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A ball of mass m=50 g is suspended by a ...

A ball of mass `m=50 g` is suspended by a weightless spring with stiffness `x=20.0 N//m`. Due to external vertical harmonic force with frequency `omega=25.0 s^(-1)` the ball performs steady `-` state oscillations with amplitude `a=1.3 cm`. In this case the displacement of the ball lags in phase behind the external force by `varphi=(3)/(4)pi. ` Find `:`
`(a)` the quality factor of the given oscillator ,
`(b)` the work performed by the external force during one oscillation period.

Text Solution

Verified by Experts

In the formula `x= a cos ( omega t - varphi)`
we have `a=(F_(0))/( m) (1)/( sqrt((omega_(0)^(2)- omega^(2))^(2)+4 beta^(2) omega^(2)))`
`tan varphi=(2 beta omega)/( omega_(0)^(2)- omega^(2))`
Thus `beta =((omega_(0)^(2)-omega^(2))tan varphi)/(2 omega)`
Hence `omega_(0)=sqrt(K//m)=20 s ^(-1)`
and `(a)` the quality factor
`Q=(pi)/( betaT)=(sqrt(omega_(0)^(2)-beta))/( 2 beta)=(1)/(2) sqrt((4 omega^(2)omega_(0)^(2))/( ( omega_(0)^(2)- omega^(2))^(2) tan^(2) varphi)-1)= 2.17`
`(b)` work done is `A = pi a F_(0) sin varphi`
`= pi m a^(2) sqrt((omega_(0)^(2)- omega^(2))^(2)+ 4 beta^(2) omega^(2)) sin varphi= pi m a^(2) xx2 beta omega`
` = pi ma^(2) ( omega_(0)^(2)- omega^(2)) tan varphi = 6 mJ`
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