A string of mass per unit length `mu` is clamped at both ends such that one end of the string is at x = 0 and the other is at `x =l`. When string vibrates in fundamental mode, amplitude of the midpoint O of the string is a, tension in a, tension in the string is T and amplitude of vibration is A. Find the total oscillation energy stored in the string.

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)`