A tightly stretched string with fixed end points x=0 and x=l is initially in a position given by y=y_0 sin³(πx/l).It is released from rest from the initial positions.Find the displacement y(x,t)?

1 Answer
Jun 20, 2018

# y(x,t) = 3/4 y_o sin(( pi x)/l) cos(( pi c t )/l) - 1/4 y_o sin(( 3 pi x)/l) cos((3 pi c t )/l)#

Explanation:

For 1-D wave equation (WE):

  • #y_("xx") = 1/c^2 y_("tt")#

Separate the variables:

  • #y(x,t) = X(x) tau (t)#

The WE then reads:

  • #X^('')/X = 1/c^2 (ddot tau)/tau = color(red)(-) k^2#

Using #- k^2# gives the correct kind of solution (sinusoidal) from physical perspective:

#{(X^('') + k^2 X = 0, X(x) = c_1 sin(k x) + c_2 cos(k x) ),(ddot tau + c^2 k^2 tau = 0, tau(t) = c_3 sin(ck t) + c_4 cos(ck t) ):}#

  • #y(x,t) = (c_1 sin(k x) + c_2 cos(k x))(c_3 sin(ck t) + c_4 cos(ck t))#

The boundary conditions are

  • #y(0,t) = 0#, fixed node

  • # y(l,t) = 0#, fixed node

  • #y_t(x,0) = 0#, string is only released at #t = 0#

  • #y(x,0) = y_0 sin^3((pi x)/l)#, initial shape

1st BC :

  • #y(0,t) = (0 + c_2 )(c_3 sin(ck t) + c_4 cos(ck t)) = 0#

  • #implies c_2 = 0#.

2nd BC :

  • #y(l,t) = (c_1 sin k l )(c_3 sin(ck t) + c_4 cos(ck t)) = 0#

The fixed nodes are at #x = 0, l#, but a physical solution envisages possible nodes at other points (harmonics) so:

  • #k = (n pi)/l, n = 1,2,....#.

The solution thusfar:

#y(x,t) = c_1 sin((n pi x)/l) (c_3 sin((n pi c t )/l) + c_4 cos((n pi c t )/l) )#

3rd BC :

#0 = y_t(x,0) = c_1 sin((n pi x)/l) (c_3 (n pi c)/l cos((n pi c t )/l) - c_4 (n pi c)/l sin((n pi c t )/l) )#

#= c_1 sin((n pi x)/l) * c_3 (n pi c)/l implies c_3 = 0 #

#implies y(x,t) = c_1c_4 sin((n pi x)/l) cos((n pi c t )/l) #

So the basis functions are #4 sin((n pi x)/l) cos((n pi c t )/l) # with superimposed general solution:

  • # y(x,t) = sum_1^oo b_n sin((n pi x)/l) cos((n pi c t )/l) qquad square#

4th BC :

  • #y(x,0) = y_0 sin^3((pi x)/l) #

By identity:

# = y_o( 3/4 sin((pi x)/l) - 1/4 sin((3 pi x)/l)) #

#= sum_1^oo b_n sin((n pi x)/l) # from #square#

Comparing co-efficients:

#b_1 sin(( pi x)/l) + b_2 sin(( 2pi x)/l) + b_3 sin(( 3 pi x)/l) + b_4 sin(( 4 pi x)/l) + .... = 3/4y_o sin((pi x)/l) - 1/4y_o sin((3 pi x)/l)#

#implies {(b_1 = 3/4 y_o),(b_2 = 0), (b_3 = - 1/4 y_o),(b_4 = 0) ,(...):}#

So #square # becomes:

# y(x,t) = 3/4 y_o sin(( pi x)/l) cos(( pi c t )/l) - 1/4 y_o sin(( 3 pi x)/l) cos((3 pi c t )/l)#