Question

Torsional vibration of a shaft is governed by the wave equation,

Torsional vibration of a shaft is governed by the wave equation, 9 where 0(x, t) is the angular displacement (angle of twist) along the shaft, x is the distance from the end of the shaft and t is time. For a shaft of length 2T that is supported by frictionless bearings at each end, the boundary conditions are 0 x(0,t) = 0x(2T, t) = 0, t> 0 Suppose that the initial angular displacement and angular velocity are O(x,0) = 6 cos(4x), 0t(x,0) = 5+3 cos(4x), 0 x2T, respectively. You may use the result that the eigenvalues of the boundary-value problem x'(0) = X'(27) = 0 X" -X 0, are n2 = 0,1,2,3, ... n 22 with corresponding eigenfunctions find the final solution e(, t). Use the method of separation of variables (It is possible to get half-marks for this question for a good but incomplete solution. However, half marks will still show as wrong with "How did I do?", so check your mark too.) 0(x,t) =

Answer 1

We let Where X is function of x only and T is function of t only

Β(α, t) XT

Then given partial differential equation becomes

XT 9XT X" A (say) X 9T

So we have two differential equations

$X''-\lambda X=0~~~~~~~~~~~~~~~(1)\\~~~~~~ T''-9\lambda T=0~~~~~~~~~~~~~~(2)$

From the given information solution of (1) is

(3) Xn (x)An cos

Using the value of $\lambda$ we have from (2)

3n T 0 2 T"

The solution of such an equation is

$T_n(t)=B_n\cos \left ({3n\over 2} \right )t+C_n\sin \left ({3n\over 2} \right )t~~~~~~~~~(4)$

From (3) and (4) we have

$\theta_{n}(x,t)=X_{n}(x)T_n(t)=A_n\cos {n\over 2}x\left (B_n\cos \left ({3n\over 2} \right )t+C_n\sin \left ({3n\over 2} \right )t \right )\\ \\ ~~~~~~~~~~~~~~~=D_n\cos ({n\over 2}x) \cos \left ({3n\over 2} \right )t+ E_n\cos ({n\over 2}x) \sin \left ({3n\over 2} \right )t$

By Superimposition principle we get

$\theta(x,t)=\sum_{n=0}^{\infty}\theta_{n}(x,t)\\ \\ ~~~~~~~~~~~~~~~=\sum_{n=0}^{\infty}\left [D_n\cos ({n\over 2}x) \cos \left ({3n\over 2} \right )t+ E_n\cos ({n\over 2}x) \sin \left ({3n\over 2} \right )t \right ]$

Put t=0 and use given initial condition gives

$6\cos(4x)=\theta(x,0)=\sum_{n=0}^{\infty}\left [D_n\cos ({n\over 2}x) \right ] \\ ~~~~~~~~~~~~~~~~~= D_0+D_1\cos{x\over 2}+\cdots +D_8 \cos(4x)+\cdots$

So $D_8=6~,~D_0=D_1=\cdots=0$

Hence

3n En cos() sin r, t)6 cos(4r) cos(12t) 2 n-0

Differentiate with respect to t gives

$\theta_{t}(x,t)=-72\cos(4x)\sin(12t)+\sum_{n=0}^{\infty}\left [ {3n\over 2}(E_n)\cos ({n\over 2}x) \cos \left ({3n\over 2} \right )t \right ]$

Put t=0 gives

$5+3\cos (4x)=\theta_{t}(x,0)\\ ~~~~~~~~~~~~~~~~~~~~~~~=0+\sum_{n=0}^{\infty}\left [ {3n\over 2}(E_n)\cos ({n\over 2}x) \right ]\\ ~~~~~~~~~~~~~~~~~~~~~~~={3\over 2}\sum_{n=1}^{\infty} nE_n\cos ({n\over 2}x)$

Which is Cosine half range series so

3 1 (53 cos (4a)) cos)dr 27T nEn 2 2

Clearly for all
n 8
, $nE_n=0$

But For we have

n = 8

2m 1 3 (5 3 cos(4r)) cos(4x)da (8Es) 27T 2 2m sin(4r 3,sin(8x) 8 4 27T 3

hence $E_8={1\over 8}$

Therefore we have the solution

$\theta(x,t)=6\cos(4x)\cos(12t)+{1\over 8} \cos (4x)\sin (12t)$