The sine wave
Apply the energy results to a harmonic wave, the sine wave of the chapter Introduction to Waves. Its displacement is \(y(x, t) = A\sin(kx - \omega t)\). Here \(A\) is the amplitude, \(k = 2\pi/\lambda\) the angular wave number and \(\omega = 2\pi f\) the angular frequency, with \(f\) the frequency. The energy formulas need the two first derivatives of \(y\).
\[ \begin{aligned} \frac{\partial y}{\partial t} &= -A\omega\cos(kx - \omega t) \\ \frac{\partial y}{\partial x} &= Ak\cos(kx - \omega t) \end{aligned} \]
The time derivative holds \(x\) fixed and brings out a factor \(-\omega\). The space derivative holds \(t\) fixed and brings out a factor \(k\). Both contain the same cosine.
\[ dK = \tfrac12\,\mu\left(\dfrac{\partial y}{\partial t}\right)^2 dx \]
Why. the kinetic energy of a segment, from lesson 3.
\[ dK = \tfrac12\,\mu\omega^2A^2\cos^2(kx - \omega t)\,dx \]
Why. put in the time derivative \(\partial y/\partial t\) found above. Squaring removes its minus sign.
\[ dU = \tfrac12\,T\left(\dfrac{\partial y}{\partial x}\right)^2 dx \]
Why. the potential energy of a segment, from lesson 3.
\[ dU = \tfrac12\,Tk^2A^2\cos^2(kx - \omega t)\,dx \]
Read it. put in the space derivative \(\partial y/\partial x\) found above.
The two lines differ only in the factor in front: \(\mu\omega^2\) for the kinetic energy and \(Tk^2\) for the potential energy. These two factors are equal. The proof uses only the wave speed, found in two ways.
\[ v = \dfrac{\omega}{k} \]
Why. the speed of a harmonic wave, from the chapter Introduction to Waves: one wavelength in one period, \(v = f\lambda = \omega/k\).
\[ v^2 = \dfrac{T}{\mu} \]
Why. the speed on a string, from lesson 1.
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Your turn. both lines give \(v^2\). Which line equates them?
\[ Tk^2 = \mu\omega^2 \]
Read it. multiply both sides by \(\mu k^2\).
In a harmonic wave on a string, \(Tk^2 = \mu\omega^2\), so \(dK = dU\). The kinetic and the potential energy of any small segment are always equal. The total energy of the segment is twice its kinetic energy, \(dE = dK + dU = 2\,dK\).
The bars of Fig 4.1 hold the key difference from the energy of SHM in the chapter Dynamics of SHM. The piece P moves up and down in SHM, like a block on a spring. But the block trades kinetic energy for potential energy, and its total stays fixed. The piece of string does not. Its kinetic and potential energy rise together to a peak as it crosses the rest line, and fall together to zero at a crest.
So the energy of one piece of string is not constant. It flows in from the string on one side and out to the string on the other. That is exactly what carrying energy means. The next lesson measures the rate of that flow.
A small piece of string carries a travelling sine wave. How do its kinetic and potential energy change as it moves from a crest down through the rest line?
A harmonic wave of frequency \(f\) passes a fixed point of a string. How many times each second does the energy per unit length at that point reach its peak?
The average energy
The energy per unit length \(\mu\omega^2A^2\cos^2(kx - \omega t)\) tells how the energy at a point \(x\) changes with time \(t\). It oscillates, but its average over time is constant. The average of \(\cos^2\) over one period is \(\tfrac12\).
\[ \left\langle \dfrac{dE}{dx} \right\rangle = \mu\omega^2A^2\left\langle \cos^2\alpha\right\rangle \]
Why. the angle brackets mean the average over one period, and \(\alpha = kx - \omega t\). The factor \(\mu\omega^2A^2\) is constant and comes out of the average.
\[ \cos^2\alpha = \tfrac12\left(1 + \cos 2\alpha\right) \]
Why. the double-angle identity for the cosine.
\[ \left\langle \cos^2\alpha\right\rangle = \tfrac12 \]
Why. over one period \(\cos 2\alpha\) is positive as long as it is negative, so its average is zero. Only the constant \(\tfrac12\) is left.
\[ \left\langle \dfrac{dE}{dx} \right\rangle = \tfrac12\,\mu\omega^2A^2 \]
Read it. put the average into step 1. This is the average energy per unit length of a harmonic wave.
