A string of mass `m` is fixed at both ends. The fundamental tone oscillations are excited with circular frequency `omega` and maximum displacement amplitude `a_(max)` . Find `:`
`(a)` the maximum kinetic energy of the string,
`(b)` the mean kinetic energy of the string averaged over one oscillation period.

Explanation

Let two waves `xi_(1) = a cos ( omega t - kx ) ` and `xi_(2)=a cos ( omega t + kx )`, superpose and as a result, we have a standing wave `(` the resultant wave `)` in the string of the form ` xi=2 a cos kx cos omegat`.
Accroding to the problem `2a =a _(m)`.
Hence the standing wave excited in the string is
`xi=a _(m) cos kx cos omegat ....(1)`
or,` ( delta xi)/( deltat)=- omega_(m) cos kx sin omegat ...(2)`
So the kinetic energy confined in the string element o length `dx`, is given by `:`
`dT=(1)/(2) ((m)/(l) dx)((deltaxi)/( deltat))^(2)`
or, ` dT=(1)/(2) ((m)/( l) dx) a_(m)^(2) omega^(2) cos ^(2) k x sin ^(2) omegat `
or, `dT=(m a_(m)^(2)omega^(2))/( 2l) sin ^(2) omega t cos ^(2) (( 2pi)/( lambda))x dx`
Hence the kinetic energy confined in the string corresponding to the fundamental tone
`T=int dT=(m a_(m)^(2) omega^(2))/( 2l) sin ^(2) omegat int _(0)^(lambda//2)cos ^(2)((2pi)/( lambda))x dx`
Because, for the fundamental tone, length of the string `l=(lambda)/( 2)`
Integrating we get, `: T=(1)/(4) m a_(m)^(2) omega^(2) sin ^(2) omegat`
Hence the sought maximum kinetic energy equals, `T_(max)=(1)/(4) m a_(m)^(2) omega^(2)`,
because for `T_(max), sin^(2) omegat =1`
`(ii)` Mean kinetic energy averaged over one oscillation period
`lt T gt ( int Tdt)/(int dt)=(1)/(4) m a_(m)^(2) omega^(2)(int _(0)^(2pi//omega)sin ^(2)omegatdt)/(int_(0)^(2 pi //omega)dt)`
or, `lt T gt =(1)/(8)m a_(m)^(2) omega^(2)`.