A string with tension T and mass per unit length `mu` is clamped down at x=0 and at x=L. at t=0, the string is at rest and displaced in the y-direction
`y(x,0)=2"sin"(2pix)/(L)+2"sin"(pix)/(L)`
Q. The string is released at t=0, and it starts to oscillate. the displacement of string at time t is

Answer

`(A^(2)pi^(2)T)/(4l)`

Explanation

`l=(lamda)/(2)rArrlamda=2l,k=(2pi)/(lamda)=(pi)/(l)`
The amplitude at a distance x from x = 0 is given
Total mechanical energy at x of length dx is
`dE=(1)/(2)(dm)A^(2)omega^(2)=(1)/(2)(mudx)(a sin kx)^(2)(2pif)^(2)`
`rArr dE=2pi^(2)muf^(2)a^(2)sin^(2)kxdx`
Here, `f=(v^(2))/(lamda^(2))=(((T)/(mu)))/((4l^(2)))and k=(pi)/(l)`
Substituting these values in equation (i) adn integrating it from `x =0` to `x = l`, we get total energy of string.
`E = (pi^(2)a^(2)T)/(4l)`