Question

A mathematical pendulum oscillates in a medium for which the logarithmic damping decrement is equal to `lambda_(0)=1.50`. What will be the logarithmic damping decrement if the resistance of the medium increases `n=2.00` time? How many times has the resistance of the medium to be increased for the oscillations to become impossible?

Answer 1

By definition of the logarithmic decrement `(lambda=beta((2pi)/(omega)))`, we get for the original decrement
`lambda_(0)`
`lambda_(0)=beta(2pi)/(sqrt(omega_(0)^(2)-beta^(2)))` and finally `lambda=(2pi n beta)/(sqrt(omega_(0)^(2)-n^(2)beta^(2)))`
Now `(beta)/(sqrt(omega_(0)^(2))-beta^(2))=(lambda_(0))/(2pi)` or `(beta)/(omega_(0))=(lambda_(0)//2pi)/(sqrt(a+((lambda_(0))/(2pi))^(2)))`
so `(lambda//2pi)/(sqrt(1+((lambda)/(2pi))^(2)))=(n(lambda_(0))/(2pi))/(sqrt(1+((lambda_(0))/(2pi))^(2)))`
Hence` (lambda)/(2pi)=(nlambda_(0)//2pi)/((sqrt(1-(n^(2)-1)lambda_(0))/(2pi))^(2))`
For critical damping ` omega_(0)=n_(c)beta`
`(1)/(n_(c))=(beta)/(omega_(0))=(lambda_(0)//2pi)/(sqrt(1+((lambda_(0))/(2pi))^(2)))` or `n_(c)=sqrt(1+((2pi)/(lambda_(0))))`