By Brian H. Chirgwin and Charles Plumpton (Auth.)

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**Example text**

This result is a special case of a very general principle (see Vol. I l l , § 9:6). If a vibration of an oscillating system is resolved into the sum of component vibrations in the normal modes, the kinetic energy of the vibration is the sum of the kinetic energies of the component vibrations in the normal modes. , there are no terms involving products of different normal coordinates. Miscellaneous Exercises I 1. Prove t h a t the Fourier expansion of the function x2 in the interval — π <; x ^ π is oo 4 â n2 + Σ (— l) w —£ oosnx.

The initial condition dy/dt = 0 at t == 0 is satisfied if D = 0 and the boundary conditions y = 0 a t x = 0 and at x — I are satisfied if ^4 = 0, Bsinal = 0. This latter condition is satisfied without y vanishing identically if al = sn, where 5 = 1 , 2 , 3 , . . Therefore y = As sin(sπχβ) cos(snctjl), (3) where As is constant. The solution (3) satisfies the boundary conditions at x = 0, x = I, but it does not satisfy the initial conditions. When t = 0, eqn. (3) gives / snx s y = As sin I — which represents a sine-wave instead of the shape with two straight segments.

1:5 FOURIER SERIES 4. If show t h a t f(x) = x for f(x) = a — x for 4a f{X) = 33 0 < x < Ja, \a < # < a , ~ ( — l ) w sin {(2ft + 1) nxja) (2n + l) 2 "rf~n=o Obtain a solution of the equation which satisfies the conditions: (i) V-> 0 as y-> oo, (ii) 7 = 0 when x = 0 and when x — a for all y > 0, (iii) V = f(x) as defined above when y = 0 a n d 0 < x < a. 5. A long bar of square cross-section has the faces x = 0, # = a and 2/ = 0 maintained a t zero temperature, a n d t h e face y = a a t a constant temperature F 0 .