Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 5 / b / ii

Table of contents
- Solution ii

Solution

Multiply the transformed equation by

sin (k_{n} x)

and integrate from zero to

L

. Two integrations by parts, using the zero endpoint values of both

y

and the sine, give

\int_{0}^{L} y_{xx} sin (k_{n} x) d x = - k_{n}^{2} \int_{0}^{L} y sin (k_{n} x) d x .

The delta source contributes

F (ω) sin (k_{n} x_{0})

. Multiplying by

2/ L

and using the coefficient definition therefore yields

(k_{n}^{2} - ω^{2}) b_{n} (ω) = \frac{2}{L} sin (k_{n} x_{0}) F (ω) .

Away from the resonant frequencies this gives

b_{n} (ω) = α_{n} \frac{F ( ω )}{k _{n}^{2} - ω ^{2}}, α_{n} = \frac{2}{L} sin (k_{n} x_{0}) .

At

ω = \pm k_{n}

the quotient must be interpreted with a pole prescription if the source excites that mode. In distributional form, arbitrary delta terms at those frequencies may also be added; they are the free homogeneous oscillations. This is the frequency-space issue addressed explicitly in part (iii).

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