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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 ? |
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Answer» Solution :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)))` |
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