Paper 4, Section II, A

Complex Methods | Part IB, 2016

(a) Show that the Laplace transform of the Heaviside step function H(ta)H(t-a) is

0H(ta)eptdt=eapp\int_{0}^{\infty} H(t-a) e^{-p t} d t=\frac{e^{-a p}}{p}

for a>0a>0.

(b) Derive an expression for the Laplace transform of the second derivative of a function f(t)f(t) in terms of the Laplace transform of f(t)f(t) and the properties of f(t)f(t) at t=0t=0.

(c) A bar of length LL has its end at x=Lx=L fixed. The bar is initially at rest and straight. The end at x=0x=0 is given a small fixed transverse displacement of magnitude aa at t=0+t=0^{+}. You may assume that the transverse displacement y(x,t)y(x, t) of the bar satisfies the wave equation with some wave speed cc, and so the tranverse displacement y(x,t)y(x, t) is the solution to the problem:

2yt2=c22yx2 for 0<x<L and t>y(x,0)=yt(x,0)=0 for 0<x<L,y(0,t)=a;y(L,t)=0 for t>0.\begin{array}{cl} \frac{\partial^{2} y}{\partial t^{2}}=c^{2} \frac{\partial^{2} y}{\partial x^{2}} & \text { for } 0<x<L \text { and } t> \\ y(x, 0)=\frac{\partial y}{\partial t}(x, 0)=0 & \text { for } 0<x<L, \\ y(0, t)=a ; y(L, t)=0 & \text { for } t>0 . \end{array}

(i) Show that the Laplace transform Y(x,p)Y(x, p) of y(x,t)y(x, t), defined as

Y(x,p)=0y(x,t)eptdtY(x, p)=\int_{0}^{\infty} y(x, t) e^{-p t} d t

is given by

Y(x,p)=asinh[pc(Lx)]psinh[pLc]Y(x, p)=\frac{a \sinh \left[\frac{p}{c}(L-x)\right]}{p \sinh \left[\frac{p L}{c}\right]}

(ii) By use of the binomial theorem or otherwise, express y(x,t)y(x, t) as an infinite series.

(iii) Plot the transverse displacement of the midpoint of the bar y(L/2,t)y(L / 2, t) against time.

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