Question

The Q factor of a sonometer wire of frequency 500 Hz is 5000. In what time will its energy reduces to 1/e to its initial value in absence of damping. Also find the resistive force constant/ Damping force .

Answer 1

Answer:

The average energy of oscillation is

$\large\rm { E(t) = E_{0} \ e^{ \frac {-ω_{0} t}{Q_{0}}}}$

It means that Q is related to the number of oscillation over which the energy fall to 1/e of its initial value $E_{0}$ . This happens in time t =τ where

$\large\rm { in \ time \ t = τ \ where \ → \frac {ω_{0} t}{Q} = 1 }$

$\large\rm { → τ = \frac {Q}{ω_{0}} = \frac {Q}{2πv_{0}}}$

$\large\rm { Therefore \ T = \frac {5000}{2π(500)} = \frac {10}{2π} = \frac {5}{π} }$

$\large\rm { = 1.59 = 1.6 \ seconds }$

$\large\rm { Because , \ Q = 5000, v_{0} = 500Hz }$