Question

A wave packet with centre frequency $\omega$ is propagating is a dispersive medium with phase velocity of $1.5 \times {10^3}\;m{s^{ - 1}}$. 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?
A. $0.25 \times {10^3}\;m{s^{ - 1}}$
B. $1.0 \times {10^3}\;m{s^{ - 1}}$
C. $0.6 \times {10^3}\;m{s^{ - 1}}$
D. $0.75 \times {10^3}\;m{s^{ - 1}}$

Answer 1

Hint: A wave with some frequency propagates in a medium with some phase velocity. After a period, the frequency increased by $2\%$ and the phase velocity decreased by $3\%$. Then using the formula for the phase velocity and the formula for the group velocity. The group velocity of the packet can be calculated.

Useful formula:
The phase velocity of the wave is given by,
${v_p} = \dfrac{\omega }{k}$
Where, ${v_p}$ is the phase velocity of the wave, $\omega$ is the frequency of the wave and $k$ is the propagation coefficient with respect to the refractive index.

The group velocity of the wave is given by,
${v_g} = \dfrac{{d\omega }}{{dk}}$
Where, ${v_g}$ is the group velocity of the wave, $d\omega$ is the first order differential of frequency of the wave and $dk$ is the first order differential of propagation coefficient with respect to refractive index.

Given data:
The phase velocity of the wave, ${v_p} = 1.5 \times {10^3}\;m{s^{ - 1}}$
Increase in frequency of the wave, $\dfrac{{d\omega }}{\omega } = 2\%$
Decrease in phase velocity of the wave, $\dfrac{{d{v_p}}}{{{v_p}}} = - 3\%$

Step by step solution:
The phase velocity of the wave is given by,
${v_p} = \dfrac{\omega }{k}\;..........................................\left( 1 \right)$
By applying partial differentiation on equation (1), we get
$\dfrac{{d{v_p}}}{{{v_p}}} = \dfrac{{d\omega }}{\omega } - \dfrac{{dk}}{k}$
Rearranging above equation, we get
$\dfrac{{dk}}{k} = \dfrac{{d\omega }}{\omega } - \dfrac{{d{v_p}}}{{{v_p}}}$
By substituting the given values in above equation, we get
$\dfrac{{dk}}{k} = 2\% - \left( { - 3\% } \right) \\\ \dfrac{{dk}}{k} = 5\% \\\$

The group velocity of the wave is given by,
${v_g} = \dfrac{{d\omega }}{{dk}}\;.........................................\left( 2 \right)$
By multiplying and dividing the equation (2) by ${v_p}$, we get
${v_g} = \dfrac{{{v_p}}}{{{v_p}}} \times \dfrac{{d\omega }}{{dk}}\;$
Substitute the value of ${v_p}$in the denominator,
${v_g} = \dfrac{{{v_p}}}{{\left( {\dfrac{\omega }{k}} \right)}} \times \dfrac{{d\omega }}{{dk}}\; \\\ {v_g} = {v_p} \times \left( {\dfrac{k}{\omega }} \right)\dfrac{{d\omega }}{{dk}}\; \\\ {v_g} = {v_p} \times \dfrac{{\left( {\dfrac{{d\omega }}{\omega }} \right)}}{{\left( {\dfrac{{dk}}{k}} \right)}}\;\;......................................\left( 3 \right) \\\$
Substitute the given values in equation (3), we get
${v_g} = \left( {1.5 \times {{10}^3}\;m{s^{ - 1}}} \right) \times \left( {\dfrac{{2\% }}{{5\% }}} \right) \\\ {v_g} = \left( {0.3 \times {{10}^3}} \right) \times 2 \\\ {v_g} = 0.6 \times {10^3}\;m{s^{ - 1}} \\\$

Hence, the option (C) is correct.

Note: The value of propagation coefficient $k$ is constant, with respect to the refractive index of the medium. The relation between the propagation coefficient and refractive index is given by $k = n{k_0}$, where n is the refractive index and ${k_0}$ is the propagation constant of the medium on which the wave propagates. The group velocity of the wave is the derivative of the phase velocity of the wave.