The velocity amplitude of a particle is equal to half the maximum value at the frequencies `omega_(1)` and `omega_(2)` of external harmonic force.
Find`:`
`(a)` the frequency corresponding to the velocity resonance,
`(b)` the damping coefficient `beta` an dthe damped oscillation frequency `omega` of the particle.

Explanation

`x=(F_(0))/(m)((omega_(0)^(2)-omega^(2)) cos omegat + 2 beta sinomegat )/(sqrt((omega^(2)- omega_(0)^(2))^(2)+ 4 beta^(2) omega^(2)))`
Then `dot(x)=(F_(0)omega)/(m)(2 beta omega cos omegat + ( omega^(2) - omega_(0)^(2)) sin omegat)/( ( omega_(0)^(2)- omega^(2))^(2) + 4 beta^(2) omega^(2))`
Thus the velocity amplitude is
`V_(0)=( F_(0 ) omega)/( m sqrt(( omega_(0)^(2)- omega^(2))^(2) + 4 beta^(2) omega^(2)))`
` = ( F_(0))/(msqrt(((omega_(0)^(2))/( omega)-omega)^(2)+ 4 beta^(2)))`
This is maximum when `omega^(2)=omega_(0)^(2)=omega_(res)^(2)`
and then `V_(0 res) =(F_(0))/(2 m beta)`
Now at half maximum `((omega_(0)^(2))/( omega)-omega)^(2)=12 beta^(2)`
or ` omega^(2)+-2 sqrt(3) beta omega - omega_(0)^(2)=0`
`omega=+- betasqrt(3)+ sqrt(omega_(0)^(2)+ 3 beta^(2))`
where we have rejected a solution with `- ve ` sign before there dical . Writing
`omega_(1)=sqrt(omega_(0)^(2)+3 beta^(2))+ betasqrt(3), omega_(2)=sqrt(omega_(0)^(2)+ 3 beta^(2))- betasqrt(3)`
we get `(a) omega_(res)=omega_(0)=sqrt(omega_(1)omega_(2))` ( Velocity resonance frequency )
(b) `beta=(| omega_(1)-omega_(2)|)/( 2 sqrt(3))` and damped oscillation frequency
`sqrt(omega_(0)^(2)-beta^(2))=sqrt(omega_(1)omega_(2)-((omega_(1)-omega_(2))^(2))/( 12))`