Question

❍The period of oscillation of the single pendulum depends on its length and acceleration for the period of oscillation. Derive the expression for period of oscillation?

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Answer 1

Answer:

The period of a simple pendulum equals 2 times π times the square root of the length of the pendulum over g, the acceleration due to gravity. 2, π, and g are constant, so the only variable is L. Therefore the period is dependent on the length.

Answer 2

Answer:

here's your answer dear

Time period T= time required to complete one oscillation.

Time period T= time required to complete one oscillation.At equilibrium T0=mg.

Time period T= time required to complete one oscillation.At equilibrium T0=mg.For small displacement θ

Time period T= time required to complete one oscillation.At equilibrium T0=mg.For small displacement θRestoring force =−mgsinθ

Time period T= time required to complete one oscillation.At equilibrium T0=mg.For small displacement θRestoring force =−mgsinθfor small θ sinθ≈θ

Time period T= time required to complete one oscillation.At equilibrium T0=mg.For small displacement θRestoring force =−mgsinθfor small θ sinθ≈θ⇒fe=−mgθ=−mg(lx)

Time period T= time required to complete one oscillation.At equilibrium T0=mg.For small displacement θRestoring force =−mgsinθfor small θ sinθ≈θ⇒fe=−mgθ=−mg(lx)acceleration a=mfe=l−g.x

Time period T= time required to complete one oscillation.At equilibrium T0=mg.For small displacement θRestoring force =−mgsinθfor small θ sinθ≈θ⇒fe=−mgθ=−mg(lx)acceleration a=mfe=l−g.xWe know for SHM, a=−w2x

Time period T= time required to complete one oscillation.At equilibrium T0=mg.For small displacement θRestoring force =−mgsinθfor small θ sinθ≈θ⇒fe=−mgθ=−mg(lx)acceleration a=mfe=l−g.xWe know for SHM, a=−w2x⇒  On comparing we get w=g/1

∴  Time period of oscillation T=w2Π=2Πl/g.