Energy and Power of a Wave Traveling Along a String — Question 1

Question 1

A traveling crest need not be a point of maximum energy density

On a uniform string, $y=A\cos(kx-\omega t)$ with $\omega^2=(F/\mu)k^2$. The small-slope kinetic and elastic energy densities are $u_K=\mu y_t^2/2$ and $u_U=F y_x^2/2$.

1. Find both densities and compare them point by point.

2. Locate maximum total energy density and explain its value at a displacement crest.

Question 1 - Solution

A traveling crest need not be a point of maximum energy density

Method and mathematical working

Differentiate the displacement before substituting into the kinetic and elastic energy densities. Use $F k^2=\mu\omega^2$ for this traveling wave.$$\begin{aligned}y_t&=A\omega\sin\Phi,\qquad y_x=-Ak\sin\Phi,\\u_K&=\tfrac12\mu A^2\omega^2\sin^2\Phi=u_U.\\[4pt]\Phi=0&:\quad u_K=u_U=0,\\\Phi=\pi/2&:\quad u_K=u_U=\tfrac12\mu A^2\omega^2,\quad u_{\mathrm{total}}=\mu A^2\omega^2.\end{aligned}$$

(a) Calculation and conclusion

With $\Phi=kx-\omega t$, $y_t=A\omega\sin\Phi$ and $y_x=-Ak\sin\Phi$. Thus $u_K=u_U=\mu A^2\omega^2\sin^2\Phi/2$, and $u=\mu A^2\omega^2\sin^2\Phi$. Equality is local for this single traveling wave.

(b) Interpretation and limits

Maxima occur at zero-displacement crossings, where $|\sin\Phi|=1$. At a crest, both transverse velocity and local slope vanish, so both densities vanish in this model. Elastic energy is associated with string extension from slope, not with displacement from a separate local spring.

Physical check

At a crest the element is instantaneously stationary and the local slope vanishes, so both densities vanish there. Maximum displacement is therefore not the location of maximum local traveling-wave energy.