# Sum of waves of different frequencies and group velocity

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 2 7 - 46 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 2.7a

A pulse composed of two frequencies, $\omega _{0}\pm {\mathrm {\Delta } }\omega$ , can be represented by factors involving the sum and difference of the two frequencies. If the two components have the same amplitudes, we can write

{\begin{aligned}A\cos \left(\kappa _{1}x-\omega _{1}t\right),{\quad }A\cos \left(\kappa _{2}x-\omega _{2}t\right),\end{aligned}} where $\omega _{1}=\omega _{0}+\Delta \omega ,\omega _{2}=\omega _{0}-\Delta \omega ,k_{0}=2\pi /\lambda _{0}=\omega _{0}/V$ ,

{\begin{aligned}\kappa _{1}&\approx \kappa _{0}+\Delta \kappa \approx \left(\omega _{0}+\Delta \omega \right)/V,\ \mathrm {and} \\\kappa _{2}&\approx \kappa _{0}-\Delta \kappa \approx \left(\omega _{0}-\Delta \omega \right)/V.\end{aligned}} Show that the composite wave is given approximately by the expression

{\begin{aligned}B\cos(\kappa _{0}x-\omega _{0}t),\end{aligned}} where $B=2A\cos \left\{{\mathrm {\Delta } }\kappa \left[x-\left({\mathrm {\Delta } }\omega /{\mathrm {\Delta } }\kappa \right)t\right]\right\}$ .

### Background

When different frequency components in a pulse have different phase velocities $V$ (the velocity with which a given frequency travels), the pulse changes shape as it moves along. The group velocity $U$ is the velocity with which the envelope of the pulse travels.

The envelope of a pulse comprises two mirror-image curves that are tangent to the waveform at the peaks and troughs, and therefore define the general shape of the pulse.

### Solution

Adding the two components and using the identity

{\begin{aligned}\cos \theta +\cos \phi =2\cos \left[{\frac {1}{2}}\left(\theta +\phi \right)\right]\cos \left[{\frac {1}{2}}\left(\theta -\phi \right)\right],\end{aligned}} we get for the composite wave

{\begin{aligned}2A\cos \left\{{\mathrm {\Delta } }\kappa \left[x-\left({\mathrm {\Delta } }\omega /{\mathrm {\Delta } }\kappa \right)t\right]\right\}\cos \left(\kappa _{0}x-\omega _{0}t\right)=B\cos \left(\kappa _{0}x-\omega _{0}t\right).\end{aligned}} ## Problem 2.7b

Why do we regard $B$ as the amplitude? Show that the envelope of the pulse is the graph of $B$ plus its reflection in the $x$ -axis.

### Solution

The solution in (a) can be written $\psi =B\psi _{0}$ . We regard $B$ as the amplitude for two reasons: (1) $B$ repeats every time that $x$ increases by $2\pi /{\mathrm {\Delta } }\kappa$ or each time that $t$ increase by $2\pi /{\mathrm {\Delta } }\omega$ . But ${\mathrm {\Delta } }\kappa \ll \kappa _{0}$ , ${\mathrm {\Delta } }\omega \ll \omega _{0}\}$ , so $B$ must repeat more slowly than $\psi _{0}$ . (2) Each time that $\psi _{0}$ attains its limiting values, $\pm 1$ , $\psi$ has the value $\pm B$ and therefore never exceeds $\left|B\right|$ ; thus the curves of $+\left|B\right|$ and $-\left|B\right|$ pass through the maxima and minima of $\psi _{0}$ and therefore constitute the envelope.

## Problem 2.7c

Show that the envelope moves with the group velocity $U$ where

 {\begin{aligned}U={\frac {{\mathrm {\Delta } }\omega }{{\mathrm {\Delta } }\kappa }}\approx {\frac {\mathrm {d} \omega }{\mathrm {d} \kappa }}\approx V-{\mathrm {\lambda } }{\frac {\mathrm {d} V}{\mathrm {d} {\mathrm {\lambda } }}}=V+\omega {\frac {\mathrm {d} V}{\mathrm {d} \omega }}\end{aligned}} (2.7a)

(see Figure 2.7a).

### Solution

If we consider the quantity $B$ as a wave superimposed on the primary wavelet $\cos \left(\kappa _{0}x-\omega _{0}t\right)$ , comparison with the basic wave type $f(x-Vt)$ of problem 2.5a shows that $\left[x-\left({\mathrm {\Delta } }\omega /{\mathrm {\Delta } }\kappa \right)t\right]$ takes the place of $(x-Vt)$ , i.e., $\left({\mathrm {\Delta } }\omega /{\mathrm {\Delta } }\kappa \right)$ is the velocity $U$ with which the envelope travels. In the limit, $U$ is given by

{\begin{aligned}U=\mathrm {d} \omega /\mathrm {d} {\kappa }=\mathrm {d} f/\mathrm {d} \left(1/{\mathrm {\lambda } }\right)=\left(\mathrm {d} f/\mathrm {d} {\mathrm {\lambda } }\right)\left[\mathrm {d} {\mathrm {\lambda } }/\mathrm {d} \left(1/{\mathrm {\lambda } }\right)\right],\end{aligned}} where $f$ is the frequency. The result is

{\begin{aligned}U=-{\mathrm {\lambda } }^{2}{\frac {\mathrm {d} f}{\mathrm {d} \lambda }}.\end{aligned}} We introduce the phase velocity $V$ by noting that $V={\mathrm {\lambda } }f$ , so

{\begin{aligned}\mathrm {d} V={\mathrm {\lambda } }\mathrm {d} f+f\mathrm {d} \lambda ,\;{\mathrm {\lambda } }{\frac {\mathrm {d} f}{\mathrm {d} {\mathrm {\lambda } }}}={\frac {\mathrm {d} V}{\mathrm {d} {\mathrm {\lambda } }}}-f.\end{aligned}} Thus,

{\begin{aligned}U=-{\mathrm {\lambda } }\left({\mathrm {\lambda } }{\frac {\mathrm {d} V}{\mathrm {d} {\mathrm {\lambda } }}}\right)={\mathrm {\lambda } }f-{\mathrm {\lambda } }{\frac {\mathrm {d} V}{\mathrm {d} {\mathrm {\lambda } }}}=V-{\mathrm {\lambda } }{\frac {\mathrm {d} V}{\mathrm {d} \lambda }}.\end{aligned}} To replace the derivative $\mathrm {d} V/\mathrm {d} {\mathrm {\lambda } }$ with $\mathrm {d} V/\mathrm {d} f$ , we find the relation between ${\mathrm {\lambda } }/\mathrm {d} {\mathrm {\lambda } }$ and $f/\mathrm {d} f$ ; to do this we let $V$ be constant so that

{\begin{aligned}V={\mathrm {\lambda } }f,\;\mathrm {d} V=0={\mathrm {\lambda } }\mathrm {d} f+f\mathrm {d} \lambda ,\end{aligned}} hence,

{\begin{aligned}{\mathrm {\lambda } }/\mathrm {d} {\mathrm {\lambda } }=-f/\mathrm {d} f,\ \mathrm {and} \ U=V+f{\frac {\mathrm {d} V}{\mathrm {d} f}}=V+\omega {\frac {\mathrm {d} V}{\mathrm {d} \omega }}.\end{aligned}} 