Question

The dispersion relation for certain type of waves is given by \(\omega=\sqrt{k^2+a^2}\), where k is the wave vector and a is a constant. Which one of the following sketches represents v_g, the group velocity?

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The Correct Option is B

To determine the correct sketch that represents the group velocity \(v_g\) for the given dispersion relation, we begin by understanding the relationship and deriving the group velocity expression.

The given dispersion relation is:

\(\omega = \sqrt{k^2 + a^2}\)

Here, \(\omega\) is the angular frequency, \(k\) is the wave vector, and \(a\) is a constant. The group velocity \(v_g\) is defined as the derivative of \(\omega\) with respect to \(k\):

\(v_g = \frac{d\omega}{dk}\)

Let's differentiate \(\omega = \sqrt{k^2 + a^2}\) with respect to \(k\):

1. Apply the chain rule for differentiation:
2. \(\omega = (k^2 + a^2)^{1/2}\)
3. \(\frac{d\omega}{dk} = \frac{1}{2}(k^2 + a^2)^{-1/2} \cdot 2k\)
4. Simplify the expression:
5. \(\frac{d\omega}{dk} = \frac{k}{\sqrt{k^2 + a^2}}\)

Hence, the expression for the group velocity is:

\(v_g = \frac{k}{\sqrt{k^2 + a^2}}\)

Analyzing the expression, we see:

• As \(k \to 0\), \(v_g \to 0\).
• As \(k\) increases, \(v_g\) approaches 1 since the function simplifies to \(v_g \approx \frac{k}{k} = 1\) as \(k^2 \gg a^2\).

This implies that the group velocity increases from 0 to 1 as \(k\) increases from 0 to very large values. The sketch that represents this behavior will show the group velocity starting from 0 and asymptotically approaching 1.

Comparing the options, the correct representation of the group velocity is found in the following sketch:

This option shows \(v_g\) starting from 0 and asymptotically approaching 1 with increasing \(k\), which matches our derived behavior for the group velocity.