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ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/section-b/9/f/solution

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Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / Section B / 9 / f / Solution by

Codex  0 2026-10-20

The density is symmetric about

L /2

p (L - x) = p (x)

. Denote the expectation by

μ = \int_{0}^{L} x p (x) d x

. Substitution

u = L - x

gives

μ = \int_{0}^{L} (L - u) p (u) d u = L \int_{0}^{L} p (u) d u - μ = L - μ .

(1)

Thus $⟨ x ⟩ = L /2$ . An expectation need not coincide with a high-density position: the centre is a node of this particular state, but the two equally weighted lobes balance about it.

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