Tie one end of a long skipping rope to a gatepost and pull it fairly tight. Give the free end one sharp flick. A hump runs along the rope, and a moment later the gatepost is tugged. No piece of rope travelled from your hand to the post. Yet the work of your wrist reached the post.
A wave carries energy
When you flick a rope, you do work on it. That work becomes the energy of the wave, and the wave carries the energy down the rope. A wave is, at heart, a way of carrying energy from one point to another. In this lesson and the next two we find how much energy a wave on a string carries. Then we find the rate at which it is carried, the power of the wave.
Kinetic energy
Every segment of the string that takes part in the wave has energy in two forms. The first is kinetic energy, because the segment is moving up and down. Watch one particle alone, and its kinetic energy is greatest when its speed is greatest. Now look at the whole string at one instant.
The chapter Introduction to Waves showed that the velocity of a particle is minus the wave speed times the slope of the string, \(v_y = -v\,\partial y/\partial x\). So at one instant the particles move fastest where the string is steepest. The kinetic energy of the string is therefore greatest near the region where its slope is greatest.
Take a small segment of mass \(dm\). Its mass is \(dm = \mu\,ds\). As in the speed derivation, a small amplitude makes the length \(ds\) very close to its horizontal projection \(dx\), so \(dm \approx \mu\,dx\). The segment moves vertically with the transverse speed \(v_y = \partial y/\partial t\).
Keep two speeds apart. The transverse speed \(v_y = \partial y / \partial t\) is the up-and-down speed of one piece of string, and it changes from moment to moment. The wave speed \(v = \sqrt{T/\mu}\) is the horizontal speed of the pattern, and it is fixed by the string.
\[ dK = \tfrac12\,(dm)\,v_y^2 \]
Why. the standard kinetic energy of a body of mass \(dm\) moving at speed \(v_y\).
\[ dm = \mu\,dx, \qquad v_y = \dfrac{\partial y}{\partial t} \]
Why. the mass of the segment for a small amplitude, and its transverse speed.
\[ dK = \tfrac12\,(\mu\,dx)\left(\dfrac{\partial y}{\partial t}\right)^2 \]
Why. substitute both into step 1. This is the kinetic energy of the segment.
\[ \dfrac{dK}{dx} = \tfrac12\,\mu\left(\dfrac{\partial y}{\partial t}\right)^2 \]
Read it. divide by the length \(dx\) of the segment. This is the kinetic energy per unit length at the point \(x\).
A string has a mass per unit length of \(0.01\,\text{kg/m}\). At some instant the string at a point moves up at \(2\,\text{m/s}\). What is the kinetic energy per unit length there, in J/m?
Potential energy
Even without a wave, the string is already stretched beyond its natural length. It stores elastic potential energy, much like a stretched rubber band. When the wave passes, each segment is stretched a little more than before. It then stores a little extra elastic potential energy, because of the wave. We call this extra energy the potential energy of the wave.
With no wave, a segment lies flat and its length is \(dx\). With the wave present, the same segment is tilted, and its length is \(ds\). Fig 3.1 shows the two cases side by side.
The tilted length follows from the angle of inclination \(\theta\) of the string: \(ds = dx\sec\theta\). A larger angle gives a longer segment. So the string is stretched most where it is most tilted. The potential energy of a wave on a string is therefore concentrated where the slope of the string is greatest.
The work done in stretching
To find the potential energy, find the work done to stretch the segment against the tension. The work to stretch something by a small length \(dL\) against a constant force \(F\) is \(dW = F\,dL\). Here the force is the tension \(T\), and the stretch is the extension \(ds - dx\).
\[ dU = T\,(ds - dx) \]
Why. the work done against a constant force \(T\) through the extension \(ds - dx\), stored as potential energy.
\[ ds^2 = dx^2 + dy^2 \]
Why. Pythagoras' theorem for the tiny right triangle whose hypotenuse is the segment, with sides \(dx\) and \(dy\).
\[ ds = \sqrt{dx^2 + dy^2} \]
Why. take the square root of both sides.
\[ ds = dx\sqrt{1 + \left(\dfrac{dy}{dx}\right)^2} \]
Read it. take \(dx^2\) out of the root as \(dx\). At one instant \(dy/dx\) is the slope \(\partial y/\partial x\).
