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show that bigilar oscillation are SHM derive the expression for periodic time of bifilar oscillation with parallel threads |
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Answer» Explanation: Time period T= time required to complete ONE OSCILLATION. At equilibrium T 0
=mg. For SMALL displacement θ Restoring force =−mgsinθ for small θ sinθ≈θ ⇒fe=−mgθ=−mg( l x
) acceleration a= m fe
= l −g
.x We know for SHM, a=−w 2 x ⇒ On COMPARING we get w= l g
∴ Time period of oscillation T= w 2Π
=2Π g l
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