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Find the angular frequency of small oscillation of block m in the arrangement shown. Rod is massless. (Assume gravity to be absent) |
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Answer» Solution :Suppose the mass `m` is displaced a distance `x` downward and let `theta` be the angular displacement of the rod. It is clear that the SPRING of force constant `k_(2)` will have a compression `=(x-(l theta)/(2))` for small `x` and `theta`. `:.` Balancing the TORQUES on (massless) rod `rArr k_(2)(x-(l theta)/(2))(l)/(2)costheta=k_(3)//theta//costheta` `rArrk_(2)(x-(l theta)/(2))=2k_(3)l thetarArrl theta=(2k_(2)x)/(4k_(3)+k_(2))` Now for mass `m` `F_(real)=-k_(1)x-k_(2)(x-(l theta)/(2))=-k_(1)x-k_(2)x+(k_(3)^(2)x)/(4k_(3)+k_(2))` (putting the value of `l theta`) `rArrm(d^(2)x)/(dt^(2))= -(4k_(1)k_(3)+k_(1)k_(2)+4k_(2)k_(1))/(4k_(3)+k_(2))x` `:. m(d^(2)x)/(dt^(2))+(4k_(1)k_(3)+k_(1)k_(2)+4k_(2)k_(1))/(4k_(3)+k_(2))x=0` `:. ` Motion is simple harmonic and `omega=sqrt((4k_(1)k_(3)+k_(1)k_(2)+4k_(2)k_(3))/((4k_(3)+k_(1))m))` |
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