= Solution
For a short length $\delta x$, the <kinetic energy> is $\tfrac12\mu y_t^2\delta x$. For small slope, its increase in length is $[\sqrt{1+y_x^2}-1]\delta x\simeq\tfrac12y_x^2\delta x$, so its <elastic potential energy> is $\tfrac12Ty_x^2\delta x$. For a <travelling wave> $y=f(x-vt)$, $y_t=-vf'$ and $y_x=f'$. Using $v^2=T/\mu$ gives
$$
\boxed{\delta K=\delta U=\frac T2(f')^2\delta x,\qquad\delta E=T(f')^2\delta x.}
$$
Writing a pulse as $f(x)=A g(x/w)$ yields $E=(TA^2/w)\int[g'(s)]^2ds$. Thus energy is proportional to amplitude squared for a fixed shape and width. Width must also be retained when comparing pulses on different strings.
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