A child ties one end of a long rope to a tree and shakes the other end up and down. Waves run along the rope to the tree. As long as her arm moves, the waves keep coming. The moment she stops, the rope soon falls still. Her arm tires, so her shaking must be feeding the rope something all the time.
Where the energy comes from
The last two lessons found the energy that a wave on a string carries. Where does this energy come from? It is supplied by the external source that makes the disturbance, such as a hand moving one end of the string up and down. The rate at which the source supplies energy to the string is the power of a wave.
In other words, power is the rate at which energy is carried past a point of the string. In a steady wave, energy passes every point at the same average rate at which the source puts it in.
A wave on a string carries a power of \(2\,\text{W}\) to the right. What does this mean?
The force across a point
Now find the power at a point \(x = x_0\) by a direct force analysis. Imagine the string cut in two at \(x_0\): a left part, where \(x < x_0\), and a right part, where \(x > x_0\). The left part exerts a force \(\vec F\) on the right part. The string is stretched, so \(\vec F\) acts along the tangent to the string at \(x_0\). The tangent makes an angle \(\theta\) with the horizontal.
Split \(\vec F\) into a horizontal and a vertical component, of sizes \(F\cos\theta\) and \(F\sin\theta\). The points of the string move only vertically, with no horizontal velocity. So the horizontal component does no work. For a string under a uniform tension \(T\), the horizontal component must be exactly \(T\). So \(F\cos\theta = T\), which gives \(F = T/\cos\theta\).
\[ |F_y| = F\sin\theta = T\tan\theta \]
Why. the size of the vertical component. Put \(F = T/\cos\theta\), and \(\sin\theta/\cos\theta = \tan\theta\).
\[ \tan\theta = \dfrac{\partial y}{\partial x} \]
Why. the tangent of the string's angle is its slope at \(x_0\). For small angles this is how the slope enters.
\[ \dfrac{\partial y}{\partial x} > 0:\ F_y < 0 \]
Why. the left part pulls the right part back towards itself, along the tangent. Where the string rises to the right, that backward tangent goes downward, as in Fig 5.2. So the pull points down and to the left.
\[ \dfrac{\partial y}{\partial x} < 0:\ F_y > 0 \]
Why. where the string falls to the right, the backward tangent goes upward. So the pull points up and to the left.
\[ F_y = -T\,\dfrac{\partial y}{\partial x} \]
Read it. one formula covers both cases. The vertical pull is opposite in sign to the slope, and its size is \(T\) times the slope.
The minus sign is not a convention added at the end. It comes from the direction of the pull. The left part always pulls the right part back towards itself, so its vertical pull points against the slope.
The power across the point
The point \(x_0\) of the string moves up and down with the velocity \(v_y = \partial y/\partial t\). The rate at which a force does work is the force times the velocity along it. Only the vertical component works, so the power passed from the left part to the right part is \(F_y\,v_y\).
In general this power can be positive or negative. Its sign depends on the direction of the wave and on the local slope. A positive \(P\) means energy passes \(x_0\) to the right. A negative \(P\) means it passes to the left.
Now use the relation of the chapter Introduction to Waves. For every wave \(y = f(x - vt)\) that travels in the \(+x\) direction, the chain rule gives \(\partial y/\partial t = -v f'\) and \(\partial y/\partial x = f'\). So \(\partial y/\partial t = -v\,\partial y/\partial x\). This holds for a pulse of any shape, not only for a sine wave.
\[ P = -T\left(\dfrac{\partial y}{\partial x}\right)\left(\dfrac{\partial y}{\partial t}\right) \]
Why. the power across \(x_0\), from the force analysis.
\[ P = -T\left(\dfrac{\partial y}{\partial x}\right)\left(-v\,\dfrac{\partial y}{\partial x}\right) \]
Why. put \(\partial y/\partial t = -v\,\partial y/\partial x\), true for every wave travelling in the \(+x\) direction.
\[ P = Tv\left(\dfrac{\partial y}{\partial x}\right)^2 \]
Read it. the two minus signs cancel. A square is never negative, so the energy of such a wave always flows in the \(+x\) direction.
This is the instantaneous power delivered across the position \(x_0\). It is largest where the string is steepest, the same places where the energy sits. Look again at Fig 5.2. The string rises to the right at \(x_0\), so the point moves down, since \(v_y = -v\,\partial y/\partial x\). The pull \(F_y\) also points down. Force and velocity agree, so the left part does positive work on the right part.
