A standing wave `xi= a sin kx. Cos omegat ` is maintained in a homogeneous rod with cross `-` sectional area `S` and density `rho`. Find the total mechanical energy confined between the sections corresponding to the adjacent displacement nodes.

Explanation

We have a standing wave given by the eqaution
` xi=a sin k x cos omegat`
So, `(deltaxi)/( deltat)=-a sin k x sin omega t ....(1)`
and `(delta xi)/( deltat)=a k cos k x cos omegat ...(2)`
The kinetic energy confined in an element of length `dx` of the rod
`dT=(1)/(2) ( rho Sdx)((deltaxi)/( deltat))^(2)=(1)/(2) rho S a^(2) omega^(2) sin ^(2) omegat sin ^(2)k x dx`
So total kinetic energy confined into rod
`T=int dT=(1)/(2)rho Sa^(2) omega^(2) sin^(2) omegatint_(0)^(gamma//2)sin^(2)((2pi)/( lambda))x dx`
or`, T=(piSa^(2) omega^(2) rho sin^(2)omegat)/( 4k) ...(3)`
The potential energy in the above rod element
`dU=int deltaU=-int _(0)^(xi)F_(xi)dxi`, where `'F_(xi)=(rhoSdx)(delta^(2)xi)/(deltat^(2))`
or, `F_(xi)=(rho S dx)omega^(2) xi`
so, `dU=omega^(2)rho S dxint _(0)^(xi)xi dxi`
or, `dU=(rho omega^(2) Sxi_(2))/( 2) dx=(rho omega^(2) Sa^(2) cos^(2) omegat sin^(2) kx dx)/(2)`
Thus the total potential energy stored in the rod `U=int dU`
or,`U= rho omega^(2) Sa^(2) cos^(2) omegatint _(0)^(gamma//2)sin ^(2) ((2pi)/( lambda)) x dx`
So, `U=(pi rho S a^(2) omega^(2)cos^(2) omegat)/( 4 k )`
To find the potential energy stored in the rod element we may adopt an easier way . We know that the potential energy dencity confined in a rod under elastic force equals `:`
`U_(D)=(1)/(2)` ( stress `xx` straing ) `=(1)/(2) sigma epsilon=(1)/(2)Y epsilon^(2)`
`=(1)/(2) rhov^(2) epsilon^(2)=(1)/(2) ( rho omega^(2))/(k^(2))epsilon^(2)`
`=(1)/(2) ( rho omega^(2))/( k^(2))((deltaxi)/( deltax))^(2)=(1)/(2) rho a^(2) omega^(2) cos^(2) omegat cos^(2) kx`
Hence the total potential energy stored in the rod
`U=int U_(D) dV=int_(0)^(gamma//2)(1)/(2) rho^(2)omega^(2) cos^(2) omega t cos^(2) k x S dx`
`=(pi rho Sa^(2) omega^(2)cos ^(2) omegat )/( 4k ) (4)`
Hence the sought mechanical energy confined in the rod between the two adjacent nodes
`E=T+U=(pi rho omega^(2) a^(2)S)/( 4k )`