1 Linear theory of the oscillation of a hanging chain Daniel Bernoulli in 1732 and Euler in 1781. m is the mass per unit length and x is measured upwards from the bottom. The tension, T mgx , and so the equation of SMALL oscillations is y 2 y y 2 y T m g or . x x x x x t 2 t 2 Motion sinusoidal with time If we write y Y x cost , this becomes g d dY 2 x cost Y cos t so that dx dx d dY 2 Y 0. x dx dx g x du Substitute u 2 so that and therefore g dx gx d x dY 2 2 d u dY 2 Y 0 Y 0 and hence or gu du 2 du g gx du gx du g 1 d dY u Y 0 . u du du d 2Y 1 dY This finally produces Y 0. du 2 u du Approach 1 2 dy 2 d y x x x 2 v 2 y 0 which has the solution Bessel’s differential equation is 2 dx dx y AJv x BYv x in which J v x is the Bessel function of the first kind and Yv x and the Bessel function of the second kind. In our case v 0 and Y0 0 is infinite so that the final solution is x Y AJ 0 u AJ 0 2 g . 1 x 2 m! n m ! and therefore 1 x J 2 x m! 0 m 2 m0 m . m n 2mn m0 The series for J n x is J n x x 2m 2 Approach 2 Try Y amu m and substitute into d 2Y 1 dY Y 0 to give u du du 2 am mm 1 m am2 0 or m 2 am am2 0 . There are two solutions: 1 m m even: a0 1, am m! 2 when m positive and am 0 when m is negative. negative. m odd: This gives values for am when mis positive and Therefore it is infinite when x 0 - no good. Thus either way we obtain Y constant 1 x m . 2 m0 m! m The if the chain is hung from a fixed point, that point must be a node. Travelling wave Return to g y 2 y and consider solution x x x t 2 y y , x t2 g t 2 x g . 2 x g 4 t 2 Then 1 x gx t and 3 2 1 1 g y x y gx gx 2 4 y y 4 hence 2 2 y y 1 y y 1 y y 2 y 4 y y 2 y 4 so that y 2 y y 2 y 4 4 3 4 4 y 4 4 y 1 3 1 1 1 1 1 4 4 y 4 4 y 1 1 1 1 1 1 1 1 2 2 4 4 4 4 4 4 3 3 4 4 y y y y 4 2 2 4 1 1 1 1 2 2 2 4 4 3 4 4 0 y y 2 2 4 4 2 y y y 0 4 2 y y y 4 1 1 2 2 4 4 4 y y 5 1 2 5y 4 4 4 y 9 16 4 5 4