= Solution
For a small-slope string element, the transverse net <tension> is $T[y_x(x+\delta x)-y_x(x)]\simeq Ty_{xx}\delta x$. Its mass is $\rho\delta x$, so <Newton's second law> gives
$$
\boxed{y_{tt}=\frac T\rho y_{xx},\qquad c^2=T/\rho.}
$$
Its 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:
$$
\boxed{dU=\frac T2y_x^2\,\delta x,\qquad dK=\frac\rho2y_t^2\,\delta x.}
$$
For a single travelling wave $y=f(x\pm ct)$, $y_t^2=c^2y_x^2$, 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 $c=2Lf$. Hence
$$
\boxed{T=\rho(2Lf)^2=16.5\,\mathrm N.}
$$
If98Hz 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 $\int_0^L\sin^2kx\,dx=L/2$, yielding
$$
\boxed{K_{\max}=\frac14\rho LA^2\omega^2=4.25\times10^{-4}\,\mathrm J,\qquad\omega=2\pi(98\,\mathrm{Hz}).}
$$
At this instant the ideal mode's elastic energy is zero; at maximum displacement all its energy is elastic. <University Physics Volume 1 by William Moebs and others>[Chapter16, transverse-string wave speed, wave energy and fixed-end normal modes]{ref}
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