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Chapter 6: Problem 19

The dispersion law is $$ \omega^{2}=g k+\frac{T k^{3}}{\rho} $$ where \(g=980, T=72\), and \(\rho=1.0\) (all in CGS units). Derive formulas for the group velocity and for the phase velocity. Show that the group velocity equals the phase velocity when gk and \(T k^{3} / \rho\) are equal and that this occurs for wavelength 1.7 \(\mathrm{cm}\) and velocity \(23.1 \mathrm{~cm} / \mathrm{sec} .\) Show that for surface-tension waves, i.e., waves with wavelength very short compared with \(1.7 \mathrm{~cm}\), the group velocity is \(1.5\) times the phase velocity. Show that for gravity waves, i.e., waves with wavelength long compared with \(1.7 \mathrm{~cm}\), the group velocity is half the phase velocity. Extend Table \(6.1\), Sec. 6,2 to include wavelength 128 meters and 256 meters. Give the wave velocities in \(\mathrm{km} / \mathrm{hr}\) as well as \(\mathrm{cm} / \mathrm{sec}\). (To observe frequencies as low as four or five per minute, go to a beach on a sheltered ocean bay on a day when there is no strong offshore wind. Then the only waves are those coming from far out at sea.)

Short Answer

Expert verified
Group velocity equals phase velocity at 1.7 cm wavelength for velocity 23.1 cm/s. Surface-tension waves have \( v_g = 1.5 v_p \); gravity waves have \( v_g = 0.5 v_p \). Extend table for 128m and 256m wavelengths.

Step by step solution

01

Express Period as Function of k

The given dispersion relation is \( \omega^2 = g k + \frac{T k^3}{\rho} \). First, take the square root to express \( \omega \) as a function of \( k \): \[ \omega = \sqrt{g k + \frac{T k^3}{\rho}}.\]
02

Calculate Phase Velocity

Phase velocity \( v_p \) is given by \( v_p = \frac{\omega}{k} \). Substitute the expression for \( \omega \) to get: \[ v_p = \frac{1}{k} \sqrt{g k + \frac{T k^3}{\rho}} = \sqrt{\frac{g}{k} + \frac{T k^2}{\rho}}.\]
03

Derive Group Velocity

The group velocity \( v_g \) is defined as the derivative of \( \omega \) with respect to \( k \): \[ v_g = \frac{d\omega}{dk}.\] Using the chain rule, \( \frac{d\omega}{dk} = \frac{1}{2\omega} \left( g + \frac{3T k^2}{\rho} \right) \). Express \( v_g \) as: \[ v_g = \frac{g + \frac{3T k^2}{\rho}}{2\sqrt{g k + \frac{T k^3}{\rho}}}. \]
04

Equate Group and Phase Velocities

Equate \( v_g \) and \( v_p \), \( \sqrt{\frac{g}{k} + \frac{T k^2}{\rho}} = \frac{g + \frac{3T k^2}{\rho}}{2\sqrt{g k + \frac{T k^3}{\rho}}} \). Solve for \( k \) where \( gk = \frac{T k^3}{\rho} \) implies \( k^2 = \frac{g \rho}{T} \). Calculate \( \lambda \) using \( \lambda = \frac{2 \pi}{k} \) to show \( \lambda \approx 1.7 \text{ cm} \).
05

Calculate Velocities for Wavelength 1.7 cm

Substitute \( k \) back into the phase velocity \( v_p \) for \( \lambda = 1.7 \text{ cm} \) to find the velocities. Using the dispersion relation, calculate the velocity to confirm it is approximately \( 23.1 \text{ cm/sec} \).
06

Surface-Tension Wave Velocities

For short wavelengths (\( \lambda \ll 1.7 \, \text{cm} \)), the term \( \frac{T k^3}{\rho} \) dominates. For \( \omega^2 = \frac{T k^3}{\rho} \), solve to show \( v_g = 1.5 v_p \) for large \( k \).
07

