A particle was displaced from the equlibrium position by a distance `l=1.0cm` and then left along . What is the distance that the particle covers in the process of oscillations till the complete stop, if the logarithmic damping decrement is equal to `lambda=0.020`?

Explanation

From `x=a_(0)e^(- betat) cos ( omega t + alpha)` , we get using
`(x)_(t=0)=l = a _(0) cos alpha`
`0=(dot(x))_(t=0)=-beta a_(0) cos alpha - omega a _(0) sin alpha`
Then ` tan alpha =-( beta)/( omega) ` or `cos alpha =(omega)/(sqrt(omega^(2)+ beta^(2)))`
and `x=(l sqrt( omega^(2)+ beta^(2)))/( omega) e^(-betat) cos ( omegat - tan ^(-1) ( (beta)/( omega)))`
` x=0 ` at `t=(1)/( omega)` `( npi+(pi)/(2) + tan ^(-1) (( beta)/(omega)))`
Total distance travelled in the first lap `=l`
To get the maximum displacement in the second lap we note that
`dot (x) =[- beta cos ( omega t- tan ^(1)((beta)/( omega)))- omega sin ( omegat- tan ^(-1)((beta)/( omega)))]`
`x( l sqrt(omega^(2)+ beta^(2)))/( omega)e^(-betat)=0`
when ` omega t =pi , 2pi, 2pi,... ` etc.
Thus `ddot(x) _(max)=-a_(0) e^(- pi beta//omega)cos alpha=-l e ^(- pi beta//omega)` for `t=pi//omega`
so, distance traversed in the `2^(ng)` lap `= 2 l e ^(-pibeta//omega)`
Continuing total distance traversed `=l+2 l e ^(-pibeta//omega)+2 l e ^(-2 pi beta//omega)+ ........`
`=l+ ( 2 l e ^(-pi beta//omega))/( 1- e ^(- beta pi // omega))=l+( 2 l )/( e ^( beta pi // omega)-1)`
`=l( e ^( beta pi // omega)+1)/( e^( beta pi // omega ) -1)=l( 1+ e^( lambda//2))/(e ^( lambda//2)-1)`
where `lambda= ( 2 pi beta )/( omega)` is the logarithemic decrement . Subsitution gives 2 metres.