Answer to Question #112433 in Physics for Sam

Question #112433
For a system shown in the figure the block has mass 1.5 kg & spring constant 8.0 N/m.Suppose the block is pulled down a distance 12 c.m & released.If the friction of force is given by -bv where b=0.23 kg/s.find the number of oscillation made by the block during the time interval required for amplitude to fall to 1/3 of initial value
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Expert's answer
2020-04-29T09:50:49-0400
ω0=km(b2m)2\omega_0=\sqrt{\frac{k}{m}-\left(\frac{b}{2m}\right)^2}

FA2F \backsim A^2

If amplitude gets reduced to one third the force will reduce to one nineth.


Faω2F \backsim a \backsim \omega^2

t=2(3)πω0=2(3)πkm(b2m)2t=\frac{2(3)\pi}{\omega_0}=\frac{2(3)\pi}{\sqrt{\frac{k}{m}-\left(\frac{b}{2m}\right)^2}}

t=2(3)π81.5(0.232(1.5))2=8.2 st=\frac{2(3)\pi}{\sqrt{\frac{8}{1.5}-\left(\frac{0.23}{2(1.5)}\right)^2}}=8.2\ s


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