Gravity Wave Velocities

For long wavelengths, (\( \lambda \gg 1.7 \, \text{cm} \)), \( gk \) is dominant, simplifying to \( \omega^2 = gk \). Hence, \( v_g = \frac{1}{2} v_p \) when \( gk \gg \frac{T k^3}{\rho} \).
08

Extend Table for Long Wavelengths

Given \( g = 980 \), find \( v_p \) and \( v_g \) for \( \lambda = 128 \text{ m and } 256 \text{ m} \), using the relation \( v_p \approx \sqrt{gL} \) where \( L \) is the wavelength. Convert velocities to \( km/hr \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Group Velocity
Group velocity is a fundamental concept in wave physics. It describes how the overall shape of a wave packet travels through space. In essence, it is the velocity at which the envelope of the wave group moves. This is particularly important because it gives us the speed at which energy or information travels along the wave.

For a dispersive medium where different frequencies travel at different speeds, group velocity (\( v_g \)) is calculated as the derivative of the angular frequency (\( \omega \)) with respect to the wave number (\( k \)):
  • \( v_g = \frac{d\omega}{dk} \)
In the given problem, after computing this derivative, we obtained:\[v_g = \frac{g + \frac{3T k^2}{\rho}}{2\sqrt{g k + \frac{T k^3}{\rho}}}\]This formula helps determine the speed at which energy in a wave packet of a fluid moves, depending on the balance of gravitational and surface tension forces.
Phase Velocity
Phase velocity is another crucial metric of wave motion. This velocity describes the speed at which individual wave crests and troughs move. Specifically, phase velocity (\( v_p \)) is the speed at which a particular phase of the wave, like a crest, travels through space.

Mathematically, it is given by the ratio of the angular frequency to the wave number:
  • \( v_p = \frac{\omega}{k} \)
For surface-tension and gravity waves, this formula becomes:\[v_p = \sqrt{\frac{g}{k} + \frac{T k^2}{\rho}}\]This shows how phase velocity depends on both gravitational constants and surface tension. For the scenario where phase and group velocities equate, both surface tension and gravity equally affect the motion, corresponding to certain critical wavelengths.
Surface-Tension Waves
Surface-tension waves exist when the wavelengths are very short relative to certain critical values, such as 1.7 cm in the exercise example. In this case, the surface tension significantly influences wave properties, making them surface-tension waves.

For such waves, the dispersion relation simplifies mainly to surface tension effects: \( \omega^2 = \frac{T k^3}{\rho} \). The group velocity becomes notably different from the phase velocity. Specifically:
  • \( v_g = 1.5 v_p \)
Indicating that the entire wave packet, carrying energy, moves faster than individual wave crests. This equality holds at very short wavelengths where surface tension dominates the wave dynamics over gravitational forces.
Gravity Waves
Gravity waves are typical in cases where waves have longer wavelengths compared with the critical wavelength, such as significantly longer than 1.7 cm. These waves are primarily influenced by gravity more than surface tension.

In this state, gravity becomes the driving force behind the wave motion, simplifying the dispersion relation to: \( \omega^2 = g k \). This means gravity predominantly determines the wave's characteristics. For such gravity waves:
  • \( v_g = \frac{1}{2} v_p \)
This indicates that the group velocity is half of the phase velocity. Hence, energy or information moves more slowly than the individual phases of the wave, a characteristic trait of gravity-dominated ocean waves, often seen in the open sea.

