Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 345 / 4 / e

Table of contents
- Solution e

Solution

Introduce the virtual-origin coordinate

ζ = z + z_{0}, z_{0} = \frac{B _{0}}{β},

so that

B = βζ

. Seek

Q = A ζ^{4/3}

and

M = C ζ^{5/3}

. Substitution gives

A = (\frac{27 E ^{2} β}{80})^{1/3}, C = (\frac{27 E β ^{2}}{100})^{1/3} .

Therefore the exact similarity solution is

Q = A ζ^{4/3}, M = C ζ^{5/3},

W = \frac{M}{Q} = (\frac{4 β}{5 E})^{1/3} ζ^{1/3}, b = \frac{Q ^{2}}{M} = \frac{3 E}{4} ζ,

and

T - T_{0} = \frac{T _{0} B}{g Q} = \frac{T _{0} β}{g A} ζ^{- 1/3} .

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