 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/section-a/4/solution

M, L, T

[v] = L T^{- 1}, [ω] = T^{- 1}, [ρ] = M L^{- 3}, [g] = L T^{- 2},

[ω^{a} ρ^{b} g^{c}] = M^{b} L^{- 3 b + c} T^{- a - 2 c} .

Equating the independent dimensional powers requires

b = 0, - 3 b + c = 1, - a - 2 c = - 1.

Consequently

a = - 1, b = 0, c = 1, v \propto \frac{g}{ω} .

The mass density drops out because there is no other mass-bearing quantity with which to combine it under the stated assumptions.

ω^{2} = g k

ω / k = g / ω

d ω / d k = g / (2 ω)

. Both obey the inferred scaling. The phrase wave speed alone does not distinguish these two velocities or fix the coefficient.

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