To find the total energy in one wavelength, multiply the average energy per unit length by the wavelength. Averaging along one wavelength at a fixed instant gives the same \(\tfrac12\), since \(\cos^2\) runs through the same values in space as in time.
Play Fig 4.2. The humps of energy ride along under the wave at the wave speed. Each hump sits under a crossing of the rest line, and the humps are half a wavelength apart. The marked point P stays where it is. Its dot on the lower graph rises and falls twice in each period, as the humps pass it.
A harmonic wave of amplitude \(0.02\,\text{m}\) and angular frequency \(100\,\text{rad/s}\) travels on a string of mass per unit length \(0.01\,\text{kg/m}\). What is its average energy per unit length, in J/m?
A wave travels along a string with linear mass density \(\mu = 0.005\,\text{kg/m}\) and tension \(T = 50\,\text{N}\). The wave has the form \(y(x, t) = 0.005\sin(4\pi x - 400\pi t)\), with \(x\) and \(y\) in metres and \(t\) in seconds. Find (a) the speed of the wave, (b) the average energy per unit length, (c) the total energy in one wavelength.
(a) The speed of the wave is \(v = \sqrt{T/\mu} = \sqrt{50/0.005} = \sqrt{10\,000} = 100\,\text{m/s}\). The wave form agrees: \(\omega/k = 400\pi/4\pi = 100\,\text{m/s}\).
(b) The amplitude is \(A = 0.005\,\text{m}\) and the angular frequency \(\omega = 400\pi\,\text{rad/s}\). So \(\langle dE/dx\rangle = \tfrac12\mu\omega^2A^2 = \tfrac12(0.005)(400\pi)^2(0.005)^2\).
That is \(0.0025 \times 160\,000\,\pi^2 \times 0.000025 = 0.01\pi^2 \approx 0.0987\,\text{J/m}\).
(c) The wavelength is \(\lambda = 2\pi/k\), and \(k = 4\pi\), so \(\lambda = 2\pi/4\pi = 0.5\,\text{m}\). The total energy in one wavelength is \(E_\lambda = \langle dE/dx\rangle\lambda = 0.0987 \times 0.5 = 0.0494\,\text{J}\).
On the same string, a wave of the same amplitude is sent with twice the frequency. What happens to the energy in one wavelength, \(E_\lambda\)?
In many gyms athletes train with battle ropes: two heavy ropes tied to a post, which they whip up and down in waves. The work of the arms goes into the energy of the waves. The energy in each metre of rope grows as \(\omega^2A^2\). Swinging twice as fast with the same arm movement puts four times the energy into each metre. That is why a fast, wide swing tires the arms so quickly.
- Thinking that the energy of a piece of string stays constant, as for a block on a spring.For a piece of string \(dK = dU\), and both change together. The piece passes its energy on.
- Using \(\langle\cos^2\rangle = 1\) or forgetting the average.The average of \(\cos^2\) over one period is \(\tfrac12\), so \(\langle dE/dx\rangle = \tfrac12\mu\omega^2A^2\).
- Writing \(\omega\) in hertz.\(\omega = 2\pi f\) in rad/s. Read it from the coefficient of \(t\) in \(\sin(kx - \omega t)\).
- Taking \(E_\lambda\) to grow as \(\omega^2\).\(\lambda\) falls as \(1/\omega\) at a fixed speed, so \(E_\lambda\) grows as \(\omega\).