Question

A longitudinal standing wave `xi a cos kx. Cos omega t ` is maintained in a homogeneous medium of density `rho`. Find the expressions for the space density of
`(a)` potential energy `w_(p)(x,t),`
`(b)` kinetic energy `w_(k)(x,t),`
Plot the space density distribution of the total energy `w` in the space between the displacement nodes at the moments `t=0` and `t=T//4`, where `T` is oscillation period.

Answer 1

Given `xi= a cos k x cos omegat`
`(a)` The potential energy density `(` per unit volume `)` is the energy of longitudinal strain. This is
`w_(p)=((1)/(2) "stress" xx "strain")=(1)/(2) E(( delta xi)/(deltax))^(2), ((delta xi)/(deltax)` is longitudinal strain )
`w_(p) =(1)/(2) Ea^(2) k^(2) sin^(2) k x cos^(2) omega t `
But `( omega^(2))/( k^(2))=(E)/( rho)` or ` E k^(2)= rho omega^(2)`
Thus `w _(p) = (1)/(2) rho a^(2) omega^(2) sin^(2) k x cos^(2) omega t `
`(b)` The kinetic energy density is
`=(1)/(2) rho((deltaxi)/( deltat))=(1)/(2) rho a^(2) omega^(2) cos ^(2) k x sin^(2) omegat` .
On plotting we get figure given in the book `(p.332)` . For example at `t=0`
`w=w_(p) + w_(k)=(1)/(2) rho a^(2) omega^(2) sin ^(2) kx `
and the displacement node are at `x=+- ( pi )/( 2k)` so we do get the figure.