 Solution

ID: past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2018/3/c/solution

m = \frac{0.2070 kg mo l ^{- 1}}{N _{A}} = 3.4373 \times 1 0^{- 25} kg .

Let

u_{n}

m \overset{u}{¨}_{n} = κ (u_{n + 1} + u_{n - 1} - 2 u_{n}) .

Insert the normal mode

u_{n} = u e^{i (kna - ω t)}

to obtain

ω^{2} = \frac{4 κ}{m} sin^{2} \frac{ka}{2} .

ω ≃ a κ / m ∣ k ∣

v_{s} = a \frac{κ}{m} = 2.1155 \times 1 0^{3} m s^{- 1} .

The relevant inertia in this chain equation is one atomic mass. The reduced mass

m /2

belongs to the relative vibration of an isolated pair and should not be substituted for

m

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