Why is KE equal to PE for waves on a string? Clarification needed For waves on a string why is Kinetic Energy said to be equal to Potential Energy? The linked answer gives no reason for it.
I knew that KE for each elementary part of wave is $(1/2)(dm)v^2$ ($v$ i.e. $\frac{dy}{dt}$ being instantaneous velocity) while PE is $T\Delta \ell$ (as in the linked answer).
 A: 
Kinetic Energy Of Plane Progressive Wave In A String:
Consider a small element of the string at a distance $x$ from origin $dx$.
K.E of that small part of the string is: $$\frac{1}{2}(\mu dx)(\frac{\partial y}{\partial t})^2$$
Potential Energy Of Plane Progressive Wave In A String:
Again consider a small element of the string at a distance $x$ from origin $dx$.
Since there is no motion in the x-direction, points on the string only move
in the y-direction. If the slope of a segment changes, the length must also
change. Since it takes work to stretch the string, there is also potential energy stored in the string. The change in length of a segment is:
$$dl = \sqrt{dx^2 + dy^2} − dx ≈ \frac{1}{2}(\frac{∂y}{∂x})^2dx$$
Hence, potential energy of that small part of the string is:
$$Tdl=T\left(\frac{1}{2}(\frac{∂y}{∂x})^2dx\right)$$

Hence, for unit length of string:
$$K_l=\frac{1}{2}(\mu)(\frac{\partial y}{\partial t})^2$$
$$U_l=T\left(\frac{1}{2}(\frac{∂y}{∂x})^2\right)$$
If $y=f(x \pm vt)$ (general solution of wave equation for travelling waves)
$$K_l=\frac{1}{2}(\mu)v^2(f')^2$$
$$U_l=\frac{1}{2}T(f')^2$$
Now using, $v^2=\frac{T}{\mu}$ we can easily see that $K_l=U_l$.
Note that this result is only true for a single travelling wave; it is not true if waves travelling in both directions are present!
Source: http://mpalffy.lci.kent.edu/Optics/Chapters/Ch1_Waves_on_a_String.pdf
A: I don't see why the KE should be equal to the PE, for a wave on a string, except for some special time (like a mass oscillating on a spring).  For example, consider a simple stationary wave on a string.  At some time, the whole string is instantly straight (at this time : PE = 0 since there is no deformation of the string, at that particular moment), while each part of the string is in motion (KE != 0).  Thus clearly KE != PE at that moment in time (instantaneously straight string while the wave is still propagating on it).  
However, KE + PE = cste (conservation of total energy), if there is no exterior agent.
This is very similar to the simpler case of a single mass oscillating on a vertical spring.  At some moment in time, you may get $K = U$, but at another moment you'll also have $U = 0$ and $K = K_{\text{max}}$, while $E \equiv K + U = \textit{cste}$ at any moment.
EDIT :  For a general wave (of wave function $y(t, x)$) on a string with fixed extremities : $y(t, 0) = y(t, L) = 0$, it is easy to use integration by parts and the wave equation to show the following :
Total kinetic energy :
\begin{equation}\tag{1}
K = \int_0^L \frac{1}{2} \Big( \frac{\partial y}{\partial t} \Big)^2 \, \lambda \, dx.
\end{equation}
Total potential energy (notice that $v^2 = T/\lambda$) :
\begin{equation}\tag{2}
U = \int_0^L \frac{v^2}{2} \Big( \frac{\partial y}{\partial x} \Big)^2 \, \lambda \, dx.
\end{equation}
Wave equation :
\begin{equation}\tag{3}
\frac{1}{v^2} \frac{\partial^2 \, y}{\partial t^2} - \frac{\partial^2 \, y}{\partial x^2} = 0.
\end{equation}
Thus integration by parts gives this relation :
\begin{equation}\tag{4}
U = K - \frac{\lambda}{4} \, \frac{d^2 \, }{d t^2} \int_0^L y^2(t, x) \, dx.
\end{equation}
So if the area below the string stays constant while the wave propagates (or more precisely the area of the function $y^2(t, x)$), the last term cancels and you do get $U = K$.  In the simple case of an impulse propagating on the string (sinusoidal wave, for example), the area is conserved, so $U = K$.  This is a special case though.  In the case of a stationary wave, the area isn't conserved and $U \ne K$.
A: Comments for a uniform harmonic string: 


*

*For a transversal xor longitudinal left- xor right-mover $f(x\pm vt)$, it is straightforward exercise to check directly that the kinetic energy density $\frac{\mu v^2}{2} (f^{\prime}(x\pm vt))^2$ matches the potential energy density, and hence
$$KE=PE.\tag{1}$$

*Eq. (1) is not necessarily true for superpositions thereof because of cross-terms, but it does hold in time average due to the virial theorem.   
