A string tied between `x = 0 and x = l` vibrates in fundamental mode. The amplitude `A`, tension `T` and mass per unit length `mu` is given. Find the total energy of the string.

Answer

A, B, D

Explanation

Here `l = (lambda)/(2)` or `lambda = 2l` Since, `k = (2pi)/(lambda) = (2pi)/(2l) = (pi)/(l)`
The amplitude of vibration at a distance `x` from `x = 0` is given by `A = a sin k x`
Mechanical energy at `x` of length `dx` is
`dE = (1)/(2)(dm)A^(2)omega^(2) = (1)/(2)(mudx)(a sin k x)^(2)(2piv)^(2)`
`= 2pi^(2)muv^(2)a^(2) sin^(2)kx dx`
Bur `v = vlambda`
:. `v = (v)/(lambda)` rArr `v^(2) = (v^(2))/(lambda^(2)) = (T//mu)/(4l^(2))` [ `because` `v = sqrt(T//mu)]`
:. `dE = 2pi^(2)mu(T//mu)/(4l^(2))a^(2) sin^(2) {((pix)/(l))x}dx`
:. Total energy of the string
`E = int dE = int_(0)^(l)2pi^(2)mu(T//mu)/(4l^(2))a^(2)sin^(2)((pix)/(l))dx`
`= (pi^(2)Ta^(2))/(4l)`