The left part of the string pulls the right part with a force along the tangent. Why does only the vertical component of this force do work?
A wave \(y = f(x + vt)\) travels in the \(-x\) direction. Then \(\partial y/\partial t = +v\,\partial y/\partial x\). What is the sign of \(P = -T(\partial y/\partial x)(\partial y/\partial t)\)?
The average power of a sine wave
Apply the power formula to the sinusoidal wave \(y(x, t) = A\sin(kx - \omega t)\). Its slope is \(\partial y/\partial x = Ak\cos(kx - \omega t)\), so the power is \(P = Tv\left(Ak\cos(kx - \omega t)\right)^2\). Like the energy, the power at a point swings between zero and a peak. To find the average power, average it over one period.
\[ \langle P\rangle = TvA^2k^2\left\langle \cos^2(kx - \omega t)\right\rangle \]
Why. the factor \(TvA^2k^2\) is constant and comes out of the average.
\[ \langle P\rangle = \tfrac12\,TvA^2k^2 \]
Why. the average of \(\cos^2\) over one period is \(\tfrac12\), as in lesson 4.
\[ P_{avg} = \tfrac12\,\mu v\omega^2A^2 \]
Read it. \(Tk^2 = \mu\omega^2\), from lesson 4.
The average power is the average energy per unit length times the wave speed: \(\tfrac12\mu\omega^2A^2 \times v\). Each second the wave carries \(v\) metres of its energy past a point. The string of the example in lesson 4 holds \(0.0987\,\text{J}\) in each metre and carries it at \(100\,\text{m/s}\), so its average power is \(9.87\,\text{W}\).
A guitar string has \(T = 80\,\text{N}\), \(\mu = 0.003125\,\text{kg/m}\) and \(v = 160\,\text{m/s}\). Find the average power needed to make a wave on it with a frequency of \(440\,\text{Hz}\), the musical note A, and an amplitude of \(1.5\,\text{mm}\).
List the values in SI units: \(\mu = 0.003125\,\text{kg/m}\), \(v = 160\,\text{m/s}\), \(A = 1.5\,\text{mm} = 1.5 \times 10^{-3}\,\text{m}\) and \(f = 440\,\text{Hz}\).
The angular frequency is \(\omega = 2\pi f = 2\pi(440\,\text{Hz}) \approx 2765\,\text{rad/s}\). Then use \(P_{avg} = \tfrac12\mu v\omega^2A^2\).
\(P_{avg} = \tfrac12(0.003125)(160)(2765)^2(1.5 \times 10^{-3})^2\,\text{W} = (0.25)(7.645 \times 10^6)(2.25 \times 10^{-6})\,\text{W} \approx 4.30\,\text{W}\).
An average power of about \(4.3\,\text{W}\) must be supplied to the string to keep making this wave. The energy is finally lost from the string, mostly as the sound we hear and a little as heat.
The amplitude of a harmonic wave on a string is doubled, while the frequency and the wave speed stay the same. How does the average power change?
A string with \(\mu = 0.01\,\text{kg/m}\) is under a tension of \(100\,\text{N}\). A harmonic wave of frequency \(50\,\text{Hz}\) and amplitude \(5\,\text{mm}\) travels on it. What is the average power, in watts?
A harmonic wave travels on a string. Which statements about its average power are true?
A whip is thick at the handle and tapers to a thin tip. A flick of the wrist sends a loop along it. The energy of the loop moves on into ever lighter pieces of the whip. With less mass to carry the same energy, each piece must move faster. Near the tip the speed passes the speed of sound in air, about \(340\,\text{m/s}\). The crack is a small sonic boom.
Now return to the child and the rope on the tree. Her arm is the source, and it feeds \(P_{avg} = \tfrac12\mu v\omega^2A^2\) into the rope. The energy leaves her hand and travels to the tree, so she must keep feeding it. Waves twice as tall need four times the power, which is why her arm tires so fast.
- Letting the horizontal pull do work.The string moves only vertically. Only \(F_y\) works, so \(P = F_y v_y\).
- Writing \(F_y = +T\,\partial y/\partial x\).The left part pulls the right part back along the tangent, so \(F_y = -T\,\partial y/\partial x\).
- Thinking \(\partial y/\partial t = -v\,\partial y/\partial x\) holds only for sine waves.It holds for every wave \(f(x - vt)\), by the chain rule.
- Taking the average power to be proportional to \(A\) or to \(f\).\(P_{avg} = \tfrac12\mu v\omega^2A^2\) grows as \(A^2\) and as \(f^2\).