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Lecture 8: Traveling & Standing Waves

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Lecture Topics

A white line formszig zags on a black background.
  • Traveling Waves
  • Boundary Conditions
  • Standing Waves
  • Sound
  • Energy in Waves

Learning Objectives

By the end of this lecture, you should:

  • define the wavelength and show why the wave repeats on this scale.
  • understand the use of the kx-ωt form of the wave solution.
  • be able to superpose waves, including how to form a standing wave.
  • determine the normal mode frequencies for continuous media with fixed ends.
  • define nodes, antinodes, fundamental, harmonics, eigenvalue, eigenstate (eigenfunction).
  • understand the boundary conditions for (over)pressure waves (sound) in an enclosure.
  • explain how to determine the speed of sound in a standing wave enclosure.
  • understand why energy is transported by waves without matter (necessarily) being transported.
  • calculate the power in a traveling wave.
  • understand that traveling waves carry momentum.
  • understand that the wave equation is different in different coordinate systems, and consequences.
  • apply boundary conditions to obtain the ratio of reflection and transmission coefficients at a boundary.

Lecture Activities

Check Yourself

  • Show by explicit differentiation that a periodic traveling wave \[y = A\sin \left[ {\frac{{2\pi }}{\lambda }(x - vt)} \right]\]satisfies the wave equation. Your steps may vary but one suggested way is shown as the answer.

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\[\begin{gathered} \frac{{\partial y}}{{\partial t}} = A\cos \left[ {\frac{{2\pi }}{\lambda }(x - vt)} \right]\left( { - \frac{{2\pi }}{\lambda }v} \right) = - A\frac{{2\pi }}{\lambda }v\cos \left[ {\frac{{2\pi }}{\lambda }(x - vt)} \right]  \\ \frac{{{\partial ^2}y}}{{\partial {t^2}}} = - A{\left( {\frac{{2\pi v}}{\lambda }} \right)^2}\sin \left[ {\frac{{2\pi }}{\lambda }(x - vt)} \right]  \\ \frac{{\partial y}}{{\partial x}} = A\cos \left[ {\frac{{2\pi }}{\lambda }(x - vt)} \right]\left( {\frac{{2\pi }}{\lambda }} \right) = A\frac{{2\pi }}{\lambda }v\cos \left[ {\frac{{2\pi }}{\lambda }(x - vt)} \right]  \\ \frac{{{\partial ^2}y}}{{\partial {x^2}}} = - A{\left( {\frac{{2\pi }}{\lambda }} \right)^2}\sin \left[ {\frac{{2\pi }}{\lambda }(x - vt)} \right]  \\ \end{gathered} \]

Comparison of the second and fourth lines shows that \[\frac{1}{{{v^2}}}\frac{{{\partial ^2}y}}{{\partial {t^2}}} = \frac{{{\partial ^2}y}}{{\partial {x^2}}}\]as the wave equation requires, so that \[y = A\sin \left[ {\frac{{2\pi }}{\lambda }(x - vt)} \right]\]is indeed a solution. As in the example of a Gaussian pulse, the chain rule has been applied in taking derivatives. This wave, however, is nominally of infinite extent in space (along x) and exists for all times.

 

  • Find under what condition the standing wave \(y = 2A\sin (kx)\cos (\omega t)\) satisfies the wave equation. Point out what might seem an inconsistency and interpret the seeming problem.

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To satisfy the wave equation we must have \[\frac{1}{{{v^2}}}\frac{{{\partial ^2}y}}{{\partial {t^2}}} = \frac{{{\partial ^2}y}}{{\partial {x^2}}}.\] Taking the derivatives, \[\begin{gathered} \frac{{\partial y}}{{\partial t}} = \omega 2A\sin (kx)\sin (\omega t)\quad \frac{{{\partial ^2}y}}{{\partial {t^2}}} = {\omega ^2}2A\sin (kx)\cos (\omega t) \ \\ \frac{{\partial y}}{{\partial x}} = k2A\cos (kx)\cos (\omega t)\quad \frac{{{\partial ^2}y}}{{\partial {x^2}}} = {k^2}2A\sin (kx)\cos (\omega t) \ \\ \end{gathered} \] So, \[\frac{{{k^2}}}{{{\omega ^2}}}\frac{{{\partial ^2}y}}{{\partial {t^2}}} = \frac{{{\partial ^2}y}}{{\partial {x^2}}}\] which is the wave equation provided \[\frac{1}{{{v^2}}} = \frac{{{k^2}}}{{{\omega ^2}}}\]or \({\omega ^2} = {k^2}{v^2}.\) The wave equation is satisfied if this condition, similar to that involving speed for traveling waves, is met. However the standing wave pattern does not move with any speed, it is stationary and oscillates in one place. The resolution of this seeming paradox is found in that the standing wave can be viewed as two oppositely moving waves, each of which does have an advancing phase, yet whose superposition does not move.

