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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. |
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Answer» `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))` |
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