91影视

Write the expression forthe wave velocity for deepwater waves.

${\mathit{v}}_{\mathbf{p}}\mathbf{=}\mathit{伪}\sqrt{\mathbf{位}}$ 鈥︹�� (1)

Here, $伪$is a constant and $位$is the wavelength.

Write the expression for phase velocity.

${\mathit{v}}_{\mathbf{p}}\mathbf{=}\frac{\mathbf{蝇}}{\mathbf{k}}$ 鈥︹�� (2)

Here$蝇$, is the angular velocity and $k$is the wave number.

(a)

Equate equations (1) and (2).

鈥︹�� (3)

$\begin{array}{c}\frac{蝇}{k}=伪\sqrt{位}\\ 蝇=k伪\sqrt{位}\end{array}$ 鈥︹�� (3)

Write the relation between the wavenumber in terms of wavelength.

$\begin{array}{c}k=\frac{2蟺}{位}\\ 位=\frac{2蟺}{k}\end{array}$

Substitute$位=\frac{2蟺}{k}$ in equation (3).

$\begin{array}{c}蝇=k伪\sqrt{\frac{2蟺}{k}}\\ 蝇=\sqrt{2蟺}伪\sqrt{k}\end{array}$

Write the equation for the group velocity.

$v_g=\frac{d蝇}{dk}$

Substitute i$蝇=\sqrt{2蟺}伪\sqrt{k}$n the above equation.

$\begin{array}{c}{v}_g=\frac{d}{dk}\left(\sqrt{2蟺}伪\sqrt{k}\right)\\ v_g=\sqrt{2蟺}伪\frac{d}{dk}\left(\sqrt{k}\right)\\ v_g=\sqrt{2蟺}伪\frac{1}{2}{\left(k\right)}^{鈭�\frac{1}{2}}\end{array}$

On further solving, the above equation becomes,

$\begin{array}{c}{v}_g=\frac{\sqrt{2蟺}伪}{2}\frac{1}{\sqrt{k}}\\ v_g=\frac{伪}{2}\sqrt{\frac{2蟺}{k}}\\ v_g=\frac{伪}{2}\sqrt{位}\\ v_g=\frac{v_p}{2}\end{array}$

Therefore, it is proved that the wave velocity of deep water waves is twice the group velocity.

(b)

From the given problem, the equation is given as:

$蠄(x,t)=A{e}^{i(px鈭�Et)/\mathrm{鈩�}}$ 鈥︹�� (4)

Write the spatial representation of a wave.

$蠄(x,t)=A{e}^{i(kx鈭�蝇t)}$ 鈥︹�� (5)

Equate equations (4) and (5).

$\begin{array}{l}A{e}^{i(px鈭�Et)/\mathrm{鈩�}}=A{e}^{i(kx鈭�蝇t)}\\ i\frac{(px鈭�Et)}{\mathrm{鈩�}}=i(kx鈭�蝇t)\\ k=\frac{p}{\mathrm{鈩�}}\\ 蝇=\frac{E}{\mathrm{鈩�}}\end{array}$

Write the expression for the wave velocity.

$v_p=\frac{蝇}{k}$

Substitute$蝇=\frac{E}{\mathrm{鈩�}}$ in the above expression.

$\begin{array}{c}{v}_p=\frac{(E/\mathrm{鈩�})}{p/\mathrm{鈩�}}\\ v_p=\frac{E}{p}\\ v_p=\frac{p^2/2m}{p}\\ v_p=\frac{\mathrm{鈩�}k}{2m}\end{array}$ 鈥︹�� (6)

Write the expression for the group velocity.

$v_g=\frac{d蝇}{dk}$

Substitute$蝇=\frac{E}{\mathrm{鈩�}}$ in the above expression.

$\begin{array}{c}{v}_g=\frac{d}{dk}\left(\frac{E}{\mathrm{鈩�}}\right)\\ =\frac{d}{dk}\left(\frac{p^2}{2m\mathrm{鈩�}}\right)\\ =\frac{d}{dk}\left(\frac{{\mathrm{鈩�}}^2{k}^2}{2m\mathrm{鈩�}}\right)\end{array}$

On further solving, the above equation becomes,

$\begin{array}{c}{v}_g=\frac{\mathrm{鈩�}}{2m}\frac{d}{dk}\left(k^2\right)\\ v_g=\frac{\mathrm{鈩�}}{2m}\left(2k\right)\\ v_g=\frac{\mathrm{鈩�}k}{m}\end{array}$ 鈥︹��. (7)

From equations (6) and (7).

$\begin{array}{c}{v}_p=\frac{\mathrm{鈩�}k}{2m}\\ v_p=\frac{v_g}{2}\end{array}$

Hence, the phase velocity is half of the group velocity.

Group velocity corresponds to the classical speed of the particle. This can be understood as follows.

Write the expression for the classical speed.

$v_c=\frac{p}{m}$

Substitute$p=\mathrm{鈩�}k$ in the above expression.

$\begin{array}{c}{v}_c=\frac{\mathrm{鈩�}k}{m}\\ v_c=v_g\end{array}$

Hence, the from the equation the wave velocity do not represent the classical speed of the particle.

Therefore, the group velocity and the wave velocity is $\frac{\mathrm{鈩�}k}{m}$and the group velocity corresponds to the classical speed.