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ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2015/paper-1/4/a/solution

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Past exam of the theoretical physics I course of the University of Cambridge / 2015 / Paper 1 / 4 / a / Solution by

Codex  0 2026-10-20

An element with unperturbed coordinate length

d x

has mass

ρ d x

and transverse speed squared

\dot{ψ}_{y}^{2} + \dot{ψ}_{z}^{2}

. Its actual length is

d s = 1 + (ψ_{y}^{'})^{2} + (ψ_{z}^{'})^{2} d x ≃ [1 + \frac{1}{2} ((ψ_{y}^{'})^{2} + (ψ_{z}^{'})^{2})] d x .

At fixed tension the extra elastic energy to this order is

F (d s - d x)

. Subtracting it from the kinetic energy gives

L = \int d x [\frac{ρ}{2} (\dot{ψ}_{y}^{2} + \dot{ψ}_{z}^{2}) - \frac{F}{2} ((ψ_{y}^{'})^{2} + (ψ_{z}^{'})^{2})] .

Here prime means

\partial_{x}

and dot means

\partial_{t}

. The approximation retains quadratic transverse energy and neglects higher-order slope effects and longitudinal motion. The resulting field equations are

ρ \ddot{ψ}_{i} = F ψ_{i}^{''}

for

i = y, z

, with wave speed

F / ρ

when

F > 0

.

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