to this order. A standing wave need not have pointwise or instantaneous kinetic–potential equality.

For a fixed-end string of length L, taking the stated note to be its fundamental gives $c=2Lf$. HenceIf98Hz were instead the nth harmonic, the required tension would be $16.5/n^2\mathrm{N}$; the usual fundamental interpretation is explicit.

For the specified standing wave, $\dot{y}=A\omega \sin kx\cos \omega t$. Every allowed nonzero fixed-end mode has ${\int }_0^L{\sin }^{2}kxdx=L/2$, yieldingAt this instant the ideal mode's elastic energy is zero; at maximum displacement all its energy is elastic. ^[ref]