Question:
The phase velocity vp 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
The phase velocity vp 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
\(v_p=C\frac{\sin(kd/2)}{(kd/2)}\)
The group velocity of these waves is
Updated On: Oct 1, 2024
- \(C[\cos(kd/2)-\frac{\sin(kd/2)}{(kd/2)}]\)
- \(C\cos(kd/2)\)
- \(C[\cos(kd/2)+\frac{\sin(kd/2)}{(kd/2)}]\)
- \(C\frac{\sin(kd/2)}{(kd/2)}\)
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The Correct Option is B
Solution and Explanation
The correct option is (B) : \(C\cos(kd/2)\).
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