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A wave packet with center frequency $\omega$ is propagating in dispersive medium with phase velocity of $1.5 \times 10^3 \mathrm{~m} / \mathrm{s}$. When the frequency $\omega$ is increased by $2 \%$, the phase velocity is found to decrease by $3 \%$. What is the group velocity of the wave packet?
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The correct answer is:
$0.6 \times 10^3 \mathrm{~m} / \mathrm{s}$
Phase velocity, $v_p=\frac{\omega}{k}$
$\begin{aligned} & \frac{d v_p}{v_p}=\frac{d \omega}{\omega}-\frac{d k}{k} \\ & \therefore \quad \frac{d k}{k}=\frac{d \omega}{\omega}-\frac{d v_p}{v_p}=2 \%-(-3 \%)=5 \%\end{aligned}$
Group velocity, $v_g=\frac{d \omega}{d k}$
$\begin{aligned} & \frac{d \omega}{d k}=v_p\left(\frac{d \omega / \omega}{d k / k}\right)=1.5 \times 10^3\left(\frac{2}{5}\right) \\ & =0.6 \times 10^3 \mathrm{~m} / \mathrm{s}\end{aligned}$
$\begin{aligned} & \frac{d v_p}{v_p}=\frac{d \omega}{\omega}-\frac{d k}{k} \\ & \therefore \quad \frac{d k}{k}=\frac{d \omega}{\omega}-\frac{d v_p}{v_p}=2 \%-(-3 \%)=5 \%\end{aligned}$
Group velocity, $v_g=\frac{d \omega}{d k}$
$\begin{aligned} & \frac{d \omega}{d k}=v_p\left(\frac{d \omega / \omega}{d k / k}\right)=1.5 \times 10^3\left(\frac{2}{5}\right) \\ & =0.6 \times 10^3 \mathrm{~m} / \mathrm{s}\end{aligned}$
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