Question

4.5 Wave Packets 27. Use a distribution of wave numbers of constant ampli- tude in a range Ak about ko: A(K) = 4, ko - Ak sk Sko + Ak = 0 otherwise and obtain Eq. 4.24 from Eq. 4.23.

Answer 1

So, what we need to do is simply evaluate the integral

y(x) = | A(k) cos(kx)dk

for the distribution of wave numbers of constant amplitude $A_0$ in the range $\Delta k$ about $k_0$ i.e.

The distribution looks like this in the wave number space:

Δk

Now, the evaluation of the Integral:

$y(x) = \int_{k=\infty}^\infty A(k) \cos (kx) dk$

Now, as there is constant wave amplitude value of $A=A_0$ at the range of from $k_0-\Delta k/2$ to $k_0+\Delta k/2$ , the above integral is reduced to:

$y(x)= \int_{k=k_0-\Delta k/2}^{k_0+\Delta k/2} A_0 \cos (kx) dk$

Here, as $A_0$ is constant throughout the integral, we can put it out of the integral, and simply evaluate the integration of $\cos kx$

$\begin{align} y(x) & = A_0 \left. \frac {\sin kx}{x} \right|_{k_0-\Delta k/2}^{k_0 + \Delta k /2} \nonumber \\ & = \frac {A_0}{x} \left[ \sin\left\{\left(k_0 + \frac {\Delta k}2 \right)x\right\} + \sin \left\{ \left( k_0 - \frac {\Delta k}{2} \right )x \right\} \right ] \nonumber \end{align}$

Now, we will use the identity

$\sin C - \sin D = 2\sin \left(\frac {C-D}{2} \right ) \cos \left(\frac {C+D}{2} \right )$

So that, we have from the integral:

$\begin{align} y(x) & = \frac {2A_0}x \sin \left(\frac {k_0 + \frac {\Delta k}{2} - k_0 + \frac {\Delta k}2}{2}x \right ) \cos \left( \frac{k_0+\frac {\Delta k}{2}+k_0 - \frac{\Delta k}{2}}{2}x \right ) \nonumber \\ & = \frac {2A_0}{x} \sin \left(\frac {\Delta k}{2}x \right ) \cos (k_0x) \nonumber \end{align}$

Thus, we have the desired result.

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