 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2014/11/d/solution

For one real mode

y_{n} = A_{n} sin (k_{n} x) cos (ω_{n} t + ϕ_{n})

, the kinetic and small-slope potential energy densities integrate to

K_{n} = \frac{ρ}{2} \int_{0}^{l} y_{n, t}^{2} d x = \frac{ρl A _{n}^{2} ω _{n}^{2}}{4} sin^{2} (ω_{n} t + ϕ_{n}),

U_{n} = \frac{T}{2} \int_{0}^{l} y_{n, x}^{2} d x = \frac{Tl A _{n}^{2} k _{n}^{2}}{4} cos^{2} (ω_{n} t + ϕ_{n}) .

Since

ρ ω_{n}^{2} = T k_{n}^{2}

, their sum is constant:

E_{n} = Tl A_{n}^{2} k_{n}^{2} /4

ϕ_{n} = 0

and

A_{1} = A

A_{2} = A /2

, hence

E_{1} = E_{2} = \frac{T π ^{2} A ^{2}}{4 l}, E_{total} = \frac{T π ^{2} A ^{2}}{2 l} .

Although the second mode's displacement amplitude is halved, its wavenumber and angular frequency are doubled, giving the same total energy. Orthogonality eliminates cross terms when the full string's energies are integrated.

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