Find the quality factor of a mathematic pendulum `l=50 cm` long if during the time interval `tau=5.2 ` min its total mechanical energy decreases `eta=4.0.10^(4)` times.

Explanation

For an undamped oscillator the mechanical energy `E=(1)/(2) m dot(x^(2))+(1)/(2) m omega_(0)^(2)x^(2)` is conserved . For a damped oscillator.
`x= a_(0) e ^(- beta t ) cos ( omegat + alpha ), omega=sqrt( omega^(2)- beta^(2))`
and `E(t)=(1)/(2) m dot ( x^(2))+ (1)/(2) m omega_(0)^(2)x^(2)`
`=(1)/(2) m a_(0)^(2)e^(-2 beta t) [ beta^(2) cos ^(2)( omegat+ alpha) + 2 beta omega cos ( omegat + alpha) xx sin ( omegat+ alpha ) + omega^(2) sin ^(2) ( omegat + alpha)]`
`+(1)/(2) m a_(0)^(2)omega_(0)^(2) e^(-2 betat) cos ^(2) ( omegat + alpha)`
`=(1)/(2) m a_(0)^(2) omega_(0)^(2) e^(-2 beta t) + (1)/(2) m a_(0)^(2)beta^(2)e ^(-2 betat) cos ( 2 omegat + 2 alpha ) + (1)/(2) m a_(0)^(2) beta omega e ^(-2 betat ) sin ( 2 omegat + 2alpha)`
If ` beta lt lt omega`, then the average of the last two terms over many oscillation about the time `t` will vanish and
`lt E(t) gt ~=(1)/(2) m a_(0)^(2) omega_(0)^(2) e^(-2 betat)`
and this is the relevant mechanical energy.
In time `tau` this decreases by a factor `(1)/(eta)` so
`e^(-2 beta tau ) =(1)/( eta)` or `tau =(1n eta)/( 2 beta)`.
`beta=(1n eta)/(2 tau)`
and `lambda=(2pi beta)/(sqrt(omega_(0)^(2)-beta^(2)))=(2pi)/(sqrt(((omega_(0))/(beta))^(2)-1))=(2pi)/(sqrt((4g tau^(2))/( l 1n ^(2) eta)))` since` omega_(0)^(2) = (g)/(l)`.
and `Q=(pi)/(lambda)=(1)/(2)sqrt((4g tau^(2))/(l1n^(2)eta)-1)~= 130`