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Most popular questions from this chapter

Assume the dispersion law is given by a single resonance, and neglect damping; i.e., assume $$ c^{2} k^{2}=\omega^{2}\left(1+\frac{\omega_{p}{ }^{2}}{\omega_{0}^{2}-\omega^{2}}\right), \quad \omega_{p}^{2}=\frac{4 \pi N e^{2}}{m} $$ where \(N\) is the number of resonating electrons per unit volume. (a) Sketch the square of the index of refraction, \(n^{2}\), versus \(\omega\), for \(0 \leqq \omega<\infty\). The important features are the value and slope at \(\omega=0\), at \(\omega\) slightly less than \(\omega_{0}\) and slightly greater than \(\omega_{0}\), at \(\omega=\sqrt{\omega_{0}^{2}+\omega_{p}^{2}}\), and at infinity. How do you interpret the region where \(n^{2}\) is negative? The region near \(\omega_{0}\) ? (b) Derive the following formula for the square of the group velocity: $$ \left(\frac{v_{g}}{c}\right)^{2}=\frac{1+\frac{\omega_{p}^{2}}{\omega_{0}^{2}-\omega^{2}}}{\left[1+\frac{\omega_{p}^{2} \omega_{0}^{2}}{\left(\omega_{0}^{2}-\omega^{2}\right)^{2}}\right]^{2}} $$ Sketch \(\left(v_{g} / c\right)^{2}\) versus \(\omega .\) Show that \(\left(v_{d} / c\right)^{2}\) is always less than unity, as required by the theory of relativity. Show that \(v_{0}{ }^{2}\) is negative in the same frequency region where \(n^{2}\) is negative. For what frequency is the group velocity greatest? What is the group velocity at that frequency?

Consider a function \(f(t)\) that is zero for negative \(t\) and equals \(\exp (-t / 2 \tau)\) for \(t \geqq 0 .\) Find its Fourier coefficients \(A(\omega)\) and \(B(\omega)\) in the continuous superposition $$ f(t)=\int_{0}^{\infty}[A(\omega) \sin \omega t+B(\omega) \cos \omega t] d \omega $$

Suppose \(\psi(t)\) is zero outside of the interval from \(t_{1}\) to \(t_{2}\), which has duration \(t_{2}-t_{1}=\Delta t\) and central value \(\frac{1}{2}\left(t_{1}+t_{2}\right)=t_{0}\). Suppose \(\psi(t)\) is equal to cos \(\omega_{0}\left(t-t_{0}\right)\) within that interval. (a) Show that \(\psi(t)\) can be Fourier-analyzed as follows: $$ \begin{aligned} \psi(t) &=\int_{0}^{\infty} B(\omega) \cos \omega\left(t-t_{0}\right) \\ \pi B(\omega) &=\frac{\sin \left[\left(\omega_{0}+\omega\right) \frac{1}{2} \Delta t\right]}{\omega_{0}+\omega}+\frac{\sin \left[\left(\omega_{0}-\omega\right) \frac{1}{2} \Delta t\right]}{\omega_{0}-\omega} \end{aligned} $$ (b) Show that if \(\Delta t\) is much shorter than the period of any frequency that we can measure or are interested in, then \(\pi B(\omega)\) has the constant value \(\Delta t\). (c) Show that if \(\Delta t\) contains many oscillations, i.e., if \(\omega_{0} \Delta t \geqslant 1\), then, for \(\omega\) sufficiently near \(\omega_{0}, B(\omega)\) is given essentially by the second term only: $$ \pi B(\omega) \approx \frac{\sin \left[\left(\omega_{D}-\omega\right) \frac{1}{2} \Delta t\right]}{\omega_{0}-\omega}, \quad\left|\omega_{0}-\omega\right| \ll\left|\omega_{0}+\omega\right| $$ (d) Sketch \(\psi(t)\) and \(B(\omega)\) for part \((c)\). This problem can help us to understand collision broadening of spectral lines. An undisturbed atom emitting almost monochromatic visible light has a mean decay time of about \(10^{-8} \mathrm{sec}\), and thus the Fourier spectrum of its radiation has a bandwidth \(\Delta \nu\) of about \(10^{8}\) cps. If the atoms are in a gas- discharge-tube light source, then it turns out that the bandwidth of the emitted light (called "linewidth" in optics) is about \(10^{9} \mathrm{cps}\), rather than \(10^{8} \mathrm{cps}\). Part of the reason for this "line broadening" is the fact that the atoms do not radiate in a free and undisturbed manner; they collide. A collision results in a sudden change in amplitude or phase constant or both. That is similar to the situation illustrated by the truncated harmonic oscillator, A given atom may spend most of its time "unexcited." Occasionally it is excited into an oscillatory motion of the optical (valence) electrons (we are speaking classically; a more accurate picture requires quantum mechanics). The atom begins to oscillate as a damped harmonic oscillation with decay time of order \(10^{-8}\) sec. However, within a time \(\Delta t\) of about \(10^{-9} \mathrm{sec}\) (in a typical gas- tube light source), it has a collision that truncates the oscillation in some random way. If one adds the light from many such sources, the bandwidth \(\Delta v\) will be given by \(\Delta p \approx(1 / \Delta t) \approx 10^{9} \mathrm{cps}\).

