Question

A point performs dampled oscillation with frequency `omega=25 s^(-1)`. Find the damping coefficient `beta` if at the initial moment the velocity of the point is equal to zero and its displacement from the equalibrium position is ` eta=1.020` times lesss than the amplitude at that moment.

Answer 1

`x=a_(0)e^(-betat)cos ( omegat+ alpha)` ltbr. Then `(dot(x))_(t=0)=-betaa_(0) cos alpha - omega a_(a) sin alpha =0`
or ` tan alpha=-(beta)/(omega^(.))`
Also `(x) _(t=0)=a_(0)cos alpha=(a_(0))/(eta)`
` sec^(2)alpha= eta^(2), tan alpha =- sqrt(eta^(2)-1)`
Thus `beta=omega sqrt(eta^(2)-1)`
( We have taken the amplitude at `t=0 ` to be `a_(0))`.