A deadweight suspended from a weightless spring extends it by `Deltax=9.8 cm`. What will be the oscillation period of the dead weight when it is pushed slightly in the vertical direction ? The logarithmic damped decrement is equal to `lambda=3.1`

Explanation

The Eqn of the dead weight is
`m ddot(x)+2 beta m dot(x) + m omega_(0)^(2)x = mg`
so `Deltax= ( g)/(omega_(0)^(2))` or ` omega_(0)^(2)=(g)/(Deltadotx)`
Now `lambda=(2pi beta)/(omega)=(2pi beta)/(sqrt(omega_(0)^(2))-beta^(2))` or ` (omega_(0))/(sqrt(omega_(0)^(2)- beta^(2)))=sqrt(1+((lambda)/(2pi))^(2))`
Thus `T=(2pi)/(sqrt(omega_(0)^(2)-beta^(2)))=(2pi)/(omega_(0))sqrt(1+((lambda)/(2pi))^(2))`
`=2pi sqrt((Deltax)/(g))sqrt(1+((lambda)/(2pi))^(2))=sqrt((Deltax)/(g)(4pi^(2)+lambda^(2)))=0.70 sec. `