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conantee
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« on: January 04, 2010, 04:47:01 AM »
59. The dispersion law for a certain type of wave motion is \omega = (c^2 k^2 +m^2)^\frac{1}{2}, where \omega is the angular frequency, k is the magnitude of the propagation vector, and c and m are constants. The group velocity of these waves approaches
(A) infinity as k \to 0 and zero as k \to \infty
(B) infinity as k \to 0 and c as k \to \infty
(C) c as k \to 0 and zero as k \to \infty
(D) zero as k \to 0 and infinity as k \to \infty
(E) zero as k \to 0 and c as k \to \infty
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conantee
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« Reply #1 on: October 11, 2010, 11:06:52 AM »
If you cannot remember formula for group velocity, at least you should remember that it somehow represents "physical" velocity, while phase velocity can be greater than, say, the speed of light in vacuum.

Hence, it is likely that (A), (B), and (D) are wrong.

You may guess (C) or (E) as an answer.

Actually, the group velocity is defined as d \omega / dk. Looking at behavior when k is very large (i.e. m is negligible), \omega \to (c^2 k^2)^{1/2} = ck. Hence, as k \to \infty, group velocity goes to c.

Note that checking limit behavior is a good strategy. Do not try tedious derivative without checking limit behavior first.

The correct answer is (E). 41 of 100 people chooses this correct choice.
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