 

  • The strings on a guitar are all roughly the same length and for proper feel of playing, have to be under about the same tension. Describe how the different notes result from the strings, referencing speed of waves and the fact that the strings are fixed at both ends.

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The normal modes in this case have frequency in Hz \({f_n} = \frac{{nv}}{{2L}}\), where n is the harmonic number, v the speed of waves on the string, and L its length. The note will basically reflect the frequency of the fundamental (n = 1), and L is the same for all strings. So v must be made to vary. For a stretched string, \(v = \sqrt {\frac{T}{\mu }}\), where the tension T is to be the same for all strings. So the only thing that can be varied is the mass density μ, explaining why the lower notes come from thicker, heavier strings.

 

  • In a previous conceptual question, it was suggested that a person could tension a rope of 100 g/m with about 100 N of force by pulling on it. If the sound from such a rope was to be audible, how much would it need to be tensioned?

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To be audible the frequency would need to be above 20 Hz, so in the fundamental \[20 = \frac{{nv}}{{2L}} = \frac{v}{2} = \frac{1}{2}\sqrt {\frac{T}{\mu }} = \frac{1}{2}\sqrt {\frac{T}{{0.1}}}\] Solving for T gives 160 N, which is not much above what could be produced by pulling alone.

 

  • Describe clearly the propagation of kinetic and potential energy in a wave propagating in a string, with attention to the displacement of the string and the propagation direction.

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Kinetic energy in the string is due to transverse motion of the string perpendicular to the direction of propagation, which is along the string. The mass at any point moving sideways results in the usual \(\frac{1}{2}m{v^2}\) kinetic energy of a moving object, which in this case is a small length of string (each small length within a wavelength moves differently). Once the string has moved it is under increased tension since it is stretched, and this gives rise to propagating potential energy.

 

  • Rework the equations for reflection and transmission in strings to be in terms of mass density (recall that the tension T is constant at the junction of the strings). Check if the behavior is as expected.

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\[R = \frac{{{v_2} - {v_1}}}{{{v_2} + {v_1}}} = \frac{{\sqrt {\frac{T}{{{\mu _2}}}} - \sqrt {\frac{T}{{{\mu _1}}}} }}{{\sqrt {\frac{T}{{{\mu _2}}}} + \sqrt {\frac{T}{{{\mu _1}}}} }} = \frac{{\sqrt {{\mu _1}{\mu _2}} \left( {\sqrt {\frac{1}{{{\mu _2}}}} - \sqrt {\frac{1}{{{\mu _1}}}} } \right)}}{{\sqrt {{\mu _1}{\mu _2}} \left( {\sqrt {\frac{1}{{{\mu _2}}}} + \sqrt {\frac{1}{{{\mu _1}}}} } \right)}} = \frac{{\sqrt {{\mu _1}} - \sqrt {{\mu _2}} }}{{\sqrt {{\mu _1}} + \sqrt {{\mu _2}} }}\].\[{T_r} = \frac{{2{v_2}}}{{{v_2} + {v_1}}} = \frac{{2\sqrt {\frac{T}{{{\mu _2}}}} }}{{\sqrt {\frac{T}{{{\mu _2}}}} + \sqrt {\frac{T}{{{\mu _1}}}} }} = \frac{{2\sqrt {{\mu _1}{\mu _2}} \left( {\sqrt {\frac{1}{{{\mu _2}}}} } \right)}}{{\sqrt {{\mu _1}{\mu _2}} \left( {\sqrt {\frac{1}{{{\mu _2}}}} + \sqrt {\frac{1}{{{\mu _1}}}} } \right)}} = \frac{{2\sqrt {{\mu _1}} }}{{\sqrt {{\mu _1}} + \sqrt {{\mu _2}} }}\] If a wave goes from a light rope to a heavy one (i.e. less mass per unit length to more mass per unit length) then the R equation gives a negative sign, and vice versa. This is as expected. The Tr equation is less intuitive in checking the general behavior.

 

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