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Damped oscillations are induced in a cirucuit whose quality factor is `Q=50` and natural oscillation frequency is `v_(0)=5.5 kHz`. How soon will the energy stored in the circuit decrease `eta= 2.0 ` times ? |
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Answer» `Q=(pi)/(betaT)=(piv)/( beta)=(omega)/( 2 beta)=(sqrt(omega_(0)^(2)-beta^(2)))/( 2 beta)` or ` (omega_(0))/( beta)=sqrt(1+ 4 Q^(2))` or `beta=( omega_(0))/( sqrt(1+Q^(2)))` Now `W=W_(0)e^(-2 beta t)` Thus energy decreases `eta` times in `(1n eta)/( 2 beta)` sec. `=1n eta(sqrt(1+4 dot (Q^(2))))/(2 omega_(0))~~(Q1n eta)/( 2 pi v_(nn))sec. = 1.033ms ` |
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