Show that for light of index \(n(\lambda)\), $$ \frac{1}{v_{g}}=\frac{1}{v_{\varphi}}-\frac{1}{c} \lambda \frac{d n(\lambda)}{d \bar{\lambda}} $$ where \(\lambda\) is the vacuum wavelength of the light.

(a) One way to produce an amplitude-modulated carrier wave is to pass a current \(I=I_{0} \cos \omega_{0} t\) oscillating at the carrier frequency \(\omega_{0}\) through a resistance \(R\) which is not constant but has a component that varies at the modulation frequency \(\omega_{\text {mod }}\), that is, \(R=R_{0}\left(1+a_{m} \cos \omega_{\text {mod }} t\right)\). (In a "carbon-granule" microphone the resistance is modulated by the motion of a diaphragm, which compresses the carbon granules that provide the resistance.) The voltage \(V=I R\) across the resistor is an amplitude-modulated carrier wave. Find the expression for \(V\) in terms of a stperposition of carrier (frequency \(\omega_{0}\) ), upper sideband (frequency \(\omega_{0}+\omega_{\text {mod }}\) ), and lower sideband (frequency \(\left.\omega_{0}-\omega_{\text {mod }}\right)\). (b) Alternatively, suppose we happen to start with two voltages, one oscillating at the carrier frequency, the other at the modulation frequency. The problem is this: How can you physically combine these two voltages, \(V_{0}=A_{0} \cos \omega_{0} t\), and \(V_{m}=A_{m} \cos \omega_{\text {mod }} t\) in such a way as to produce an amplitude-modulated carrier wave? First, suppose you merely superpose the two voltages, i.e., you put them both on the broadcasting antenna. Will this work? (c) Next, suppose that the voltages in part \((b)\), after being superposed, are then applied to the input of a voltage amplifier. (For example they may be applied between control grid and cathode of a radio tube.) Suppose that the amplifier is a linear amplifier, i.e., its output (for example the plate-to- cathode voltage of the tube) is proportional to the input. Will this work? (d) Finally, suppose that the amplifier output has both a linear and a quadratic component, as follows: $$ V_{\text {out }}=A_{1} V_{\text {in }}+A_{2}\left(V_{\text {in }}\right)^{2} $$ Let \(V_{\text {in }}=V_{0}+V_{m}\) as defined in part \((b)\). Show that because of the nonlinear quadratic term \(A_{2}\left(V_{i \mathrm{in}}\right)^{2}\) the amplifier output includes, among other things, an amplitudemodulated carrier wave, with modulation amplitude proportional to \(A_{m} .\) (e) The amplitude-modulated carrier wave in \((d)\) contributes Fourier components with frequencies \(\omega_{0}, \omega_{0}+\omega_{\mathrm{mad}}\), and \(\omega_{0}-\omega_{\text {mod }}\). What other frequency components are there in \(V_{\text {eut }}\) ? Make a diagram showing a complete frequency spectrum of the amplifier output. Describe how you could get rid of these other (undesired) components, using bandpass filters. Suppose that \(\omega_{\text {rued }}\) is small compared with \(\omega_{0}\). How selective do the filters have to be?
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