# 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).

• If you define PE=0 when the string is at rest then any change in PE must be balanced by an equal but opposite change in KE in order to conserve energy... – lemon Apr 18 '17 at 8:29
• @Farcher That answer doesn't answer my query. – user139621 Apr 18 '17 at 9:02
• I have the proof for the example given in the linked question. – Farcher Apr 18 '17 at 21:15

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 : $$\tag{1} K = \int_0^L \frac{1}{2} \Big( \frac{\partial y}{\partial t} \Big)^2 \, \lambda \, dx.$$ Total potential energy (notice that $v^2 = T/\lambda$) : $$\tag{2} U = \int_0^L \frac{v^2}{2} \Big( \frac{\partial y}{\partial x} \Big)^2 \, \lambda \, dx.$$ Wave equation : $$\tag{3} \frac{1}{v^2} \frac{\partial^2 \, y}{\partial t^2} - \frac{\partial^2 \, y}{\partial x^2} = 0.$$ Thus integration by parts gives this relation : $$\tag{4} U = K - \frac{\lambda}{4} \, \frac{d^2 \, }{d t^2} \int_0^L y^2(t, x) \, dx.$$ 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 $U \ne K$.

• Did you see the answer I linked? Is that wrong, then? – user139621 Apr 18 '17 at 13:00
• I guess that answer you linked is false. KE isn't always equal to PE. Also, don't confuse energy density and energy. KE and PE are integrals on the string's length. – Cham Apr 18 '17 at 13:05
• See the edit to my answer. – Cham Apr 18 '17 at 13:51
• For the example given in the link the kinetic energy is equal to the potential energy and there is no interchange between the two. There is proof in the link. – Farcher Apr 18 '17 at 21:18

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!

• Take note that $y = f(x \pm vt)$ is NOT the general solution of the wave equation, since it only describes traveling waves. The general solution is a superposition of such waves, and can be formulated as this : $$y(t, x) = \frac{1}{2}\big( f(x - vt) + f(x + vt) \big) + \frac{1}{2 v} \int_{x - vt}^{x + vt} g(u) \, du,$$ where $g(u)$ describes the initial derivative (or "velocity") of the wave. – Cham May 2 '17 at 0:19
• Would this be also the case for a single traveling wave on, say, a spring? Or is $KE = U$ specific to a string? – Julia Kim Mar 14 at 15:44

Comments for a uniform harmonic string:

1. 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 $$KE=PE.\tag{1}$$

2. 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.