# 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, ${\displaystyle \omega _{0}\pm {\mathrm {\Delta } }\omega }$, 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

{\displaystyle {\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 ${\displaystyle \omega _{1}=\omega _{0}+\Delta \omega ,\omega _{2}=\omega _{0}-\Delta \omega ,k_{0}=2\pi /\lambda _{0}=\omega _{0}/V}$,

{\displaystyle {\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

{\displaystyle {\begin{aligned}B\cos(\kappa _{0}x-\omega _{0}t),\end{aligned}}}

where ${\displaystyle B=2A\cos \left\{{\mathrm {\Delta } }\kappa \left[x-\left({\mathrm {\Delta } }\omega /{\mathrm {\Delta } }\kappa \right)t\right]\right\}}$.

### Background

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

{\displaystyle {\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

{\displaystyle {\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 ${\displaystyle B}$ as the amplitude? Show that the envelope of the pulse is the graph of ${\displaystyle B}$ plus its reflection in the ${\displaystyle x}$-axis.

### Solution

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

## Problem 2.7c

Show that the envelope moves with the group velocity ${\displaystyle U}$ where

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

{\displaystyle {\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}}}

Figure 2-7a  Illustrating group and phase velocity.

where ${\displaystyle f}$ is the frequency. The result is

{\displaystyle {\begin{aligned}U=-{\mathrm {\lambda } }^{2}{\frac {\mathrm {d} f}{\mathrm {d} \lambda }}.\end{aligned}}}

We introduce the phase velocity ${\displaystyle V}$ by noting that ${\displaystyle V={\mathrm {\lambda } }f}$, so

{\displaystyle {\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,

{\displaystyle {\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 ${\displaystyle \mathrm {d} V/\mathrm {d} {\mathrm {\lambda } }}$ with ${\displaystyle \mathrm {d} V/\mathrm {d} f}$, we find the relation between ${\displaystyle {\mathrm {\lambda } }/\mathrm {d} {\mathrm {\lambda } }}$ and ${\displaystyle f/\mathrm {d} f}$; to do this we let ${\displaystyle V}$ be constant so that

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

hence,

{\displaystyle {\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}}}