Question

A ball of mass `m` when suspended by a spring stretches the latter by `Delta l`. Due to external vertical lforce varying according to a harmonic law with amplitude `F_(0)` the ball performs forced oscillations. The logarighmic damping decrement if equal to `lambda`. Neglecting the mass the spring, find the angular frequency of the external force at which the displacemetn amplitude of the ball is maximum. What is the magnitude of that amplitude ?

Answer 1

For the spring `mg= k Delta l`
where `k` is its stiffness coefficient. Thus
`omega_(0)^(2)=(k)/(m)=(g)/(Deltal)`
The equation of motion of the ball is
`ddot(x) + 2 beta dot (x) + omega_(0)^(2)x=(F_(0))/(m) cos omegat`
Here, `y=(2pibeta)/( sqrt(omega_(0)^(2)-beta^(2)))` or ` ( beta)/( omega)=( lambda // 2pi )/( sqrt(1+ ( lambda//2pi)^(2)))`
To find the solution of the above equation we look for the solution of the auxiliary equation
`cancel(ddot(x))+2 betacancel(dot(x))+ omega_(0)^(2)cancel(z)=(F_(0))/(m)e^(iomegat)`
Clearyl we can tek Re `cancel(z)=x.`Now we look for a particular integral for `cancel(z)` of the form
`cancel ( z)=A e^(iomegat)`
Thus, substitution gives `A` and we get
`cancel(z) =((F_(0)//m)e^(iomegat))/(omega_(0)^(2)-omega^(2)+2 i betaomega)`
so taking the realpart
`x=((F_(0)//m)[(omega_(0)^(2)-omega^(2))cos omegat+ 2 beta omega sin t])/( ( omega_(0)^(2)-omega^(2))^(2)+ 4 beta^(2)omega^(2))`
`=(F_(0))/( m)(cos ( omegat-varphi))/(sqrt((omega_(0)^(2)-omega^(2))^(2)+ 4 beta^(2) omega^(2))), varphi= tan ^(-1) .( 2beta omegat)/( omega_(0)^(2)- omegat^(2))`
The amplitude of this oscillation is maximum when the denominator is minimum.
This happens when
`omega^(4)-2 omega_(0)^(2) omega^(2)+ 4 beta^(2) omega^(2)+omega_(0)^(4)=( omega^(2)-omega_(0)^(2)+ 2 beta^(2))+ 2 beta ^(2) omega^(2) - 4 beta ^(4)` is minimum `i.e.` for `omega^(2)=omega_(0)^(2)- 2 beta^(2)`
Thus` omega_(res)^(2)=omega_(0)^(2)(1-(2beta^(2))/(omega_(0)^(2)))`
`=(g)/(Deltal)[1-(2((lambda)/( 2pi))^(2))/(1+((lambda)/( 2pi))^(2))]=(g)/(Deltal)(1-((lambda)/(2pi))^(2))/(1-((lambda)/(2pi))^(2))`
and `a_(res)=(F_(0)//m)/(sqrt(4 beta^(2) omega_(0)^(2)- 4 beta^(4)))=(F_(0)//m)/(2 beta sqrt(omega_(0)^(2)-beta^(2)))=(F_(0)//m)/(2 beta^(2)). ( lambda)/( 2pi)`
`=(F_(0))/( 2 m omega_(0)^(2)). (1+ ((lambda)/(2pi))^(2))/( lambda//2 pi)=(F_(0)Deltallambda)/( 4 pi m g ) ( 1+ (4pi ^(2))/( lambda^(2)))`