The binomial approximation
Now we need one mathematical tool, the binomial approximation. For any small number \(z\), \((1 + z)^n \approx 1 + nz\). Here the small number is \(z = (\partial y/\partial x)^2\), the square of the slope. The amplitude is small, so the slope is small, and its square is smaller still. The power is \(n = \tfrac12\), a square root.
\[ \sqrt{1 + \left(\dfrac{\partial y}{\partial x}\right)^2} \approx 1 + \tfrac12\left(\dfrac{\partial y}{\partial x}\right)^2 \]
Why. the binomial approximation with \(z = (\partial y/\partial x)^2\) and \(n = \tfrac12\).
\[ ds \approx dx\left(1 + \tfrac12\left(\dfrac{\partial y}{\partial x}\right)^2\right) \]
Why. substitute this into the expression for \(ds\).
\[ ds - dx \approx \tfrac12\left(\dfrac{\partial y}{\partial x}\right)^2 dx \]
Why. subtract \(dx\) from step 2 to get the extension. The \(dx\) terms cancel and only the small extra length is left.
\[ dU = T \cdot \tfrac12\left(\dfrac{\partial y}{\partial x}\right)^2 dx \]
Why. put the extension into \(dU = T\,(ds - dx)\), the work done against the tension.
\[ dU = \tfrac12\,T\left(\dfrac{\partial y}{\partial x}\right)^2 dx \]
Why. tidy the factors. This is the elastic potential energy stored in the segment by its stretching.
\[ \dfrac{dU}{dx} = \tfrac12\,T\left(\dfrac{\partial y}{\partial x}\right)^2 \]
Read it. divide by the length \(dx\) of the segment. This is the potential energy per unit length at the point \(x\).
At a point the slope of a string is \(0.1\). By the binomial approximation, what is \(ds/dx = \sqrt{1 + (0.1)^2}\)?
A string under a tension of \(50\,\text{N}\) carries a wave. At some instant its slope at a point is \(0.02\). What is the potential energy of the wave per unit length there, in J/m?
Where the energy sits
Put the two results together. The kinetic energy is greatest where the particles move fastest, which is where the slope is greatest. The potential energy is greatest where the stretch is greatest, which is also where the slope is greatest. So both kinds of energy crowd into the same regions of the string.
The region of the string with the greatest slope has the greatest kinetic energy and the greatest potential energy. That region moves along the string with the wave speed \(v\), so the wave carries its energy with it. The total energy of a small segment is its kinetic energy plus its potential energy.
The two energies are in fact equal in every small segment. The relation of the chapter Introduction to Waves, \(\partial y/\partial t = -v\,\partial y/\partial x\), shows it in three lines. It holds for every wave of the form \(y = f(x - vt)\), not only for sine waves.
\[ \left(\dfrac{\partial y}{\partial t}\right)^2 = v^2\left(\dfrac{\partial y}{\partial x}\right)^2 \]
Why. square the relation \(\partial y/\partial t = -v\,\partial y/\partial x\) for a wave travelling in the \(+x\) direction. The minus sign disappears.
\[ \tfrac12\,\mu\left(\dfrac{\partial y}{\partial t}\right)^2 = \tfrac12\,\mu v^2\left(\dfrac{\partial y}{\partial x}\right)^2 \]
Why. multiply by \(\tfrac12\mu\). The left side is now the kinetic energy per unit length.
\[ \mu v^2 = T \]
Why. the speed on a string is \(v^2 = T/\mu\).
\[ \tfrac12\,\mu\left(\dfrac{\partial y}{\partial t}\right)^2 = \tfrac12\,T\left(\dfrac{\partial y}{\partial x}\right)^2 \]
Read it. put \(\mu v^2 = T\) into step 2. The left side is the kinetic energy per unit length and the right side the potential energy per unit length, so \(dK = dU\).
A sinusoidal wave travels along a string. At what point of the wave is the kinetic energy of a string element greatest?
The total energy per unit length at a point is the sum of the kinetic and the potential parts. It is the energy per unit length, written \(dE/dx\).
Look at the crest in Fig 3.2. The string there is flat for an instant, so it is not stretched beyond its tension. The particle at the crest is also at rest for an instant. Both energies vanish together. At the shaded crossings the particles race through the rest line and the string is tilted most. Both energies peak together.
A student says, "The potential energy of a wave is greatest at a crest." The reason given is that the string is displaced most there. What is wrong with this?
John William Strutt, the third Lord Rayleigh, worked on sound in his own laboratory at his country house in England. In 1877 he published the first volume of The Theory of Sound. In it he wrote the energy of a vibrating string as two parts. The kinetic part came from the motion of each piece. The potential part came from the tension times the extra length of the tilted string. That is the calculation of this lesson.
Now return to the rope tied to the gatepost. Every piece the hump passes holds energy in two equal parts, one of motion and one of stretch. Both are packed where the rope is steepest, on the sides of the hump. The hump moves at the wave speed, and it carries that packet of energy to the post.
- Using the wave speed \(v\) in the kinetic energy of a segment.The segment moves at the transverse speed \(\partial y/\partial t\). The wave speed enters only through \(\partial y/\partial t = -v\,\partial y/\partial x\).
- Placing the potential energy of the wave where the displacement is largest.It depends on the slope, \(\tfrac12 T(\partial y/\partial x)^2\). It is zero at a crest.
- Dropping the half in the binomial step.\(\sqrt{1 + z} \approx 1 + \tfrac12 z\).
- Counting the energy of the tension that exists with no wave.The potential energy of the wave is only the extra energy from the extra stretch \(ds - dx\).