Question

Torsional vibration of a shaft is governed by the wave equation, = 4 where ex, ) is the angular displacement (angle of twist) along the shaft, x is the distance from the endc the shaft and is time. For a shaft of length 4x that is supported by frictionless bearings at each end, the boundary conditions are Ox(O.t) = 0x(4r, f) = 0, 1>0. Suppose that the initial angular displacement and angular velocity are Ox, 0) = 6 cos(x), 0x, 0) = 5+3 cos (4x), 0x< 4r respectively You may use the result that the eigenvalues of the boundary-value problem x" - 2x 0 x'(0) = x'(42) = 0 are n0, 1,2,3... with corresponding eigenfunctions nx Xn =COs Use the method of separation of variables to find the final solution (x, 1) (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.) θ(r ) =

Answer 1

Observe that with the given problem is

  $\frac{\partial^2\theta}{\partial t^2}= 4\frac{\partial^2\theta}{\partial x^2}$ (1)

with

  $\theta_x(0, t)=\theta_x(4\pi, t)=0$ (2)

We proceed by the method of separation of variables. So, put

   (*)

Βα,t ΧαTΐ)

Putting this value in equation (1) to obtain

$4X''(x)T(t)= X(x)T''(t)\implies \frac{4X''(x)}{X(x)}= \frac{T''(t)}{T(t)}$

Now since X is a function of x and T is of t, both the fractions are equal means that

$\frac{X''(x)}{X(x)}= \frac{T''(t)}{4T(t)}= \lambda$ (3)

where $\lambda$ is some constant. From above equation, we get

$\frac{X''(x)}{X(x)}= \lambda \implies X''(x)-\lambda X(x)=0$ (4)

and

  $X'(0)=X'(4\pi)=0$. (5)

Boundary value system given by (4) and (5) is same as the given system. So, for this system  eigenvalues are

  $\lambda_n=-\frac{n^2}{4}$

and eigen functions are

  $X_n = \cos \frac{nx}{4}$.

Further, solving (3) for t, we get

  $T''(t)-4\lambda T(t)=0$

Substituting $\lambda$ in above (for each n), we get

$T_n(t)= A_n\cos 2\times \frac{n}{4}t+B_n\sin 2\times \frac{n}{4}t= A_n \cos \frac{n}{2}t+B_n\sin\frac{n}{2}t$

Putting these values of X_n(x) and T_n(t) in (*) after using super-position principle to get

$\theta(x, t)= \sum_{n\geq 0}X_n(x)T_n(t)= \sum_{n\geq 0}\cos \frac{nx}{4}\bigg(A_n \cos \frac{nt}{2}+B_n \sin \frac{nt}{2}\bigg)$ (6)

Now

  $\theta(x, 0)= 6\cos x= \sum_{n\geq 0}A_n\cos \frac{nx}{4}$

Above means that

  $A_4=6 \ \text{and all other}\ A_i's=0$ (7)

Also, we have

$\theta_t(x, 0)= \sum_{n\geq 0}X(x)T'(t)= \sum_{n\geq 0}\cos \frac{nx}{4}\bigg(-\frac{nA_n}{2}\sin \frac{nt}{2}+\frac{nB_n}{2}\cos \frac{nt}{2}\bigg)$ (8)

Now we need to apply the fourth boundary condition which is

$\theta_t(x, 0)= 5+3\cos 4x$

Putting t=0 in (8) to get

$\theta_t(x, 0)= \sum_{n\geq 0}\cos \frac{nx}{4}\times \frac{nB_n}{2}$

Compare this to get Bn and use (7) to get the solution (6).