Is Wave Velocity Always Double the Group Velocity in Deep Water?

[pt176900]

Problem 9.22 from Griffiths:

Show that in deep water (where the depth is greater than the wavelength) the wave velocity is twice that of the group velocity. where Vw = w/k and Vg = dw/dk

where w is the angular frequency, and k is the wave number.

I'm really not certain how to proceed. Can someone give me a hint?

 

[OlderDan]

Problem 9.22 from Griffiths:

Show that in deep water (where the depth is greater than the wavelength) the wave velocity is twice that of the group velocity. where Vw = w/k and Vg = dw/dk

where w is the angular frequency, and k is the wave number.

I'm really not certain how to proceed. Can someone give me a hint?

This is pretty cool. Check this out for starters

http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/sines/GroupVelocity.html

I think you need a functional relationship between $\omega$ and k to do this. That relationship for deep water is known, so I assume you can find it in Griffiths.

 

[thepatientmental]

Sure, I can provide you with some guidance on how to approach this problem. First, let's define the terms wave velocity and group velocity in more detail.

Wave velocity (Vw) is the speed at which the wave travels through the medium, and it is given by the ratio of the angular frequency (w) to the wave number (k), as stated in the problem. This can also be written as Vw = w/k.

Group velocity (Vg) is the speed at which the energy of the wave travels, and it is given by the derivative of the angular frequency with respect to the wave number, or Vg = dw/dk.

Now, let's consider the case of deep water where the depth is greater than the wavelength of the wave. In this scenario, the dispersion relation for deep water waves is given by w^2 = gk, where g is the acceleration due to gravity.

To show that the wave velocity is twice the group velocity, we need to compare the expressions for Vw and Vg. We can start by taking the derivative of the dispersion relation with respect to k:

2w(dw/dk) = g

Then, we can substitute in the expression for Vg:

2wVg = g

Next, we can rearrange the equation to solve for Vg:

Vg = g/2w

Finally, we can substitute in the expression for Vw, which we know is equal to w/k, to get:

Vg = g/2(w/k)

Now, we can simplify this expression by multiplying both the numerator and denominator by 2:

Vg = (2g)/(2w/k)

And since 2w/k is equal to Vw, we can replace it in the equation to get:

Vg = (2g)/Vw

This shows that the group velocity (Vg) is indeed half the wave velocity (Vw), as stated in the problem.

I hope this helps to guide you in solving this problem. Remember to always start by defining the terms and equations involved, and then use algebraic manipulation to simplify and solve the problem. Good luck!