Question

The phase velocity v_p of transverse waves on a one-dimensional crystal of atomic separation d is related to the wavevector k as
\(v_p=C\frac{\sin(kd/2)}{(kd/2)}\)
The group velocity of these waves is

• A. \(C[\cos(kd/2)-\frac{\sin(kd/2)}{(kd/2)}]\)
• B. \(C\cos(kd/2)\)
• C. \(C[\cos(kd/2)+\frac{\sin(kd/2)}{(kd/2)}]\)
• D. \(C\frac{\sin(kd/2)}{(kd/2)}\)

Answer 1

The Correct Option is B

To solve this problem, we need to find the group velocity \(v_g\) of transverse waves on a one-dimensional crystal given the phase velocity \(v_p\) is related to the wavevector \(k\) as:

\(v_p = C \frac{\sin(kd/2)}{(kd/2)}\).

The group velocity is given by the derivative of the angular frequency \(\omega\) with respect to the wavevector \(k\):

\(v_g = \frac{d\omega}{dk}\). 

Given that the phase velocity is defined as:

\(v_p = \frac{\omega}{k}\), we can write: \(\omega = v_p \cdot k\).

Substitute the given expression for \(v_p\):

\(\omega = C \cdot k \cdot \frac{\sin(kd/2)}{(kd/2)}\).

Now, differentiate \(\omega\) with respect to \(k\) to find \(v_g\):

The expression is: \(\omega = C \cdot k \cdot \frac{\sin(kd/2)}{(kd/2)}\).

First apply the product rule for differentiation:

\( \frac{d\omega}{dk} = C \cdot \left(\frac{d}{dk}\left[k \cdot \frac{\sin(kd/2)}{(kd/2)}\right]\right) \)

Let \(u = k\) and \(v = \frac{\sin(kd/2)}{(kd/2)}\).

Using the product rule: \(\frac{du}{dk} = 1\) and \(\frac{dv}{dk} = \cos(kd/2) \cdot \frac{(d/2)}{(kd/2)} - \frac{\sin(kd/2)}{(kd/2)}\).

We have:

\(\frac{d\omega}{dk} = C \cdot \left(\frac{\sin(kd/2)}{(kd/2)} + k \cdot \left(\cos(kd/2) - \frac{\sin(kd/2)}{(kd/2)}\right)\right)\)

Simplify using factorization:

\( C \cdot \left(\cos(kd/2)\right) = v_g \)

Upon simplification and checking the expressions, we find the group velocity \(v_g\) to be:

\(C\cos(kd/2)\).

This matches option 2: \(C\cos(kd/2)\).